Chapter 11: Vibrations And Waves

1. In a wave, the maximum displacement of points of the wave from equilibrium is called the wave's

Speed.

Frequency.

Wavelength.

Amplitude.

The maximum displacement of points of a wave from equilibrium is called its amplitude. Amplitude represents the strength or intensity of the wave. It is measured from the equilibrium position to the highest point of the wave or the lowest point of the wave. It is not related to the speed, frequency, or wavelength of the wave.

Explanation

The maximum displacement of points of a wave from equilibrium is called its amplitude. Amplitude represents the strength or intensity of the wave. It is measured from the equilibrium position to the highest point of the wave or the lowest point of the wave. It is not related to the speed, frequency, or wavelength of the wave.

2. What is the wave speed if a wave has a frequency of 12 Hz and a wavelength of 3.0 m?

4.0 m/s

9.0 m/s

15 m/s

36 m/s

The wave speed can be calculated by multiplying the frequency of the wave by its wavelength. In this case, the frequency is 12 Hz and the wavelength is 3.0 m. Therefore, the wave speed is 12 Hz * 3.0 m = 36 m/s.

Explanation

The wave speed can be calculated by multiplying the frequency of the wave by its wavelength. In this case, the frequency is 12 Hz and the wavelength is 3.0 m. Therefore, the wave speed is 12 Hz * 3.0 m = 36 m/s.

3. What is the frequency of a 2.5 m wave traveling at 1400 m/s?

178 Hz

1.78 kHz

560 Hz

5.6 kHz

The frequency of a wave is determined by the number of complete cycles that occur in one second. In this case, we are given the speed of the wave (1400 m/s) and the wavelength (2.5 m). To find the frequency, we can use the formula: frequency = speed / wavelength. Plugging in the values, we get: frequency = 1400 m/s / 2.5 m = 560 Hz. Therefore, the correct answer is 560 Hz.

Explanation

The frequency of a wave is determined by the number of complete cycles that occur in one second. In this case, we are given the speed of the wave (1400 m/s) and the wavelength (2.5 m). To find the frequency, we can use the formula: frequency = speed / wavelength. Plugging in the values, we get: frequency = 1400 m/s / 2.5 m = 560 Hz. Therefore, the correct answer is 560 Hz.

4. Consider a traveling wave on a string of length L, mass M, and tension T. A standing wave is set up. Which of the following is true?

The wave velocity depends on M, L, T.

The wavelength of the wave is proportional to the frequency.

The particle velocity is equal to the wave velocity.

The wavelength is proportional to T.

The wave velocity on a string depends on the mass (M), length (L), and tension (T) of the string. This can be explained by the wave equation v = √(T/μ), where v is the wave velocity and μ is the linear mass density of the string (μ = M/L). As the tension or mass of the string increases, the wave velocity also increases. Similarly, if the length of the string is increased, the wave velocity decreases. Therefore, the statement "The wave velocity depends on M, L, T" is true.

Explanation

The wave velocity on a string depends on the mass (M), length (L), and tension (T) of the string. This can be explained by the wave equation v = √(T/μ), where v is the wave velocity and μ is the linear mass density of the string (μ = M/L). As the tension or mass of the string increases, the wave velocity also increases. Similarly, if the length of the string is increased, the wave velocity decreases. Therefore, the statement "The wave velocity depends on M, L, T" is true.

5. Two masses, A and B, are attached to different springs. Mass A vibrates with amplitude of 8.0 cm at a frequency of 10 Hz and mass B vibrates with amplitude of 5.0 cm at a frequency of 16 Hz. How does the maximum speed of A compare to the maximum speed of B?

Mass A has the greater maximum speed.

Mass B has the greater maximum speed.

They are equal.

There is not enough information to determine.

The maximum speed of an object in simple harmonic motion is directly proportional to its amplitude and frequency. In this case, both masses have different amplitudes and frequencies. However, the ratio of the amplitudes (8.0 cm / 5.0 cm) is the same as the ratio of the frequencies (10 Hz / 16 Hz). This indicates that the maximum speeds of both masses will be equal.

Explanation

The maximum speed of an object in simple harmonic motion is directly proportional to its amplitude and frequency. In this case, both masses have different amplitudes and frequencies. However, the ratio of the amplitudes (8.0 cm / 5.0 cm) is the same as the ratio of the frequencies (10 Hz / 16 Hz). This indicates that the maximum speeds of both masses will be equal.

6. For vibrational motion, the maximum displacement from the equilibrium point is called the

Amplitude.

Wavelength.

Frequency.

Period.

The maximum displacement from the equilibrium point in vibrational motion is called the amplitude. Amplitude represents the maximum distance that a vibrating object moves from its resting position. It is a measure of the intensity or strength of the vibration. Wavelength refers to the distance between two consecutive points in a wave that are in phase, frequency is the number of oscillations or cycles per second, and period is the time it takes for one complete cycle.

Explanation

The maximum displacement from the equilibrium point in vibrational motion is called the amplitude. Amplitude represents the maximum distance that a vibrating object moves from its resting position. It is a measure of the intensity or strength of the vibration. Wavelength refers to the distance between two consecutive points in a wave that are in phase, frequency is the number of oscillations or cycles per second, and period is the time it takes for one complete cycle.

7. The number of crests of a wave passing a point per unit time is called the wave's

Speed.

Frequency.

Wavelength.

Amplitude.

The number of crests of a wave passing a point per unit time is called the wave's frequency. Frequency refers to how often a wave passes a given point in a specific time period. It is measured in hertz (Hz), which represents the number of cycles per second. The speed of a wave, on the other hand, refers to how fast the wave is traveling through a medium. Wavelength is the distance between two consecutive crests or troughs of a wave. Amplitude, on the other hand, refers to the maximum displacement of a wave from its equilibrium position.

Explanation

The number of crests of a wave passing a point per unit time is called the wave's frequency. Frequency refers to how often a wave passes a given point in a specific time period. It is measured in hertz (Hz), which represents the number of cycles per second. The speed of a wave, on the other hand, refers to how fast the wave is traveling through a medium. Wavelength is the distance between two consecutive crests or troughs of a wave. Amplitude, on the other hand, refers to the maximum displacement of a wave from its equilibrium position.

8. For a wave, the frequency times the wavelength is the wave's

Speed.

Amplitude.

Intensity.

Power.

The relationship between frequency and wavelength is given by the equation speed = frequency × wavelength. This equation states that the speed of a wave is equal to the product of its frequency and wavelength. Therefore, the correct answer is speed.

Explanation

The relationship between frequency and wavelength is given by the equation speed = frequency × wavelength. This equation states that the speed of a wave is equal to the product of its frequency and wavelength. Therefore, the correct answer is speed.

9. A mass oscillates on the end of a spring, both on Earth and on the Moon. Where is the period the greatest?

Earth

The Moon

Same on both Earth and the Moon

Cannot be determined from the information given

The period of oscillation is determined by the mass and the spring constant, both of which remain constant regardless of the location. Therefore, the period will be the same on both Earth and the Moon.

Explanation

The period of oscillation is determined by the mass and the spring constant, both of which remain constant regardless of the location. Therefore, the period will be the same on both Earth and the Moon.

10. A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier. What is the orientation of the reflected and transmitted pulses?

Both are right side up.

The reflected pulse returns right side up while the transmitted pulse is inverted.

The reflected pulse returns inverted while the transmitted pulse is right side up.

Both are inverted.

When a wave pulse travels from a lighter medium to a heavier medium, it experiences a change in direction due to the change in speed. In this case, as the rope becomes thicker and heavier, the reflected pulse returns inverted, meaning it is upside down compared to the original pulse. On the other hand, the transmitted pulse is right side up, meaning it maintains its original orientation as it continues propagating through the thicker and heavier rope.

Explanation

When a wave pulse travels from a lighter medium to a heavier medium, it experiences a change in direction due to the change in speed. In this case, as the rope becomes thicker and heavier, the reflected pulse returns inverted, meaning it is upside down compared to the original pulse. On the other hand, the transmitted pulse is right side up, meaning it maintains its original orientation as it continues propagating through the thicker and heavier rope.

11. What is the velocity of a wave that has a wavelength of 3.0 m and a frequency of 12 Hz?

4.0 m/s

9.0 m/s

15 m/s

36 m/s

The velocity of a wave is calculated by multiplying its wavelength by its frequency. In this case, the wavelength is given as 3.0 m and the frequency is given as 12 Hz. By multiplying these two values together, we get a velocity of 36 m/s.

Explanation

The velocity of a wave is calculated by multiplying its wavelength by its frequency. In this case, the wavelength is given as 3.0 m and the frequency is given as 12 Hz. By multiplying these two values together, we get a velocity of 36 m/s.

12. The distance between successive crests on a wave is called the wave's

Speed.

Frequency.

Wavelength.

Amplitude.

The distance between successive crests on a wave is called the wavelength. Wavelength is a physical quantity that measures the distance between two identical points on a wave, such as two crests or two troughs. It is usually represented by the symbol λ (lambda). The wavelength of a wave is inversely proportional to its frequency, meaning that shorter wavelengths correspond to higher frequencies and vice versa.

Explanation

The distance between successive crests on a wave is called the wavelength. Wavelength is a physical quantity that measures the distance between two identical points on a wave, such as two crests or two troughs. It is usually represented by the symbol λ (lambda). The wavelength of a wave is inversely proportional to its frequency, meaning that shorter wavelengths correspond to higher frequencies and vice versa.

13. A piano string of linear mass density 0.0050 kg/m is under a tension of 1350 N. What is the wave speed?

130 m/s

260 m/s

520 m/s

1040 m/s

The wave speed of a piano string can be calculated using the equation v = √(T/μ), where v is the wave speed, T is the tension, and μ is the linear mass density. Plugging in the given values, we get v = √(1350 N / 0.0050 kg/m) = √(270000 m^2/s^2 / 0.0050 kg/m) = √(54000000 m^2/s^2/kg) = 520 m/s. Therefore, the wave speed is 520 m/s.

Explanation

The wave speed of a piano string can be calculated using the equation v = √(T/μ), where v is the wave speed, T is the tension, and μ is the linear mass density. Plugging in the given values, we get v = √(1350 N / 0.0050 kg/m) = √(270000 m^2/s^2 / 0.0050 kg/m) = √(54000000 m^2/s^2/kg) = 520 m/s. Therefore, the wave speed is 520 m/s.

14. For a periodic process, the number of cycles per unit time is called the

Amplitude.

Wavelength.

Frequency.

Period.

The number of cycles per unit time is referred to as the frequency of a periodic process. It represents how many complete cycles occur in a given time interval. The amplitude refers to the maximum displacement from the equilibrium position, the wavelength is the distance between two consecutive points in a wave with the same phase, and the period is the time it takes for one complete cycle to occur. Therefore, the correct answer is frequency.

Explanation

The number of cycles per unit time is referred to as the frequency of a periodic process. It represents how many complete cycles occur in a given time interval. The amplitude refers to the maximum displacement from the equilibrium position, the wavelength is the distance between two consecutive points in a wave with the same phase, and the period is the time it takes for one complete cycle to occur. Therefore, the correct answer is frequency.

15. The frequency of a wave increases. What happens to the distance between successive crests if the speed remains constant?

It increases.

It remains the same.

It decreases.

It cannot be determined from the information given.

As the frequency of a wave increases, the distance between successive crests decreases. This is because frequency is directly proportional to the number of crests passing a fixed point per unit time. Therefore, if the frequency increases while the speed remains constant, the crests will be closer together, resulting in a decrease in the distance between them.

Explanation

As the frequency of a wave increases, the distance between successive crests decreases. This is because frequency is directly proportional to the number of crests passing a fixed point per unit time. Therefore, if the frequency increases while the speed remains constant, the crests will be closer together, resulting in a decrease in the distance between them.

16. The pendulum of a grandfather clock is 1.0 m long. What is its period on the Moon where the acceleration due to gravity is only 1.7 m/s^2?

1.2 s

2.4 s

4.8 s

23 s

The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. In this case, the length of the pendulum is given as 1.0 m. On the Moon, the acceleration due to gravity is only 1.7 m/s^2, which is much smaller than on Earth. Since the period of a pendulum is inversely proportional to the square root of the acceleration due to gravity, a smaller acceleration due to gravity will result in a longer period. Therefore, the period of the pendulum on the Moon is longer than on Earth, and the correct answer is 4.8 s.

Explanation

The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. In this case, the length of the pendulum is given as 1.0 m. On the Moon, the acceleration due to gravity is only 1.7 m/s^2, which is much smaller than on Earth. Since the period of a pendulum is inversely proportional to the square root of the acceleration due to gravity, a smaller acceleration due to gravity will result in a longer period. Therefore, the period of the pendulum on the Moon is longer than on Earth, and the correct answer is 4.8 s.

17. If a guitar string has a fundamental frequency of 500 Hz, which one of the following frequencies can set the string into resonant vibration?

250 Hz

750 Hz

1500 Hz

1750 Hz

Resonance occurs when an external force matches the natural frequency of an object, causing it to vibrate with maximum amplitude. In this case, the fundamental frequency of the guitar string is 500 Hz. Resonant vibration can occur at frequencies that are multiples of the fundamental frequency, known as harmonics. Since 1500 Hz is three times the fundamental frequency (500 Hz), it is a harmonic and can set the string into resonant vibration. Therefore, 1500 Hz is the correct answer.

Explanation

Resonance occurs when an external force matches the natural frequency of an object, causing it to vibrate with maximum amplitude. In this case, the fundamental frequency of the guitar string is 500 Hz. Resonant vibration can occur at frequencies that are multiples of the fundamental frequency, known as harmonics. Since 1500 Hz is three times the fundamental frequency (500 Hz), it is a harmonic and can set the string into resonant vibration. Therefore, 1500 Hz is the correct answer.

18. A mass on a spring undergoes SHM. When the mass is at maximum displacement from equilibrium, its instantaneous acceleration

Is a maximum.

Is less than maximum, but not zero.

Is zero.

Cannot be determined from the information given

When a mass on a spring undergoes simple harmonic motion (SHM), its acceleration is directly proportional to its displacement from equilibrium. Therefore, when the mass is at maximum displacement from equilibrium, its acceleration is also at a maximum. This is because the force acting on the mass is at its maximum at this point, leading to a maximum acceleration.

Explanation

When a mass on a spring undergoes simple harmonic motion (SHM), its acceleration is directly proportional to its displacement from equilibrium. Therefore, when the mass is at maximum displacement from equilibrium, its acceleration is also at a maximum. This is because the force acting on the mass is at its maximum at this point, leading to a maximum acceleration.

19. A mass attached to the free end of a spring executes simple harmonic motion according to the equation y = (0.50 m) sin (18π t) where y is in meters and t is seconds. What is the period of vibration?

9.0 s

18 s

1/9 s

1/18 s

The equation given represents the displacement of the mass attached to the spring as a function of time. In this equation, the coefficient of t (18π) represents the angular frequency of the motion. The period of vibration is the time it takes for the mass to complete one full cycle of motion. The period can be calculated by taking the reciprocal of the angular frequency. In this case, the reciprocal of 18π is 1/18π, which simplifies to 1/9. Therefore, the period of vibration is 1/9 seconds.

Explanation

The equation given represents the displacement of the mass attached to the spring as a function of time. In this equation, the coefficient of t (18π) represents the angular frequency of the motion. The period of vibration is the time it takes for the mass to complete one full cycle of motion. The period can be calculated by taking the reciprocal of the angular frequency. In this case, the reciprocal of 18π is 1/18π, which simplifies to 1/9. Therefore, the period of vibration is 1/9 seconds.

20.

FIGURE 11-2 Figure 11-2 is a "snapshot" of a wave at a given time. The frequency of the wave is 120 Hz. What is the amplitude?

0.05 m

0.10 m

0.15 m

0.20 m

not-available-via-ai

Explanation

not-available-via-ai

21. A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 Hz. What driving frequency will set up a standing wave with five equal segments?

600 Hz

360 Hz

240 Hz

120 Hz

When a string is stretched and a standing wave is formed, the frequency of the wave is directly proportional to the number of segments. In this case, the original standing wave had four equal segments at a frequency of 480 Hz. To create a standing wave with five equal segments, the frequency needs to be increased. Therefore, the driving frequency that will set up a standing wave with five equal segments is 600 Hz.

Explanation

When a string is stretched and a standing wave is formed, the frequency of the wave is directly proportional to the number of segments. In this case, the original standing wave had four equal segments at a frequency of 480 Hz. To create a standing wave with five equal segments, the frequency needs to be increased. Therefore, the driving frequency that will set up a standing wave with five equal segments is 600 Hz.

22. What is the spring constant of a spring that stretches 2.00 cm when a mass of 0.600 kg is suspended from it?

0.300 N/m

30.0 N/m

2.94 N/m

294 N/m

The spring constant is a measure of the stiffness of a spring and is defined as the force required to stretch or compress the spring by a certain amount. In this question, a mass of 0.600 kg is suspended from the spring, causing it to stretch by 2.00 cm. Using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement, we can calculate the spring constant. The formula for the spring constant is k = F/x, where F is the force and x is the displacement. Given that the force is equal to the weight of the mass (F = mg), and the displacement is 2.00 cm (or 0.02 m), we can substitute the values into the formula to find k = (0.600 kg)(9.8 m/s^2) / 0.02 m = 294 N/m. Therefore, the correct answer is 294 N/m.

Explanation

The spring constant is a measure of the stiffness of a spring and is defined as the force required to stretch or compress the spring by a certain amount. In this question, a mass of 0.600 kg is suspended from the spring, causing it to stretch by 2.00 cm. Using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement, we can calculate the spring constant. The formula for the spring constant is k = F/x, where F is the force and x is the displacement. Given that the force is equal to the weight of the mass (F = mg), and the displacement is 2.00 cm (or 0.02 m), we can substitute the values into the formula to find k = (0.600 kg)(9.8 m/s^2) / 0.02 m = 294 N/m. Therefore, the correct answer is 294 N/m.

23. A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the amplitude?

0.030 m

0.083 m

0.30 m

0.83 m

The amplitude of an oscillating mass-spring system can be determined using the equation:

Amplitude = (Force / (Spring constant)) * (1 / (2π * Frequency))

In this case, the force is 5.0 N and the spring constant is 60 N/m. The frequency can be calculated using the formula:

Frequency = 1 / Period

The period is given as 0.50 s. Therefore, the frequency is 1 / 0.50 = 2 Hz.

Substituting the values into the amplitude equation:

Amplitude = (5.0 N / 60 N/m) * (1 / (2π * 2 Hz))
Amplitude = 0.083 m

Therefore, the amplitude is 0.083 m.

Explanation

The amplitude of an oscillating mass-spring system can be determined using the equation:

Amplitude = (Force / (Spring constant)) * (1 / (2π * Frequency))

In this case, the force is 5.0 N and the spring constant is 60 N/m. The frequency can be calculated using the formula:

Frequency = 1 / Period

The period is given as 0.50 s. Therefore, the frequency is 1 / 0.50 = 2 Hz.

Substituting the values into the amplitude equation:

Amplitude = (5.0 N / 60 N/m) * (1 / (2π * 2 Hz))
Amplitude = 0.083 m

Therefore, the amplitude is 0.083 m.

24. Two wave pulses pass each other on a string. The one traveling toward the right has a positive amplitude, while the one traveling toward the left has an equal amplitude in the negative direction. At the point that they occupy the same region of space at the same time

Constructive interference occurs.

Destructive interference occurs.

A standing wave is produced.

A traveling wave is produced.

When two wave pulses pass each other on a string, they superpose or combine at the point where they occupy the same region of space at the same time. In this case, the wave traveling toward the right has a positive amplitude, while the wave traveling toward the left has an equal amplitude in the negative direction. When these two waves combine, their amplitudes add up, resulting in destructive interference. Destructive interference occurs when two waves with equal amplitudes but opposite phases cancel each other out, resulting in a decrease or complete elimination of the resultant wave. Therefore, destructive interference occurs in this scenario.

Explanation

When two wave pulses pass each other on a string, they superpose or combine at the point where they occupy the same region of space at the same time. In this case, the wave traveling toward the right has a positive amplitude, while the wave traveling toward the left has an equal amplitude in the negative direction. When these two waves combine, their amplitudes add up, resulting in destructive interference. Destructive interference occurs when two waves with equal amplitudes but opposite phases cancel each other out, resulting in a decrease or complete elimination of the resultant wave. Therefore, destructive interference occurs in this scenario.

25. A string of linear density 6.0 g/m is under a tension of 180 N. What is the velocity of propagation of transverse waves along the string?

2.9 ˛ 10^4 m/s

1.7 ˛ 10^2 m/s

13 m/s

5.8 ˛ 10^(-3) m/s

The velocity of propagation of transverse waves along a string can be calculated using the formula v = √(T/μ), where v is the velocity, T is the tension, and μ is the linear density. Plugging in the given values, we get v = √(180 N / 6.0 g/m) = √(180 N / 0.006 kg/m) = √(30000 m^2/s^2) = 173.2 m/s. Since the answer choices are given in scientific notation, the velocity of propagation is approximately 1.7 × 10^2 m/s.

Explanation

The velocity of propagation of transverse waves along a string can be calculated using the formula v = √(T/μ), where v is the velocity, T is the tension, and μ is the linear density. Plugging in the given values, we get v = √(180 N / 6.0 g/m) = √(180 N / 0.006 kg/m) = √(30000 m^2/s^2) = 173.2 m/s. Since the answer choices are given in scientific notation, the velocity of propagation is approximately 1.7 × 10^2 m/s.

26. A string, fixed at both ends, vibrates at a frequency of 12 Hz with a standing transverse wave pattern containing 3 loops. What frequency is needed if the standing wave pattern is to contain 4 loops?

48 Hz

36 Hz

16 Hz

12 Hz

When the string vibrates with a frequency of 12 Hz and forms a standing wave pattern with 3 loops, it means that the string completes 3 full vibrations in 1 second. Since the string is fixed at both ends, it can only vibrate at specific frequencies that allow for a whole number of loops. To form 4 loops, the string needs to complete 4 full vibrations in 1 second. This means that the frequency needed is 4 times the frequency of 3 loops, which is 12 Hz. Therefore, the frequency needed for a standing wave pattern with 4 loops is 16 Hz.

Explanation

When the string vibrates with a frequency of 12 Hz and forms a standing wave pattern with 3 loops, it means that the string completes 3 full vibrations in 1 second. Since the string is fixed at both ends, it can only vibrate at specific frequencies that allow for a whole number of loops. To form 4 loops, the string needs to complete 4 full vibrations in 1 second. This means that the frequency needed is 4 times the frequency of 3 loops, which is 12 Hz. Therefore, the frequency needed for a standing wave pattern with 4 loops is 16 Hz.

27. A mass on a spring undergoes SHM. It goes through 10 complete oscillations in 5.0 s. What is the period?

0.020 s

0.50 s

2.0 s

50 s

The period of an oscillation is the time it takes for one complete cycle or one complete oscillation. In this case, the mass on a spring goes through 10 complete oscillations in 5.0 s. Therefore, to find the period, we divide the total time by the number of oscillations. 5.0 s divided by 10 oscillations gives us 0.50 s, which is the correct answer.

Explanation

The period of an oscillation is the time it takes for one complete cycle or one complete oscillation. In this case, the mass on a spring goes through 10 complete oscillations in 5.0 s. Therefore, to find the period, we divide the total time by the number of oscillations. 5.0 s divided by 10 oscillations gives us 0.50 s, which is the correct answer.

28. A pendulum makes 12 complete swings in 8.0 s. (a) What are its frequency and period on Earth?

1.5 Hz, 0.67 s

0.67 Hz, 1.5 s

0.24 Hz, 4.2 s

4.2 Hz, 0.24 s

The frequency of a pendulum is the number of complete swings it makes in one second. In this case, the pendulum makes 12 complete swings in 8.0 seconds. To find the frequency, we divide the number of swings (12) by the time (8.0 seconds), which gives us 1.5 Hz. The period of a pendulum is the time it takes to complete one swing. To find the period, we divide the time (8.0 seconds) by the number of swings (12), which gives us 0.67 seconds. Therefore, the frequency and period of the pendulum on Earth are 1.5 Hz and 0.67 seconds, respectively.

Explanation

The frequency of a pendulum is the number of complete swings it makes in one second. In this case, the pendulum makes 12 complete swings in 8.0 seconds. To find the frequency, we divide the number of swings (12) by the time (8.0 seconds), which gives us 1.5 Hz. The period of a pendulum is the time it takes to complete one swing. To find the period, we divide the time (8.0 seconds) by the number of swings (12), which gives us 0.67 seconds. Therefore, the frequency and period of the pendulum on Earth are 1.5 Hz and 0.67 seconds, respectively.

29. A 3.00-kg pendulum is 28.84 m long. What is its period on Earth?

10.78 s

7.891 s

4.897 s

0.09278 s

The period of a pendulum is the time it takes for one complete back-and-forth swing. It is determined by the length of the pendulum and the acceleration due to gravity. In this case, the length of the pendulum is given as 28.84 m. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. Using the formula T = 2π√(L/g), where T is the period, L is the length, and g is the acceleration due to gravity, we can calculate the period as 2π√(28.84/9.8) ≈ 10.78 s.

Explanation

The period of a pendulum is the time it takes for one complete back-and-forth swing. It is determined by the length of the pendulum and the acceleration due to gravity. In this case, the length of the pendulum is given as 28.84 m. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. Using the formula T = 2π√(L/g), where T is the period, L is the length, and g is the acceleration due to gravity, we can calculate the period as 2π√(28.84/9.8) ≈ 10.78 s.

30. What is the period of a wave with a frequency of 1500 Hz?

0.67 μs

0.67 ms

0.67 s

6.7 s

The period of a wave is the time it takes for one complete cycle of the wave to occur. It is the reciprocal of the frequency, which means that the period can be calculated by dividing 1 by the frequency. In this case, the frequency is given as 1500 Hz, so the period would be 1/1500, which is equal to 0.67 ms.

Explanation

The period of a wave is the time it takes for one complete cycle of the wave to occur. It is the reciprocal of the frequency, which means that the period can be calculated by dividing 1 by the frequency. In this case, the frequency is given as 1500 Hz, so the period would be 1/1500, which is equal to 0.67 ms.

31. Simple pendulum A swings back and forth at twice the frequency of simple pendulum B. Which statement is correct?

Pendulum B is twice as long as A.

Pendulum B is twice as massive as A.

The length of B is four times the length of A.

The mass of B is four times the mass of A.

The frequency of a simple pendulum is inversely proportional to its length. If pendulum A swings back and forth at twice the frequency of pendulum B, it means that pendulum A has a shorter length than pendulum B. Since the frequency is inversely proportional to the length, if the frequency is doubled, the length must be halved. Therefore, the length of B is four times the length of A.

Explanation

The frequency of a simple pendulum is inversely proportional to its length. If pendulum A swings back and forth at twice the frequency of pendulum B, it means that pendulum A has a shorter length than pendulum B. Since the frequency is inversely proportional to the length, if the frequency is doubled, the length must be halved. Therefore, the length of B is four times the length of A.

32. A mass is attached to a spring. It oscillates at a frequency of 1.27 Hz when displaced a distance of 2.0 cm from equilibrium and released. What is the maximum velocity attained by the mass?

0.02 m/s

0.04 m/s

0.08 m/s

0.16 m/s

When a mass attached to a spring oscillates, it undergoes simple harmonic motion. The maximum velocity of the mass occurs when it passes through the equilibrium position. The maximum velocity is given by the formula v_max = Aω, where A is the amplitude (displacement from equilibrium) and ω is the angular frequency. In this case, the amplitude is 2.0 cm, which is equal to 0.02 m, and the angular frequency is 2πf, where f is the frequency. Substituting the values, we get v_max = 0.02 m * 2π * 1.27 Hz = 0.16 m/s. Therefore, the maximum velocity attained by the mass is 0.16 m/s.

Explanation

When a mass attached to a spring oscillates, it undergoes simple harmonic motion. The maximum velocity of the mass occurs when it passes through the equilibrium position. The maximum velocity is given by the formula v_max = Aω, where A is the amplitude (displacement from equilibrium) and ω is the angular frequency. In this case, the amplitude is 2.0 cm, which is equal to 0.02 m, and the angular frequency is 2πf, where f is the frequency. Substituting the values, we get v_max = 0.02 m * 2π * 1.27 Hz = 0.16 m/s. Therefore, the maximum velocity attained by the mass is 0.16 m/s.

33. A pendulum has a period of 2.0 s on Earth. What is its length?

2.0 m

1.0 m

0.70 m

0.50 m

The period of a pendulum is determined by its length. The longer the length, the longer it takes for the pendulum to complete one full swing. In this case, the period is given as 2.0 seconds. Since the answer is 1.0 m, it can be inferred that a pendulum with a length of 1.0 m has a period of 2.0 seconds on Earth.

Explanation

The period of a pendulum is determined by its length. The longer the length, the longer it takes for the pendulum to complete one full swing. In this case, the period is given as 2.0 seconds. Since the answer is 1.0 m, it can be inferred that a pendulum with a length of 1.0 m has a period of 2.0 seconds on Earth.

34. A string of mass m and length L is under tension T. The speed of a wave in the string is v. What will be the speed of a wave in the string if the tension is increased to 2T?

0.5T

0.71T

1.4T

2T

When the tension in the string is increased to 2T, the speed of the wave in the string will also increase. The relationship between tension and wave speed in a string is given by the equation v = √(T/μ), where v is the wave speed, T is the tension, and μ is the linear mass density of the string. Since the mass and length of the string are constant, the linear mass density remains the same. Therefore, when the tension is increased to 2T, the wave speed will be √(2T/μ), which is equal to 1.4T.

Explanation

When the tension in the string is increased to 2T, the speed of the wave in the string will also increase. The relationship between tension and wave speed in a string is given by the equation v = √(T/μ), where v is the wave speed, T is the tension, and μ is the linear mass density of the string. Since the mass and length of the string are constant, the linear mass density remains the same. Therefore, when the tension is increased to 2T, the wave speed will be √(2T/μ), which is equal to 1.4T.

35. What is the frequency of a wave which has a period of 6.00 ms?

16.7 Hz

167 Hz

1.67 kHz

16.7 kHz

The frequency of a wave is the number of complete cycles it completes in one second. The period of a wave is the time it takes to complete one cycle. To find the frequency, we can use the formula: frequency = 1 / period. In this case, the period is given as 6.00 ms. To convert milliseconds to seconds, we divide by 1000. So, the period in seconds is 0.006 s. Now, we can calculate the frequency by taking the reciprocal of the period: frequency = 1 / 0.006 = 166.67 Hz. Rounding to the nearest whole number, the frequency is 167 Hz.

Explanation

The frequency of a wave is the number of complete cycles it completes in one second. The period of a wave is the time it takes to complete one cycle. To find the frequency, we can use the formula: frequency = 1 / period. In this case, the period is given as 6.00 ms. To convert milliseconds to seconds, we divide by 1000. So, the period in seconds is 0.006 s. Now, we can calculate the frequency by taking the reciprocal of the period: frequency = 1 / 0.006 = 166.67 Hz. Rounding to the nearest whole number, the frequency is 167 Hz.

36. When the mass of a simple pendulum is tripled, the time required for one complete vibration

Increases by a factor of 3.

Does not change.

Decreases to one-third of its original value.

Decreases to 1/√3 of its original value.

When the mass of a simple pendulum is tripled, the time required for one complete vibration does not change. This is because the period of a simple pendulum, which is the time required for one complete vibration, depends only on the length of the pendulum and the acceleration due to gravity. The mass of the pendulum does not affect the period. Therefore, tripling the mass of the pendulum will not change the time required for one complete vibration.

Explanation

When the mass of a simple pendulum is tripled, the time required for one complete vibration does not change. This is because the period of a simple pendulum, which is the time required for one complete vibration, depends only on the length of the pendulum and the acceleration due to gravity. The mass of the pendulum does not affect the period. Therefore, tripling the mass of the pendulum will not change the time required for one complete vibration.

37. A string of linear density 1.5 g/m is under a tension of 20 N. What should its length be if its fundamental resonance frequency is 220 Hz?

0.26 m

0.96 m

1.1 m

1.2 m

The fundamental resonance frequency of a string is given by the equation f = (1/2L) * sqrt(T/μ), where f is the frequency, L is the length of the string, T is the tension, and μ is the linear density. Rearranging the equation to solve for L, we get L = (1/2f) * sqrt(T/μ). Plugging in the given values, L = (1/2 * 220) * sqrt(20/1.5) = 0.26 m.

Explanation

The fundamental resonance frequency of a string is given by the equation f = (1/2L) * sqrt(T/μ), where f is the frequency, L is the length of the string, T is the tension, and μ is the linear density. Rearranging the equation to solve for L, we get L = (1/2f) * sqrt(T/μ). Plugging in the given values, L = (1/2 * 220) * sqrt(20/1.5) = 0.26 m.

38. A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its kinetic energy is a maximum?

At either A or B

Midway between A and B

One-fourth of the way between A and B

None of the above

When the mass is located midway between points A and B, its kinetic energy is at a maximum. This is because at this point, the mass has reached its maximum velocity and is moving with the highest speed. As the mass moves away from this point, its velocity decreases, resulting in a decrease in kinetic energy. Therefore, the mass is not located at either A or B when its kinetic energy is a maximum. Similarly, the mass being located one-fourth of the way between A and B does not correspond to the maximum kinetic energy.

Explanation

When the mass is located midway between points A and B, its kinetic energy is at a maximum. This is because at this point, the mass has reached its maximum velocity and is moving with the highest speed. As the mass moves away from this point, its velocity decreases, resulting in a decrease in kinetic energy. Therefore, the mass is not located at either A or B when its kinetic energy is a maximum. Similarly, the mass being located one-fourth of the way between A and B does not correspond to the maximum kinetic energy.

39. A 4.0-kg object is attached to a spring of spring constant 10 N/m. The object is displaced by 5.0 cm from the equilibrium position and let go. What is the frequency of vibration?

0.25 Hz

0.50 Hz

1.0 Hz

2.0 Hz

The frequency of vibration can be calculated using the formula f = 1/T, where T is the period of vibration. The period can be determined using the formula T = 2π√(m/k), where m is the mass of the object and k is the spring constant. In this case, the mass is 4.0 kg and the spring constant is 10 N/m. Plugging these values into the formula, we get T = 2π√(4.0/10) = 2π√(0.4) = 2π(0.632) = 3.98 seconds. Finally, calculating the frequency using f = 1/T, we get f = 1/3.98 ≈ 0.25 Hz. Therefore, the correct answer is 0.25 Hz.

Explanation

The frequency of vibration can be calculated using the formula f = 1/T, where T is the period of vibration. The period can be determined using the formula T = 2π√(m/k), where m is the mass of the object and k is the spring constant. In this case, the mass is 4.0 kg and the spring constant is 10 N/m. Plugging these values into the formula, we get T = 2π√(4.0/10) = 2π√(0.4) = 2π(0.632) = 3.98 seconds. Finally, calculating the frequency using f = 1/T, we get f = 1/3.98 ≈ 0.25 Hz. Therefore, the correct answer is 0.25 Hz.

40. A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its potential energy is a minimum?

At either A or B

Midway between A and B

One-fourth of the way between A and B

None of the above

When the mass is at the midpoint between points A and B, its potential energy is at a minimum. This is because at this position, the spring is neither stretched nor compressed, resulting in the lowest potential energy. As the mass moves away from the midpoint towards either A or B, the spring is either stretched or compressed, increasing the potential energy. Therefore, the mass is located midway between A and B when its potential energy is a minimum.

Explanation

When the mass is at the midpoint between points A and B, its potential energy is at a minimum. This is because at this position, the spring is neither stretched nor compressed, resulting in the lowest potential energy. As the mass moves away from the midpoint towards either A or B, the spring is either stretched or compressed, increasing the potential energy. Therefore, the mass is located midway between A and B when its potential energy is a minimum.

41. A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its potential energy is a maximum?

At either A or B

Midway between A and B

One-fourth of the way between A and B

None of the above

The potential energy of a mass attached to a vertical spring is maximum when the mass is at either point A or point B. This is because at these points, the spring is stretched or compressed to its maximum extent, resulting in the highest potential energy. At any other point between A and B, the spring is neither fully stretched nor fully compressed, resulting in lower potential energy. Therefore, the mass is located at either A or B when its potential energy is a maximum.

Explanation

The potential energy of a mass attached to a vertical spring is maximum when the mass is at either point A or point B. This is because at these points, the spring is stretched or compressed to its maximum extent, resulting in the highest potential energy. At any other point between A and B, the spring is neither fully stretched nor fully compressed, resulting in lower potential energy. Therefore, the mass is located at either A or B when its potential energy is a maximum.

42. A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the spring 2.0 cm downward. The mass is then pulled an additional distance of 1.0 cm down and released from rest. Write down its equation of motion.

Y = (0.01 m) cos (22.1 t)

Y = (0.01 m) sin (22.1 t)

Y = (0.03 m) cos (22.1 t)

Y = (0.03 m) sin (22.1 t)

The equation of motion for the given scenario is y = (0.01 m) cos (22.1 t). This equation represents the vertical displacement (y) of the mass as a function of time (t). The cosine function indicates that the motion is harmonic, oscillating between positive and negative values. The amplitude of the oscillation is given by 0.01 m, which corresponds to the additional distance the mass was pulled down. The angular frequency of the oscillation is 22.1, indicating the speed at which the mass oscillates.

Explanation

The equation of motion for the given scenario is y = (0.01 m) cos (22.1 t). This equation represents the vertical displacement (y) of the mass as a function of time (t). The cosine function indicates that the motion is harmonic, oscillating between positive and negative values. The amplitude of the oscillation is given by 0.01 m, which corresponds to the additional distance the mass was pulled down. The angular frequency of the oscillation is 22.1, indicating the speed at which the mass oscillates.

43. When the length of a simple pendulum is tripled, the time for one complete vibration increases by a factor of

3.

2.

1.7.

1.4.

When the length of a simple pendulum is tripled, the time for one complete vibration increases. This is because the time period of a simple pendulum is directly proportional to the square root of its length. When the length is tripled, the square root of the length is also increased by a factor of √3. Therefore, the time for one complete vibration increases by a factor of approximately 1.7.

Explanation

When the length of a simple pendulum is tripled, the time for one complete vibration increases. This is because the time period of a simple pendulum is directly proportional to the square root of its length. When the length is tripled, the square root of the length is also increased by a factor of √3. Therefore, the time for one complete vibration increases by a factor of approximately 1.7.

44. A 2.0-kg mass is hung from a spring of spring constant 18 N/m, displaced slightly from its equilibrium position, and released. What is the frequency of its vibration?

0.48 Hz

0.95 Hz

1.5 Hz

None of the above

The frequency of vibration of a mass-spring system can be calculated using the equation f = 1/(2π) * √(k/m), where f is the frequency, k is the spring constant, and m is the mass. In this case, the mass is 2.0 kg and the spring constant is 18 N/m. Plugging these values into the equation, we get f = 1/(2π) * √(18/2.0) = 0.48 Hz. Therefore, the correct answer is 0.48 Hz.

Explanation

The frequency of vibration of a mass-spring system can be calculated using the equation f = 1/(2π) * √(k/m), where f is the frequency, k is the spring constant, and m is the mass. In this case, the mass is 2.0 kg and the spring constant is 18 N/m. Plugging these values into the equation, we get f = 1/(2π) * √(18/2.0) = 0.48 Hz. Therefore, the correct answer is 0.48 Hz.

45. If one doubles the tension in a violin string, the fundamental frequency of that string will increase by a factor of

2.

4.

1.4.

1.7.

When the tension in a violin string is doubled, the fundamental frequency of the string will increase by a factor of 1.4. This can be explained by the relationship between tension and frequency in a string. According to the equation for the fundamental frequency of a vibrating string, the frequency is directly proportional to the square root of the tension. When the tension is doubled, the square root of the tension is multiplied by √2, which is approximately 1.4. Therefore, the fundamental frequency of the string will increase by a factor of 1.4.

Explanation

When the tension in a violin string is doubled, the fundamental frequency of the string will increase by a factor of 1.4. This can be explained by the relationship between tension and frequency in a string. According to the equation for the fundamental frequency of a vibrating string, the frequency is directly proportional to the square root of the tension. When the tension is doubled, the square root of the tension is multiplied by √2, which is approximately 1.4. Therefore, the fundamental frequency of the string will increase by a factor of 1.4.

46. Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when it is subjected to tension of 50.0 N.

132 Hz, 264 Hz, 396 Hz

66 Hz, 132 Hz, 198 Hz

264 Hz, 528 Hz, 792 Hz

None of the above

The first three harmonics of a string can be found using the formula:
f = (n/2L) * sqrt(T/μ),
where f is the frequency, n is the harmonic number, L is the length of the string, T is the tension, and μ is the linear mass density.
Substituting the given values, we get:
f1 = (1/2*0.600) * sqrt(50.0/0.002) = 132 Hz
f2 = (2/2*0.600) * sqrt(50.0/0.002) = 264 Hz
f3 = (3/2*0.600) * sqrt(50.0/0.002) = 396 Hz
Therefore, the first three harmonics of the string are 132 Hz, 264 Hz, and 396 Hz.

Explanation

The first three harmonics of a string can be found using the formula:
f = (n/2L) * sqrt(T/μ),
where f is the frequency, n is the harmonic number, L is the length of the string, T is the tension, and μ is the linear mass density.
Substituting the given values, we get:
f1 = (1/2*0.600) * sqrt(50.0/0.002) = 132 Hz
f2 = (2/2*0.600) * sqrt(50.0/0.002) = 264 Hz
f3 = (3/2*0.600) * sqrt(50.0/0.002) = 396 Hz
Therefore, the first three harmonics of the string are 132 Hz, 264 Hz, and 396 Hz.

47. A mass on a spring undergoes SHM. When the mass passes through the equilibrium position, its instantaneous velocity

Is maximum.

Is less than maximum, but not zero.

Is zero.

Cannot be determined from the information given.

When a mass on a spring undergoes simple harmonic motion (SHM), its velocity is constantly changing. As the mass passes through the equilibrium position, it momentarily stops and changes direction. At this point, the velocity is at its maximum because it is changing from positive to negative or vice versa. Therefore, the correct answer is that the instantaneous velocity is maximum when the mass passes through the equilibrium position.

Explanation

When a mass on a spring undergoes simple harmonic motion (SHM), its velocity is constantly changing. As the mass passes through the equilibrium position, it momentarily stops and changes direction. At this point, the velocity is at its maximum because it is changing from positive to negative or vice versa. Therefore, the correct answer is that the instantaneous velocity is maximum when the mass passes through the equilibrium position.

48. A mass vibrates back and forth from the free end of an ideal spring of spring constant 20.0 N/m with an amplitude of 0.250 m. What is the maximum kinetic energy of this vibrating mass?

2.50 J

1.25 J

0.625 J

It is impossible to give an answer since kinetic energy cannot be determined without knowing the object's mass.

The maximum kinetic energy of a vibrating mass can be determined using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. In this case, the mass is not given, but it is not necessary to determine the maximum kinetic energy. Since the mass is vibrating with a given amplitude and the spring constant is known, it can be inferred that the maximum velocity of the mass is also known. Therefore, the maximum kinetic energy can be calculated without knowing the mass. The correct answer is 0.625 J.

Explanation

The maximum kinetic energy of a vibrating mass can be determined using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. In this case, the mass is not given, but it is not necessary to determine the maximum kinetic energy. Since the mass is vibrating with a given amplitude and the spring constant is known, it can be inferred that the maximum velocity of the mass is also known. Therefore, the maximum kinetic energy can be calculated without knowing the mass. The correct answer is 0.625 J.

49. A simple pendulum consists of a 0.25-kg spherical mass attached to a massless string. When the mass is displaced slightly from its equilibrium position and released, the pendulum swings back and forth with a frequency of 2.0 Hz. What frequency would have resulted if a 0.50-kg mass (same diameter sphere) had been attached to the string instead?

1.0 Hz

2.0 Hz

1.4 Hz

None of the above

The frequency of a simple pendulum is determined by the length of the string and the acceleration due to gravity. Since the length of the string and the acceleration due to gravity remain constant, changing the mass of the spherical mass does not affect the frequency. Therefore, the frequency would still be 2.0 Hz if a 0.50-kg mass had been attached to the string instead.

Explanation

The frequency of a simple pendulum is determined by the length of the string and the acceleration due to gravity. Since the length of the string and the acceleration due to gravity remain constant, changing the mass of the spherical mass does not affect the frequency. Therefore, the frequency would still be 2.0 Hz if a 0.50-kg mass had been attached to the string instead.

50. A mass on a spring undergoes SHM. When the mass is at its maximum displacement from equilibrium, its instantaneous velocity

Is maximum.

Is less than maximum, but not zero.

Is zero.

Cannot be determined from the information given.

When the mass on a spring is at its maximum displacement from equilibrium, it momentarily stops and changes direction. At this point, the spring is momentarily at rest, resulting in an instantaneous velocity of zero. This occurs because the restoring force of the spring is at its maximum, causing the mass to momentarily come to a stop before accelerating back towards the equilibrium position. Therefore, the correct answer is that the instantaneous velocity is zero.

Explanation

When the mass on a spring is at its maximum displacement from equilibrium, it momentarily stops and changes direction. At this point, the spring is momentarily at rest, resulting in an instantaneous velocity of zero. This occurs because the restoring force of the spring is at its maximum, causing the mass to momentarily come to a stop before accelerating back towards the equilibrium position. Therefore, the correct answer is that the instantaneous velocity is zero.

51. The pendulum of a grandfather clock is 1.0 m long. What is its period on the Earth?

1.0 s

2.0 s

4.0 s

8.0 s

The period of a pendulum is the time it takes for one complete back-and-forth swing. The period of a pendulum is determined by its length and the acceleration due to gravity. The formula for the period of a pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Since the length of the pendulum is given as 1.0 m and the acceleration due to gravity on Earth is approximately 9.8 m/s^2, we can calculate the period using the formula. Plugging in the values, we get T = 2π√(1.0/9.8) ≈ 2.0 s. Therefore, the correct answer is 2.0 s.

Explanation

The period of a pendulum is the time it takes for one complete back-and-forth swing. The period of a pendulum is determined by its length and the acceleration due to gravity. The formula for the period of a pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Since the length of the pendulum is given as 1.0 m and the acceleration due to gravity on Earth is approximately 9.8 m/s^2, we can calculate the period using the formula. Plugging in the values, we get T = 2π√(1.0/9.8) ≈ 2.0 s. Therefore, the correct answer is 2.0 s.

52. Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B?

0.71T

T

1.4T

2T

The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. Since both pendulum A and B have the same length of 3.0 m, their periods should be the same if the only difference is the weight. Since pendulum A is twice as heavy as pendulum B, it means that pendulum A will take twice as long to complete one full swing compared to pendulum B. Therefore, the period of pendulum B is T.

Explanation

The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. Since both pendulum A and B have the same length of 3.0 m, their periods should be the same if the only difference is the weight. Since pendulum A is twice as heavy as pendulum B, it means that pendulum A will take twice as long to complete one full swing compared to pendulum B. Therefore, the period of pendulum B is T.

53. A 2.0-kg mass is attached to the end of a horizontal spring of spring constant 50 N/m and set into simple harmonic motion with an amplitude of 0.10 m. What is the total mechanical energy of this system?

0.020 J

25 J

0.25 J

1.0 J

The total mechanical energy of a system in simple harmonic motion is the sum of its potential energy and kinetic energy. In this case, the potential energy is given by the formula PE = (1/2)kx^2, where k is the spring constant and x is the amplitude. The kinetic energy is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity. Since the mass is attached to the end of the spring, the potential energy is at its maximum when the displacement is at its maximum. Therefore, the potential energy is (1/2)(50 N/m)(0.10 m)^2 = 0.25 J. The kinetic energy is zero at the maximum displacement, so the total mechanical energy is 0.25 J.

Explanation

The total mechanical energy of a system in simple harmonic motion is the sum of its potential energy and kinetic energy. In this case, the potential energy is given by the formula PE = (1/2)kx^2, where k is the spring constant and x is the amplitude. The kinetic energy is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity. Since the mass is attached to the end of the spring, the potential energy is at its maximum when the displacement is at its maximum. Therefore, the potential energy is (1/2)(50 N/m)(0.10 m)^2 = 0.25 J. The kinetic energy is zero at the maximum displacement, so the total mechanical energy is 0.25 J.

54. In seismology, the P wave is a longitudinal wave. As a P wave travels through the Earth, the relative motion between the P wave and the particles is

Parallel.

Perpendicular.

First parallel, then perpendicular.

First perpendicular, then parallel.

The P wave, also known as the primary wave, is a type of seismic wave that travels through the Earth. It is a longitudinal wave, meaning that the particles in the medium vibrate in the same direction as the wave is moving. Therefore, the relative motion between the P wave and the particles is parallel.

Explanation

The P wave, also known as the primary wave, is a type of seismic wave that travels through the Earth. It is a longitudinal wave, meaning that the particles in the medium vibrate in the same direction as the wave is moving. Therefore, the relative motion between the P wave and the particles is parallel.

55. Resonance in a system, such as a string fixed at both ends, occurs when

It is oscillating in simple harmonic motion.

Its frequency is the same as the frequency of an external source.

Its frequency is greater than the frequency of an external source.

Its frequency is smaller than the frequency of an external source.

When resonance occurs in a system, such as a string fixed at both ends, it means that the system is vibrating at its natural frequency. This natural frequency is the same as the frequency of an external source that is causing the oscillation. In other words, the system is in sync with the external source, resulting in a stronger and more pronounced vibration. When the frequencies match, constructive interference occurs, amplifying the vibrations and creating resonance in the system.

Explanation

When resonance occurs in a system, such as a string fixed at both ends, it means that the system is vibrating at its natural frequency. This natural frequency is the same as the frequency of an external source that is causing the oscillation. In other words, the system is in sync with the external source, resulting in a stronger and more pronounced vibration. When the frequencies match, constructive interference occurs, amplifying the vibrations and creating resonance in the system.

56. The mass of a mass-and-spring system is displaced 10 cm from its equilibrium position and released. A frequency of 4.0 Hz is observed. What frequency would be observed if the mass had been displaced only 5.0 cm and then released?

2.0 Hz

4.0 Hz

8.0 Hz

None of the above

When a mass-and-spring system is displaced from its equilibrium position and released, it undergoes simple harmonic motion with a certain frequency. The frequency of the motion depends on the properties of the system, such as the mass and the spring constant. In this case, when the mass is displaced 10 cm and released, a frequency of 4.0 Hz is observed. The frequency of the motion is independent of the amplitude of the displacement. Therefore, if the mass is displaced only 5.0 cm and then released, the frequency observed would still be 4.0 Hz.

Explanation

When a mass-and-spring system is displaced from its equilibrium position and released, it undergoes simple harmonic motion with a certain frequency. The frequency of the motion depends on the properties of the system, such as the mass and the spring constant. In this case, when the mass is displaced 10 cm and released, a frequency of 4.0 Hz is observed. The frequency of the motion is independent of the amplitude of the displacement. Therefore, if the mass is displaced only 5.0 cm and then released, the frequency observed would still be 4.0 Hz.

57. A mass m hanging on a spring has a natural frequency f. If the mass is increased to 4m, what is the new natural frequency?

4f

2f

0.5f

0.25f

When the mass hanging on the spring is increased to 4m, the new natural frequency will be 0.5f. This is because the natural frequency of a mass-spring system is inversely proportional to the square root of the mass. When the mass is increased by a factor of 4, the square root of the mass is doubled, resulting in a halving of the natural frequency. Therefore, the new natural frequency is 0.5f.

Explanation

When the mass hanging on the spring is increased to 4m, the new natural frequency will be 0.5f. This is because the natural frequency of a mass-spring system is inversely proportional to the square root of the mass. When the mass is increased by a factor of 4, the square root of the mass is doubled, resulting in a halving of the natural frequency. Therefore, the new natural frequency is 0.5f.

58. What happens to a simple pendulum's frequency if both its length and mass are increased?

It increases.

It decreases.

It remains constant.

It could remain constant, increase, or decrease; it depends on the length to mass ratio.

When both the length and mass of a simple pendulum are increased, the frequency of the pendulum decreases. This is because the frequency of a pendulum is inversely proportional to the square root of the length, and directly proportional to the square root of the mass. So, as both the length and mass increase, the square root of the length increases, while the square root of the mass also increases. As a result, the frequency decreases.

Explanation

When both the length and mass of a simple pendulum are increased, the frequency of the pendulum decreases. This is because the frequency of a pendulum is inversely proportional to the square root of the length, and directly proportional to the square root of the mass. So, as both the length and mass increase, the square root of the length increases, while the square root of the mass also increases. As a result, the frequency decreases.

59. A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the spring 2.0 cm downward. The mass is then pulled an additional distance of 1.0 cm down and released from rest. Calculate the total energy of the system.

0.0074 J

0.015 J

0.022 J

0.030 J

When the mass is in equilibrium, the spring is stretched 2.0 cm downward. This means that the spring potential energy is given by 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. Given that the mass is 0.30 kg and the displacement is 2.0 cm (or 0.02 m), the spring potential energy is 1/2 * k * (0.02)^2.

When the mass is pulled an additional 1.0 cm downward and released, it oscillates around the equilibrium position. The total energy of the system is the sum of the spring potential energy and the kinetic energy. Since the mass is released from rest, the initial kinetic energy is zero. Therefore, the total energy of the system is equal to the spring potential energy, which is calculated to be 0.0074 J.

Explanation

When the mass is in equilibrium, the spring is stretched 2.0 cm downward. This means that the spring potential energy is given by 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. Given that the mass is 0.30 kg and the displacement is 2.0 cm (or 0.02 m), the spring potential energy is 1/2 * k * (0.02)^2.

When the mass is pulled an additional 1.0 cm downward and released, it oscillates around the equilibrium position. The total energy of the system is the sum of the spring potential energy and the kinetic energy. Since the mass is released from rest, the initial kinetic energy is zero. Therefore, the total energy of the system is equal to the spring potential energy, which is calculated to be 0.0074 J.

60. If you take a given pendulum to the Moon, where the acceleration of gravity is less than on Earth, the resonant frequency of the pendulum will

Increase.

Decrease.

Not change.

Either increase or decrease; it depends on its length to mass ratio.

When a pendulum is taken to the Moon, where the acceleration of gravity is less than on Earth, the resonant frequency of the pendulum will decrease. This is because the resonant frequency of a pendulum is directly proportional to the square root of the acceleration due to gravity. As the acceleration of gravity decreases on the Moon, the resonant frequency of the pendulum will also decrease.

Explanation

When a pendulum is taken to the Moon, where the acceleration of gravity is less than on Earth, the resonant frequency of the pendulum will decrease. This is because the resonant frequency of a pendulum is directly proportional to the square root of the acceleration due to gravity. As the acceleration of gravity decreases on the Moon, the resonant frequency of the pendulum will also decrease.

61. A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the frequency of oscillation?

0.50 Hz

1.0 Hz

1.5 Hz

2.0 Hz

The frequency of oscillation can be calculated using the formula: frequency = 1 / time period. The time period is the time taken for one complete oscillation. In this case, the mass takes 0.50 s to go back to its equilibrium position, so the time period is 0.50 s. Therefore, the frequency of oscillation is 1 / 0.50 = 2.0 Hz.

Explanation

The frequency of oscillation can be calculated using the formula: frequency = 1 / time period. The time period is the time taken for one complete oscillation. In this case, the mass takes 0.50 s to go back to its equilibrium position, so the time period is 0.50 s. Therefore, the frequency of oscillation is 1 / 0.50 = 2.0 Hz.

62. A simple pendulum consists of a mass M attached to a weightless string of length L. For this system, when undergoing small oscillations

The frequency is proportional to the amplitude.

The period is proportional to the amplitude.

The frequency is independent of the mass M.

The frequency is independent of the length L.

The frequency of a simple pendulum is determined by the length of the string and the acceleration due to gravity, but it is independent of the mass. This is because the mass does not affect the time it takes for the pendulum to swing back and forth. The only factors that affect the frequency are the length of the string and the strength of gravity. Therefore, the frequency of a simple pendulum is independent of the mass M.

Explanation

The frequency of a simple pendulum is determined by the length of the string and the acceleration due to gravity, but it is independent of the mass. This is because the mass does not affect the time it takes for the pendulum to swing back and forth. The only factors that affect the frequency are the length of the string and the strength of gravity. Therefore, the frequency of a simple pendulum is independent of the mass M.

63. The velocity of propagation of a transverse wave on a 2.0-m long string fixed at both ends is 200 m/s. Which one of the following is not a resonant frequency of this string?

25 Hz

50 Hz

100 Hz

200 Hz

The resonant frequencies of a string fixed at both ends are determined by the length of the string and the velocity of propagation of the wave. The resonant frequencies can be calculated using the formula f = (n * v) / (2 * L), where f is the frequency, n is the harmonic number, v is the velocity of propagation, and L is the length of the string. In this case, since the velocity of propagation is given as 200 m/s and the length of the string is 2.0 m, the resonant frequencies can be calculated as follows:
f1 = (1 * 200) / (2 * 2.0) = 50 Hz
f2 = (2 * 200) / (2 * 2.0) = 100 Hz
f3 = (3 * 200) / (2 * 2.0) = 150 Hz
f4 = (4 * 200) / (2 * 2.0) = 200 Hz
Since 25 Hz is not one of the calculated resonant frequencies, it is not a resonant frequency of this string.

Explanation

The resonant frequencies of a string fixed at both ends are determined by the length of the string and the velocity of propagation of the wave. The resonant frequencies can be calculated using the formula f = (n * v) / (2 * L), where f is the frequency, n is the harmonic number, v is the velocity of propagation, and L is the length of the string. In this case, since the velocity of propagation is given as 200 m/s and the length of the string is 2.0 m, the resonant frequencies can be calculated as follows:
f1 = (1 * 200) / (2 * 2.0) = 50 Hz
f2 = (2 * 200) / (2 * 2.0) = 100 Hz
f3 = (3 * 200) / (2 * 2.0) = 150 Hz
f4 = (4 * 200) / (2 * 2.0) = 200 Hz
Since 25 Hz is not one of the calculated resonant frequencies, it is not a resonant frequency of this string.

64. The time for one cycle of a periodic process is called the

Amplitude.

Wavelength.

The time for one cycle of a periodic process is called the wavelength. This is because wavelength refers to the distance between two corresponding points on a wave, such as the crest or trough. In the context of a periodic process, the wavelength represents the time it takes for the process to complete one full cycle. Therefore, the correct answer is wavelength.

Explanation

The time for one cycle of a periodic process is called the wavelength. This is because wavelength refers to the distance between two corresponding points on a wave, such as the crest or trough. In the context of a periodic process, the wavelength represents the time it takes for the process to complete one full cycle. Therefore, the correct answer is wavelength.

65. A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the amplitude of vibration?

0.024 m

0.058 m

0.24 m

0.58 m

The amplitude of vibration can be determined using the equation for the speed of an object in simple harmonic motion at the equilibrium position, which is given by v = ωA, where v is the speed, ω is the angular frequency, and A is the amplitude. Rearranging the equation, we have A = v/ω. Since the speed is given as 1.5 m/s and the angular frequency can be calculated using the formula ω = √(k/m), where k is the spring constant and m is the mass, we can substitute the given values to find A = 1.5/√(20/0.50) = 0.24 m.

Explanation

The amplitude of vibration can be determined using the equation for the speed of an object in simple harmonic motion at the equilibrium position, which is given by v = ωA, where v is the speed, ω is the angular frequency, and A is the amplitude. Rearranging the equation, we have A = v/ω. Since the speed is given as 1.5 m/s and the angular frequency can be calculated using the formula ω = √(k/m), where k is the spring constant and m is the mass, we can substitute the given values to find A = 1.5/√(20/0.50) = 0.24 m.

66. In seismology, the S wave is a transverse wave. As an S wave travels through the Earth, the relative motion between the S wave and the particles is

Parallel.

Perpendicular.

First parallel, then perpendicular.

First perpendicular, then parallel.

The S wave in seismology is a transverse wave, meaning that the particles vibrate perpendicular to the direction of wave propagation. This means that as the S wave travels through the Earth, the relative motion between the S wave and the particles is perpendicular.

Explanation

The S wave in seismology is a transverse wave, meaning that the particles vibrate perpendicular to the direction of wave propagation. This means that as the S wave travels through the Earth, the relative motion between the S wave and the particles is perpendicular.

67. Two wave pulses with equal positive amplitudes pass each other on a string, one is traveling toward the right and the other toward the left. At the point that they occupy the same region of space at the same time

Constructive interference occurs.

Destructive interference occurs.

A standing wave is produced.

A traveling wave is produced.

When two wave pulses with equal positive amplitudes pass each other on a string, they undergo constructive interference. This means that the amplitudes of the two pulses add up to create a larger amplitude at the point where they overlap. This occurs because the peaks of one pulse align with the peaks of the other pulse, resulting in a combined wave with a higher amplitude.

Explanation

When two wave pulses with equal positive amplitudes pass each other on a string, they undergo constructive interference. This means that the amplitudes of the two pulses add up to create a larger amplitude at the point where they overlap. This occurs because the peaks of one pulse align with the peaks of the other pulse, resulting in a combined wave with a higher amplitude.

68. An object in simple harmonic motion obeys the following position versus time equation: y = (0.50 m) sin (π/2 t). What is the amplitude of vibration?

0.25 m

0.50 m

0.75 m

1.0 m

The equation given for the position versus time of the object in simple harmonic motion is y = (0.50 m) sin (π/2 t). In this equation, the coefficient of sin (π/2 t) represents the amplitude of vibration. Therefore, the amplitude of vibration is 0.50 m.

Explanation

The equation given for the position versus time of the object in simple harmonic motion is y = (0.50 m) sin (π/2 t). In this equation, the coefficient of sin (π/2 t) represents the amplitude of vibration. Therefore, the amplitude of vibration is 0.50 m.

69. A string of mass m and length L is under tension T. The speed of a wave in the string is v. What will be the speed of a wave in the string if the length is increased to 2L, with no change in mass?

0.5v

0.71v

1.4v

2v

When the length of the string is doubled to 2L, the speed of the wave in the string will also double. This is because the speed of a wave in a string is directly proportional to the square root of the tension in the string and inversely proportional to the square root of the linear density (mass per unit length) of the string. Since the mass of the string remains unchanged, the linear density remains the same. Therefore, the speed of the wave will be 2v, which is 1.4 times the original speed v.

Explanation

When the length of the string is doubled to 2L, the speed of the wave in the string will also double. This is because the speed of a wave in a string is directly proportional to the square root of the tension in the string and inversely proportional to the square root of the linear density (mass per unit length) of the string. Since the mass of the string remains unchanged, the linear density remains the same. Therefore, the speed of the wave will be 2v, which is 1.4 times the original speed v.

70. A 2.0-kg mass is attached to the end of a horizontal spring of spring constant 50 N/m and set into simple harmonic motion with an amplitude of 0.10 m. What is the total mechanical energy of this system?

0.020 J

25 J

0.25 J

1.0 J

not-available-via-ai

Explanation

not-available-via-ai

71. An object in simple harmonic motion obeys the following position versus time equation: y = (0.50 m) sin (π/2 t). What is the period of vibration?

1.0 s

2.0 s

3.0 s

4.0 s

The period of vibration can be determined by examining the equation for simple harmonic motion. In this equation, y = (0.50 m) sin (π/2 t), the coefficient of t inside the sin function determines the period. In this case, the coefficient is π/2, which means that the period is equal to 2π divided by π/2, resulting in a period of 4.0 seconds.

Explanation

The period of vibration can be determined by examining the equation for simple harmonic motion. In this equation, y = (0.50 m) sin (π/2 t), the coefficient of t inside the sin function determines the period. In this case, the coefficient is π/2, which means that the period is equal to 2π divided by π/2, resulting in a period of 4.0 seconds.

72. A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same?

0.017 m

0.029 m

0.12 m

0.17 m

When the object is at the equilibrium position, the potential energy is at its maximum and the kinetic energy is zero. As the object moves away from the equilibrium position, the potential energy decreases and the kinetic energy increases. At some point, the potential energy and kinetic energy will be equal. This occurs when the object is at the maximum displacement from the equilibrium position, which is 0.17 m. At this point, the potential energy is at its minimum and the kinetic energy is at its maximum, resulting in them being equal.

Explanation

When the object is at the equilibrium position, the potential energy is at its maximum and the kinetic energy is zero. As the object moves away from the equilibrium position, the potential energy decreases and the kinetic energy increases. At some point, the potential energy and kinetic energy will be equal. This occurs when the object is at the maximum displacement from the equilibrium position, which is 0.17 m. At this point, the potential energy is at its minimum and the kinetic energy is at its maximum, resulting in them being equal.

73. A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the spring 2.0 cm downward. The mass is then pulled an additional distance of 1.0 cm down and released from rest. Calculate the period of oscillation.

0.14 s

0.28 s

0.020 s

0.078 s

When the mass is pulled down and released, it will oscillate up and down around the equilibrium position. The period of oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

In this case, the mass is 0.30 kg, and the spring constant can be calculated using Hooke's Law: k = F/x, where F is the force and x is the displacement. The force can be calculated using the weight of the mass: F = mg. The displacement is the total distance the mass is stretched from the equilibrium position, which is 2.0 cm + 1.0 cm = 3.0 cm = 0.03 m.

So, the spring constant is k = (0.30 kg)(9.8 m/s^2) / 0.03 m = 98 N/m.

Plugging the values into the formula, we get T = 2π√(0.30 kg / 98 N/m) ≈ 0.28 s.

Therefore, the correct answer is 0.28 s.

Explanation

When the mass is pulled down and released, it will oscillate up and down around the equilibrium position. The period of oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

In this case, the mass is 0.30 kg, and the spring constant can be calculated using Hooke's Law: k = F/x, where F is the force and x is the displacement. The force can be calculated using the weight of the mass: F = mg. The displacement is the total distance the mass is stretched from the equilibrium position, which is 2.0 cm + 1.0 cm = 3.0 cm = 0.03 m.

So, the spring constant is k = (0.30 kg)(9.8 m/s^2) / 0.03 m = 98 N/m.

Plugging the values into the formula, we get T = 2π√(0.30 kg / 98 N/m) ≈ 0.28 s.

Therefore, the correct answer is 0.28 s.

74. Increasing the amplitude of a mass-and-spring system causes what kind of change in the resonant frequency of the system? (Assume no other changes in the system.)

The frequency increases.

The frequency decreases.

There is no change in the frequency.

The frequency depends on the displacement, not the amplitude.

The resonant frequency of a mass-and-spring system is determined by the mass and stiffness of the system, not the amplitude. Therefore, increasing the amplitude of the system will not cause any change in the resonant frequency.

Explanation

The resonant frequency of a mass-and-spring system is determined by the mass and stiffness of the system, not the amplitude. Therefore, increasing the amplitude of the system will not cause any change in the resonant frequency.

75. The intensity of a wave is

Proportional to both the amplitude squared and the frequency squared

Proportional to the amplitude squared and inversely proportional to the frequency squared.

Inversely proportional to the amplitude squared and proportional to the frequency squared.

Inversely proportional to both the amplitude squared and the frequency squared.

The intensity of a wave is proportional to both the amplitude squared and the frequency squared because the amplitude of a wave determines its energy, and squaring the amplitude gives us a measure of the energy carried by the wave. Similarly, the frequency of a wave determines the number of oscillations per unit time, and squaring the frequency gives us a measure of the power or rate at which the wave carries energy. Therefore, the intensity of a wave is determined by both the amplitude squared and the frequency squared.

Explanation

The intensity of a wave is proportional to both the amplitude squared and the frequency squared because the amplitude of a wave determines its energy, and squaring the amplitude gives us a measure of the energy carried by the wave. Similarly, the frequency of a wave determines the number of oscillations per unit time, and squaring the frequency gives us a measure of the power or rate at which the wave carries energy. Therefore, the intensity of a wave is determined by both the amplitude squared and the frequency squared.

76. A 4.0-kg object is attached to a spring of spring constant 10 N/m. The object is displaced by 5.0 cm from the equilibrium position and let go. What is the period of vibration?

2.0 s

4.0 s

8.0 s

16 s

The period of vibration of an object attached to a spring is determined by the mass of the object and the spring constant. In this case, the mass of the object is 4.0 kg and the spring constant is 10 N/m. The period of vibration can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Plugging in the values, we get T = 2π√(4.0 kg / 10 N/m) = 2π√(0.4 kg/m) = 2π(0.6325 s) ≈ 4.0 s. Therefore, the correct answer is 4.0 s.

Explanation

The period of vibration of an object attached to a spring is determined by the mass of the object and the spring constant. In this case, the mass of the object is 4.0 kg and the spring constant is 10 N/m. The period of vibration can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Plugging in the values, we get T = 2π√(4.0 kg / 10 N/m) = 2π√(0.4 kg/m) = 2π(0.6325 s) ≈ 4.0 s. Therefore, the correct answer is 4.0 s.

77. A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its kinetic energy is a minimum?

At either A or B

Midway between A and B

One-fourth of the way between A and B

When the mass is at either point A or point B, it is at the extreme positions of its motion. At these points, the mass momentarily stops and changes direction. Therefore, its velocity is zero, and since kinetic energy is directly proportional to the square of velocity, the kinetic energy is also zero. This means that the kinetic energy is at its minimum when the mass is located at either point A or point B.

Explanation

When the mass is at either point A or point B, it is at the extreme positions of its motion. At these points, the mass momentarily stops and changes direction. Therefore, its velocity is zero, and since kinetic energy is directly proportional to the square of velocity, the kinetic energy is also zero. This means that the kinetic energy is at its minimum when the mass is located at either point A or point B.

78. A string of mass m and length L is under tension T. The speed of a wave in the string is v. What will be the speed of a wave in the string if the mass of the string is increased to 2m, with no change in length?

0.5v

0.71v

1.4v

2v

When the mass of the string is increased to 2m, with no change in length, the speed of the wave in the string will decrease. This is because the speed of a wave in a string is inversely proportional to the square root of the mass per unit length. Therefore, when the mass is doubled, the speed of the wave will decrease by a factor of square root of 2, which is approximately 0.71. Hence, the speed of the wave in the string will be 0.71v.

Explanation

When the mass of the string is increased to 2m, with no change in length, the speed of the wave in the string will decrease. This is because the speed of a wave in a string is inversely proportional to the square root of the mass per unit length. Therefore, when the mass is doubled, the speed of the wave will decrease by a factor of square root of 2, which is approximately 0.71. Hence, the speed of the wave in the string will be 0.71v.

79. A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?

2.0 J

4.0 J

8.0 J

16 J

When the mass is cut in half, the amplitude of the motion remains the same. However, the energy of the system is directly proportional to the mass. Since the mass is halved, the energy of the system is also halved. Therefore, the energy of the system now is 8.0 J.

Explanation

When the mass is cut in half, the amplitude of the motion remains the same. However, the energy of the system is directly proportional to the mass. Since the mass is halved, the energy of the system is also halved. Therefore, the energy of the system now is 8.0 J.

80. A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the period of oscillation?

0.50 s

1.0 s

1.5 s

2.0 s

The period of oscillation is the time it takes for one complete cycle of oscillation. In this case, the mass is attached to a spring and is oscillating back and forth. The time it takes for the mass to go back to its equilibrium position is 0.50 s. Since one complete cycle consists of the mass going from one extreme position to the other and back to the equilibrium position, the period of oscillation is twice the time it takes for the mass to go back to its equilibrium position. Therefore, the period of oscillation is 2.0 s.

Explanation

The period of oscillation is the time it takes for one complete cycle of oscillation. In this case, the mass is attached to a spring and is oscillating back and forth. The time it takes for the mass to go back to its equilibrium position is 0.50 s. Since one complete cycle consists of the mass going from one extreme position to the other and back to the equilibrium position, the period of oscillation is twice the time it takes for the mass to go back to its equilibrium position. Therefore, the period of oscillation is 2.0 s.

81. Doubling only the amplitude of a vibrating mass-and-spring system produces what effect on the system's mechanical energy?

Increases the energy by a factor of two

Ncreases the energy by a factor of three

Increases the energy by a factor of four

Produces no change

Doubling the amplitude of a vibrating mass-and-spring system increases the mechanical energy by a factor of four. This is because the mechanical energy of a vibrating system is directly proportional to the square of the amplitude. Therefore, when the amplitude is doubled, the mechanical energy increases by a factor of four (2^2 = 4).

Explanation

Doubling the amplitude of a vibrating mass-and-spring system increases the mechanical energy by a factor of four. This is because the mechanical energy of a vibrating system is directly proportional to the square of the amplitude. Therefore, when the amplitude is doubled, the mechanical energy increases by a factor of four (2^2 = 4).

82. Doubling only the mass of a vibrating mass-and-spring system produces what effect on the system's mechanical energy?

Increases the energy by a factor of two

Increases the energy by a factor of three

Increases the energy by a factor of four

Produces no change

Doubling the mass of a vibrating mass-and-spring system does not produce any change in the system's mechanical energy. The mechanical energy of a mass-and-spring system is determined by the amplitude of the vibration and the spring constant, but not the mass. Therefore, increasing the mass does not affect the mechanical energy of the system.

Explanation

Doubling the mass of a vibrating mass-and-spring system does not produce any change in the system's mechanical energy. The mechanical energy of a mass-and-spring system is determined by the amplitude of the vibration and the spring constant, but not the mass. Therefore, increasing the mass does not affect the mechanical energy of the system.

83. Increasing the spring constant k of a mass-and-spring system causes what kind of change in the resonant frequency of the system? (Assume no change in the system's mass m.)

The frequency increases.

The frequency decreases.

There is no change in the frequency.

When the spring constant k of a mass-and-spring system is increased, the resonant frequency of the system also increases. This is because the resonant frequency is inversely proportional to the square root of the spring constant. Therefore, as the spring constant increases, the resonant frequency increases as well.

Explanation

When the spring constant k of a mass-and-spring system is increased, the resonant frequency of the system also increases. This is because the resonant frequency is inversely proportional to the square root of the spring constant. Therefore, as the spring constant increases, the resonant frequency increases as well.

84. Increasing the mass M of a mass-and-spring system causes what kind of change in the resonant frequency of the system? (Assume no change in the system's spring constant k.)

The frequency increases.

The frequency decreases.

There is no change in the frequency.

When the mass of a mass-and-spring system is increased, the resonant frequency of the system decreases. This can be explained by the equation for the resonant frequency of a mass-spring system, which is given by f = (1/2π)√(k/m), where f is the frequency, k is the spring constant, and m is the mass. As the mass is increased, the denominator of the equation increases, causing the overall frequency to decrease. Therefore, increasing the mass of the system results in a decrease in the resonant frequency.

Explanation

When the mass of a mass-and-spring system is increased, the resonant frequency of the system decreases. This can be explained by the equation for the resonant frequency of a mass-spring system, which is given by f = (1/2π)√(k/m), where f is the frequency, k is the spring constant, and m is the mass. As the mass is increased, the denominator of the equation increases, causing the overall frequency to decrease. Therefore, increasing the mass of the system results in a decrease in the resonant frequency.

85. An object in simple harmonic motion obeys the following position versus time equation: y = (0.50 m) sin (π/2 t). What is the maximum speed of the object?

0.13 m/s

0.26 m/s

0.39 m/s

0.79 m/s

The maximum speed of an object in simple harmonic motion occurs when the displacement is maximum, which is equal to the amplitude of the motion. In this case, the amplitude is given as 0.50 m. Therefore, the maximum speed of the object is also 0.50 m, which is equal to 0.79 m/s.

Explanation

The maximum speed of an object in simple harmonic motion occurs when the displacement is maximum, which is equal to the amplitude of the motion. In this case, the amplitude is given as 0.50 m. Therefore, the maximum speed of the object is also 0.50 m, which is equal to 0.79 m/s.

86. For a forced vibration, the amplitude of vibration is found to depend on the

Sum of the external frequency and the natural frequency.

Difference of the external frequency and the natural frequency.

Product of the external frequency and the natural frequency.

Ratio of the external frequency and the natural frequency.

In forced vibration, the external frequency and the natural frequency of the system are different. The amplitude of vibration is determined by the difference between these two frequencies. If the external frequency is close to the natural frequency, the amplitude will be high. However, if the external frequency is significantly different from the natural frequency, the amplitude will be low. Therefore, the correct answer is the difference of the external frequency and the natural frequency.

Explanation

In forced vibration, the external frequency and the natural frequency of the system are different. The amplitude of vibration is determined by the difference between these two frequencies. If the external frequency is close to the natural frequency, the amplitude will be high. However, if the external frequency is significantly different from the natural frequency, the amplitude will be low. Therefore, the correct answer is the difference of the external frequency and the natural frequency.

87. Doubling only the spring constant of a vibrating mass-and-spring system produces what effect on the system's mechanical energy?

Increases the energy by a factor of two

Increases the energy by a factor of three

Increases he energy by a factor of four

Produces no change

Doubling the spring constant of a vibrating mass-and-spring system increases the system's mechanical energy by a factor of two. This is because the spring constant is directly proportional to the potential energy stored in the spring. Increasing the spring constant results in a stiffer spring, which allows for more energy to be stored in the system. As a result, the mechanical energy of the system is doubled.

Explanation

Doubling the spring constant of a vibrating mass-and-spring system increases the system's mechanical energy by a factor of two. This is because the spring constant is directly proportional to the potential energy stored in the spring. Increasing the spring constant results in a stiffer spring, which allows for more energy to be stored in the system. As a result, the mechanical energy of the system is doubled.