Chapter 11: Vibrations And Waves

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

    • Speed.
    • Frequency.
    • Wavelength.
    • Amplitude.
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Chapter 11: Vibrations And Waves - Quiz
About This Quiz

Explore the fundamentals of vibrations and waves in this engaging quiz from Chapter 11. Test your understanding of key concepts like frequency, amplitude, and simple harmonic motion (SHM), and master the principles that govern vibrational and wave phenomena in physics.

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  • 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

    Correct Answer
    A. 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.

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  • 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

    Correct Answer
    A. 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.

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  • 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.

    Correct Answer
    A. The wave velocity depends on M, L, T.
    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.

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  • 5. 

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

    • Amplitude.

    • Wavelength.

    • Frequency.

    • Period.

    Correct Answer
    A. Amplitude.
    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.

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  • 6. 

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

    • Speed.

    • Frequency.

    • Wavelength.

    • Amplitude.

    Correct Answer
    A. Frequency.
    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.

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  • 7. 

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

    • Speed.

    • Amplitude.

    • Intensity.

    • Power.

    Correct Answer
    A. 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.

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  • 8. 

    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

    Correct Answer
    A. 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.

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  • 9. 

    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.

    Correct Answer
    A. They are 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.

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  • 10. 

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

    • Amplitude.

    • Wavelength.

    • Frequency.

    • Period.

    Correct Answer
    A. 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.

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  • 11. 

    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

    Correct Answer
    A. 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.

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  • 12. 

    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

    Correct Answer
    A. 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.

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  • 13. 

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

    • Speed.

    • Frequency.

    • Wavelength.

    • Amplitude.

    Correct Answer
    A. Wavelength.
    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.

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  • 14. 

    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.

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

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  • 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.

    Correct Answer
    A. It decreases.
    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.

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  • 16. 

    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

    Correct Answer
    A. Is a maximum.
    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.

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  • 17. 
    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? - ProProfs

    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

    Correct Answer
    A. 0.10 m
  • 18. 

    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

    Correct Answer
    A. 1500 Hz
    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.

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  • 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

    Correct Answer
    A. 1/9 s
    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 (18Ï€) 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 18Ï€ is 1/18Ï€, which simplifies to 1/9. Therefore, the period of vibration is 1/9 seconds.

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  • 20. 

    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

    Correct Answer
    A. 0.083 m
    Explanation
    The amplitude of an oscillating mass-spring system can be determined using the equation:

    Amplitude = (Force / (Spring constant)) * (1 / (2Ï€ * 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 / (2Ï€ * 2 Hz))
    Amplitude = 0.083 m

    Therefore, the amplitude is 0.083 m.

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  • 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

    Correct Answer
    A. 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.

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  • 22. 

    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

    Correct Answer
    A. 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.

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  • 23. 

    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.

    Correct Answer
    A. Destructive interference occurs.
    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.

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  • 24. 

    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

    Correct Answer
    A. 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.

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  • 25. 

    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

    Correct Answer
    A. 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.

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  • 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

    Correct Answer
    A. 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.

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  • 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

    Correct Answer
    A. 0.50 s
    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.

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  • 28. 

    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 Ë› 10^4 m/s

    • 1.7 Ë› 10^2 m/s

    • 13 m/s

    • 5.8 Ë› 10^(-3) m/s

    Correct Answer
    A. 1.7 Ë› 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.

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  • 29. 

    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.

    Correct Answer
    A. 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.

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  • 30. 

    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

    Correct Answer
    A. 1.5 Hz, 0.67 s
    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.

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  • 31. 

    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

    Correct Answer
    A. 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.

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  • 32. 

    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

    Correct Answer
    A. 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.

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  • 33. 

    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

    Correct Answer
    A. 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.

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  • 34. 

    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

    Correct Answer
    A. 1.0 m
    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.

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  • 35. 

    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

    Correct Answer
    A. Midway between A and B
    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.

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  • 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.

    Correct Answer
    A. Does not change.
    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.

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  • 37. 

    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

    Correct Answer
    A. 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.

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  • 38. 

    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

    Correct Answer
    A. 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.

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  • 39. 

    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

    Correct Answer
    A. Midway between A and B
    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.

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  • 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 maximum?

    • At either A or B

    • Midway between A and B

    • One-fourth of the way between A and B

    • None of the above

    Correct Answer
    A. At either A or B
    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.

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  • 41. 

    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

    Correct Answer
    A. 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.

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  • 42. 

    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.

    Correct Answer
    A. 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.

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  • 43. 

    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)

    Correct Answer
    A. Y = (0.01 m) cos (22.1 t)
    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.

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  • 44. 

    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.

    Correct Answer
    A. Is maximum.
    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.

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  • 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.

    Correct Answer
    A. 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.

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  • 46. 

    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.

    Correct Answer
    A. 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.

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  • 47. 

    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

    Correct Answer
    A. 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.

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  • 48. 

    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

    Correct Answer
    A. 132 Hz, 264 Hz, 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.

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  • 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

    Correct Answer
    A. 2.0 Hz
    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.

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Quiz Review Timeline (Updated): Mar 20, 2023 +

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  • Mar 20, 2023
    Quiz Edited by
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  • Sep 25, 2012
    Quiz Created by
    Drtaylor
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