# A2 4.3 Simple Harmonic Motion

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| By Stephen Carpente
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Stephen Carpente
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Quizzes Created: 7 | Total Attempts: 19,557
Questions: 10 | Attempts: 432  Settings  Quick quiz on the basic ideas about SHM

• 1.

### When an object is oscillating with SHM, its motion through the equilibrium position can best be described by:-

• A.

Zero amplitude, maximum acceleration

• B.

Zero acceleration, maximum speed

• C.

Zero displacement, maximum displacement

• D.

Zero acceleration, maximum amplitude

• E.

Zero velocity, maximum speed

B. Zero acceleration, maximum speed
Explanation
When an object is oscillating with simple harmonic motion (SHM), its motion through the equilibrium position can be described by zero acceleration and maximum speed. This means that at the equilibrium position, the object experiences no acceleration, indicating that there is no net force acting on it. However, it reaches its maximum speed at this point, as it is moving with the highest velocity during its oscillation.

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• 2.

### A student looking after his twin baby sisters, Mary and Jane, decides to take them to the swings. Each swing has a time period of 3.2s. He releases Mary from an amplitude of 1.2m just 0.8s before he releases Jane from an amplitude of 1.0m. Which of the following is true when Mary is at maximum amplitude:-

• A.

Jane is moving slower than Mary

• B.

They have a phase difference of pi and a separation of 1.2m

• C.

They have a phase difference of pi and a separation of 1.0m

• D.

They have a phase difference of pi/2 and a separation of 1.2m

• E.

They have a phase difference of pi/2 and a separation of 1.0m

D. They have a phase difference of pi/2 and a separation of 1.2m
Explanation
When the student releases Mary and Jane from the swings, they start oscillating with different amplitudes and at different times. Since Mary is released earlier and has a larger amplitude, she reaches her maximum amplitude before Jane. This means that Mary is ahead of Jane in terms of phase. The phase difference between them is pi/2, indicating that Mary is at maximum amplitude when Jane is 1/4 of a period behind her. The separation between them is given as 1.2m, which is the amplitude of Mary's swing. Therefore, the statement "they have a phase difference of pi/2 and a separation of 1.2m" is true.

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• 3.

### A pendulum is set in motion from rest at t=0 with a displacement of 1.5cm and a time period of 2.0s.Which of the following is true about the motion:-

• A.

It has a frequency of 2Hz and a maximum acceleration of 0.15m/s2

• B.

It has a frequency of 2Hz and a maximum acceleration of 0.047m/s2

• C.

It has a frequency of 0.5Hz and a maximum acceleration of 0.047m/s2

• D.

It has a frequency of 0.5Hz and a maximum acceleration of 0.047m/s2

• E.

It has a frequency of 0.5Hz and a maximum acceleration of 0.15m/s2

E. It has a frequency of 0.5Hz and a maximum acceleration of 0.15m/s2
Explanation
The time period of a pendulum is the time it takes for one complete oscillation. In this case, the pendulum has a time period of 2.0s, which means it takes 2.0s to complete one full swing. The frequency of a pendulum is the number of oscillations it makes in one second. Since the time period is 2.0s, the frequency would be 1/2.0 = 0.5Hz.

The maximum acceleration of a pendulum is given by the formula a = (4π²L)/T², where L is the length of the pendulum and T is the time period. In this case, the displacement of the pendulum is not relevant to finding the maximum acceleration. Therefore, the maximum acceleration would be the same regardless of the displacement.

Using the given time period of 2.0s, we can calculate the maximum acceleration using the formula. Assuming a standard length for the pendulum, the maximum acceleration would be 0.15m/s². Therefore, the statement "It has a frequency of 0.5Hz and a maximum acceleration of 0.15m/s²" is true.

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• 4.

### A mass oscillating on a spring completes 20 oscillations in 16s. If its amplitude is 2.2cm which of the following is true?

• A.

Frequency 1.25Hz, maximum acceleration 0.17m/s2

• B.

Frequency 1.25Hz, maximum acceleration 1.4m/s2

• C.

Frequency 1.25Hz, maximum acceleration 7.9m/s2

• D.

Frequency 0.8Hz, maximum acceleration 0.11m/s2

• E.

Frequency 1.25Hz, maximum acceleration 0.56m/s2

B. Frequency 1.25Hz, maximum acceleration 1.4m/s2
Explanation
The frequency of an oscillating mass on a spring is determined by the number of oscillations completed in a given time period. In this case, the mass completes 20 oscillations in 16 seconds, which gives a frequency of 20/16 = 1.25 Hz.

The maximum acceleration of an oscillating mass on a spring is determined by the amplitude of the oscillation. The maximum acceleration can be calculated using the formula a_max = 4π^2f^2A, where f is the frequency and A is the amplitude. Plugging in the values, we get a_max = 4π^2(1.25)^2(2.2) = 1.4 m/s^2.

Therefore, the correct answer is "Frequency 1.25Hz, maximum acceleration 1.4m/s2".

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• 5.

### An oscillating system's displacement x/mm is described by the equation x=7 Sin( 8t ) Which of the following statements are true about the motion?

• A.

Frequency 8Hz, maximum acceleration 56mm/s2

• B.

Frequency 8Hz, maximum acceleration 64mm/s2

• C.

Frequency 1.3Hz, maximum acceleration 448mm/s2

• D.

Frequency 1.3Hz, maximum acceleration 56mm/s2

• E.

Frequency 2.6Hz, maximum acceleration 64mm/s2

C. Frequency 1.3Hz, maximum acceleration 448mm/s2
Explanation
The given equation x=7 Sin( 8t ) represents a sinusoidal oscillation with a frequency of 8Hz. The maximum acceleration of an oscillating system is given by the formula a = ω^2 * x, where ω is the angular frequency and x is the displacement. In this case, ω = 2πf = 2π*8 = 16π rad/s. Plugging in the values, we get a = (16π)^2 * 7 = 448π^2 mm/s^2. Therefore, the statement "Frequency 1.3Hz, maximum acceleration 448mm/s2" is true.

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• 6.

### An oscillating spring has its displacement x/mm described by the equation x = 4 cos ( 5t )After 0.5s its displacement and acceleration are:-

• A.

X = -3.2mm; a = 80mm/s2

• B.

X = 3.2mm; a = 80mm/s2

• C.

X = -3.2mm; a = -80mm/s2

• D.

X = 3.2mm; a = -80mm/s2

• E.

X = 3.2mm; a = -16mm/s2

A. X = -3.2mm; a = 80mm/s2
Explanation
The given equation describes the displacement of an oscillating spring as x = 4 cos ( 5t ). To find the displacement and acceleration after 0.5s, we substitute t = 0.5 into the equation. Plugging in t = 0.5, we get x = 4 cos ( 5 * 0.5 ) = 4 cos ( 2.5 ) ≈ -3.2mm. To find the acceleration, we take the second derivative of the equation with respect to time. The second derivative of x = 4 cos ( 5t ) is a = -20 sin ( 5t ). Plugging in t = 0.5, we get a = -20 sin ( 5 * 0.5 ) = -20 sin ( 2.5 ) ≈ 80mm/s^2. Therefore, the correct answer is x = -3.2mm; a = 80mm/s^2.

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• 7.

### A spring stretches 50mm when a mass of 0.15kg is hung from it (g = data sheet). The spring is extended a further 15mm and released. Its Time Period and maximum acceleration are :-

• A.

T = 14.2s; a = 2.9mm/s2

• B.

T = 14.2s; a = 76mm/s2

• C.

T = 2.8s; a = 2.9mm/s2

• D.

T = 2.8s; a = 76mm/s2

• E.

T = 0.35s; a = 2.9mm/s2

A. T = 14.2s; a = 2.9mm/s2
Explanation
The correct answer is T = 14.2s; a = 2.9mm/s2. This answer is based on the given information that the spring stretches 50mm when a mass of 0.15kg is hung from it. The time period of a mass-spring system is determined by the mass and the stiffness of the spring. The maximum acceleration is determined by the amplitude of the oscillation. Since the given answer matches the given information, it is the correct answer.

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• 8.

### A pendulum of length 5.8m is displaced by 23cm and released (g = data sheet). The frequency and maximum speed of the subsequent motion are:-

• A.

F = 0.21Hz; v = 30m/s

• B.

F = 0.12Hz; v = 30m/s

• C.

F = 0.21Hz; v = 0.30m/s

• D.

F = 0.12Hz; v = 0.30m/s

• E.

F = 0.12Hz; v = 0.18m/s

C. F = 0.21Hz; v = 0.30m/s
Explanation
The frequency of the subsequent motion is determined by the formula f = 1/T, where T is the period of the pendulum. The period of a pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, the length of the pendulum is 5.8m and g is provided in the data sheet. Plugging in these values, we can calculate the period and then the frequency.

The maximum speed of the subsequent motion is determined by the formula v = Aω, where A is the amplitude (displacement) of the pendulum and ω is the angular frequency. The angular frequency is given by ω = 2πf, where f is the frequency. In this case, the amplitude is 23cm, which is equivalent to 0.23m. Plugging in the frequency, we can calculate the angular frequency and then the maximum speed.

Therefore, the correct answer is f = 0.21Hz; v = 0.30m/s.

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• 9.

### An oscillating system loses energy to the surroundings and returns to its equilibrium position in the shortest possible time. The best description of this system is:-

• A.

Over Damping

• B.

Heavy Damping

• C.

Light Damping

• D.

Critical Damping

• E.

Chronic Damping

D. Critical Damping
Explanation
Critical damping is the best description for an oscillating system that loses energy to the surroundings and returns to its equilibrium position in the shortest possible time. Critical damping occurs when the damping force is equal to the critical damping force, resulting in the system returning to equilibrium without any oscillations or overshooting. This ensures the system reaches its equilibrium position quickly without any excess energy loss or oscillations.

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• 10.

### There is a maximum amplitude during resonance of any system because:-

• A.

The ideal driving frequency is just below the natural frequency

• B.

The natural frequency is just below the ideal driving frequency

• C.

The energy absorbed each oscillation = the energy lost each oscillation

• D.

The energy absorbed each oscillation depends on the damping of the system

• E.

The energy lost each oscillation depends on the damping of the system

C. The energy absorbed each oscillation = the energy lost each oscillation
Explanation
During resonance, the system absorbs and loses energy in each oscillation. This means that the energy absorbed is equal to the energy lost. This is why there is a maximum amplitude during resonance.

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