Question 1

An object is attached to a spring, and its frequency of oscillation is measured. Then another object is connected to the first object, and the resulting mass is nine times the original value. By what factor is the frequency of oscillation changed?
A) 9
B) 3
C) 1/3
D) 1/9

Accepted Answer

The frequency of oscillation is inversely proportional to the square root of the mass. When the mass is increased by a factor of nine, the frequency will be decreased by a factor of the square root of nine, which is three. Therefore, the frequency is changed by a factor of 1/3.

Question 2

When an object is moving in simple harmonic motion, which of the following is at a minimum when the displacement from equilibrium is zero?
A) the magnitude of the velocity
B) the magnitude of the acceleration
C) the kinetic energy
D) the total mechanical energy

Accepted Answer

When the displacement from equilibrium is zero, the object is at its equilibrium position and its velocity is at its maximum. Therefore, the magnitude of the velocity is not at a minimum. However, the acceleration of the object is zero at the equilibrium position since there is no net force acting on it. Therefore, the magnitude of the acceleration is at a minimum. Kinetic energy and total mechanical energy are both constant throughout simple harmonic motion and are not affected by the position of the object, so they are not at a minimum when the displacement from equilibrium is zero.

Question 3

A 0.20-kg object is oscillating on a spring with a spring constant of k = 30 N/m. What is the potential energy of the system when the object's displacement is 0.040 m, exactly half the maximum amplitude?
A) 0.96 J
B) 0.024 J
C) 0.012 J
D) zero

Accepted Answer

The potential energy of a spring system can be calculated using the equation:

U = 0.5 * k * x^2

where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium.

At half the maximum amplitude, the object's displacement is 0.040 m / 2 = 0.020 m.

Substituting the values into the equation, we get:

U = 0.5 * 30 N/m * (0.020 m)^2 = 0.024 J

Therefore, the answer is B, 0.024 J.

Question 4

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where the acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would the pendulum frequency be changed?
A) about 6.0
B) about 2.5
C) about 0.41
D) about 0.17

Accepted Answer

The frequency of a pendulum is proportional to the square root of the acceleration due to gravity. Since gravity on the Moon is 1/6th of that on Earth, the frequency would be reduced by a factor of √(1/6), which is approximately 0.41.

Question 5

A simple pendulum has a period of 1.0 s. What is the pendulum length? (g = 9.8 m/s²) A) 0.25 m B) 0.39 m C) 0.99 m D) 1.0 m

Accepted Answer

The answer of A simple pendulum has a period of...

Question 6

What are the base SI units for the torsion constant  ?
A) 
B) 
C) 
D)

Accepted Answer

The answer of What are the base SI units for...

Question 7

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 5.0 cm. If a timer is started when its displacement is a maximum (hence x = 5 cm when t = 0), what is the acceleration magnitude when t = 3 s? A) 31.6 m/s² B) 15.8 m/s² C) 10.5 m/s² D) zero

Accepted Answer

The answer of A mass on a spring vibrates in...

Question 8

The position of a 0.64-kg mass undergoing simple harmonic motion is given by x = (0.120 m) cos ( $\pi$t/16). What is the maximum net force on the mass as it oscillates? A) 0.98 *10^-³ N B) 7.4 *10^-³ N C) 3.0 * 10^-³ N D) 4.7 N

Accepted Answer

The answer of The position of a 0.64-kg mass undergoing...

Question 9

A 0.40-kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal, frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)
A) 17 m/s
B) 35 cm/s
C) 3.5 cm/s
D) 6.9 cm/s

Accepted Answer

The answer of A 0.40-kg object, attached to a spring...

Question 10

A simple pendulum of length 2.00 m has a mass of 100 g attached. It is drawn back 30.0$\degree$ and then released. What is the maximum speed of the mass?
A) 1.14 m/s
B) 4.58 m/s
C) 2.29 m/s
D) 1.62 m/s

Accepted Answer

The answer of A simple pendulum of length 2.00 m...

Question 11

The motion of a piston in an automobile engine is nearly simple harmonic. If the 1-kg piston travels back and forth over a distance of 12.0 cm, what is its maximum speed when the engine is running at 3000 rpm?
A) 4.71 m/s
B) 9.42 m/s
C) 44.0m/s
D) 22.0 m/s

Accepted Answer

The answer of The motion of a piston in an...

Question 12

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and amplitude of 8.0 cm. If the mass of the object is 0.25 kg, what is the spring constant?
A) 40 N/m
B) 87 N/m
C) 126 N/m
D) 158 N/m

Accepted Answer

The answer of A mass on a spring vibrates in...

Question 13

For a mass suspended on a spring in the vertical direction, the time for one complete oscillation will depend on:
A) the value for g (the acceleration due to gravity).
B) the distance the mass was originally pulled down.
C) the maximum speed of the oscillating mass.
D) the time doesn't depend on any of the above.

Accepted Answer

The time for one complete oscillation of a mass suspended on a spring (the period) depends on the mass of the object and the spring constant (Hooke's Law), not on the acceleration due to gravity, the initial displacement, or the maximum speed of the oscillation.

Question 14

By what factor should the length of a simple pendulum be changed if the period of vibration were to be quadrupled?
A) 16
B) 4.0
C) 0.25
D) 1/16

Accepted Answer

The answer of By what factor should the length of...

Question 15

An object moving in simple harmonic motion has an amplitude of 0.020 m and a maximum acceleration of 20 m/s². What is the frequency of the system? A) 0.40 Hz B) 5.0 Hz C) 10.1 Hz D) 32 Hz

Accepted Answer

The maximum acceleration of a simple harmonic motion can be expressed as a = -ω^2x_max, where ω is the angular frequency and x_max is the amplitude. Rearranging this equation gives ω = sqrt(a/(-x_max)) = sqrt(20/(-0.020)) = 10 rad/s. The frequency can be found by dividing the angular frequency by 2π, so f = ω/(2π) = 10/(2π) ≈ 1.59 Hz. Therefore, the closest choice is B, which is 5.0 Hz (the only choice within a reasonable range of the calculated frequency).

Question 16

A simple pendulum has a mass of 0.50 kg and a length of 1.0 m. It is displaced through an angle of 30$\degree$ and then released. After a time, the maximum angle of swing is only 10$\degree$. How much energy has been lost to friction?
A) 2.0 J
B) 1.6 J
C) 0.66 J
D) 0.58 J

Accepted Answer

The answer of A simple pendulum has a mass of...

Question 17

By what factor must one change the weight suspended vertically from a spring coil in order to double its period of simple harmonic motion?
A) 4
B) 2
C) 1/2
D) 1/4

Accepted Answer

The period of simple harmonic motion is given by:
T = 2π√(m/k)
Where T is the period, m is the mass of the suspended weight, and k is the spring constant.
To double the period, we need to make T=2T. Therefore:
2T = 2π√(m/k)
T = π√(m/k)
(2π√(m/k)) / (π√(m/k)) = √(m/k) / √(m/k/4) = 2
Therefore, we need to change the weight suspended by a factor of 2^2 = 4 to double its period of simple harmonic motion.

Question 18

The kinetic energy of the bob on a simple pendulum swinging in simple harmonic motion has its maximum value when the displacement from equilibrium is at what point in its swing?
A) zero displacement
B) 1/4 the amplitude
C) 1/2 the amplitude
D) equal the amplitude

Accepted Answer

The kinetic energy of a pendulum is maximum at the equilibrium position (zero displacement) because this is where its speed is highest. At the equilibrium point, all the potential energy has been converted into kinetic energy.

Question 19

Four identical springs each have the same spring constant k. If these four springs are attached end to end forming a spring four times the length of one of the original springs, what will be the spring constant of the combination?
A) k
B) k/4
C) 4k
D) 1.41k

Accepted Answer

When four identical springs are attached end to end, they behave like a single equivalent spring with a spring constant that is one-fourth the spring constant of each individual spring. Therefore, the spring constant of the combination is k/4.

Question 20

Halving the mass of the bob on a simple pendulum will cause a change in the frequency of the pendulum swing by what factor?
A) 4.0
B) 2.0
C) 1.0
D) 0.50

Accepted Answer

The answer of Halving the mass of the bob on...

Quiz 11: Harmonic Motion and Elasticity

An object is attached to a spring, and its frequency of oscillation is measured. Then another object is connected to the first object, and the resulting mass is nine times the original value. By what factor is the frequency of oscillation changed?

When an object is moving in simple harmonic motion, which of the following is at a minimum when the displacement from equilibrium is zero?

A 0.20-kg object is oscillating on a spring with a spring constant of k = 30 N/m. What is the potential energy of the system when the object's displacement is 0.040 m, exactly half the maximum amplitude?

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where the acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would the pendulum frequency be changed?

A simple pendulum has a period of 1.0 s. What is the pendulum length? (g = 9.8 m/s²)

What are the base SI units for the torsion constant ?

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 5.0 cm. If a timer is started when its displacement is a maximum (hence x = 5 cm when t = 0) , what is the acceleration magnitude when t = 3 s?

The position of a 0.64-kg mass undergoing simple harmonic motion is given by x = (0.120 m) cos ( $\pi$ t/16) . What is the maximum net force on the mass as it oscillates?

A 0.40-kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal, frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)

A simple pendulum of length 2.00 m has a mass of 100 g attached. It is drawn back 30.0 $\degree$ and then released. What is the maximum speed of the mass?

The motion of a piston in an automobile engine is nearly simple harmonic. If the 1-kg piston travels back and forth over a distance of 12.0 cm, what is its maximum speed when the engine is running at 3000 rpm?

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and amplitude of 8.0 cm. If the mass of the object is 0.25 kg, what is the spring constant?

For a mass suspended on a spring in the vertical direction, the time for one complete oscillation will depend on:

By what factor should the length of a simple pendulum be changed if the period of vibration were to be quadrupled?

An object moving in simple harmonic motion has an amplitude of 0.020 m and a maximum acceleration of 20 m/s². What is the frequency of the system?

A simple pendulum has a mass of 0.50 kg and a length of 1.0 m. It is displaced through an angle of 30 $\degree$ and then released. After a time, the maximum angle of swing is only 10 $\degree$ . How much energy has been lost to friction?

By what factor must one change the weight suspended vertically from a spring coil in order to double its period of simple harmonic motion?

The kinetic energy of the bob on a simple pendulum swinging in simple harmonic motion has its maximum value when the displacement from equilibrium is at what point in its swing?

Four identical springs each have the same spring constant k. If these four springs are attached end to end forming a spring four times the length of one of the original springs, what will be the spring constant of the combination?

Halving the mass of the bob on a simple pendulum will cause a change in the frequency of the pendulum swing by what factor?