Question 1

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

Accepted Answer

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

Question 2

A mass-spring system on a horizontal frictionless surface is set in simple harmonic motion with amplitude A.The mass is then doubled and the system is again set into simple harmonic motion with the same amplitude.Which of the following is true about the total mechanical energy of the system due to doubling the mass?

Accepted Answer

The total mechanical energy of a mass-spring system only depends on the amplitude and equilibrium position of the system, not the mass. Therefore, doubling the mass will not change the total mechanical energy of the system.

Question 3

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

Accepted Answer

The period of a simple pendulum is given by:

T=2π√(L/g)

where L is the length of the pendulum, g is the acceleration due to gravity.

The maximum speed of the mass occurs at the bottom of the swing, where all the potential energy is converted to kinetic energy. Therefore, the maximum speed of the mass is given by:

v_max=√(2gL(1-cosθ))

where θ is the angle the pendulum is released from.

Substituting the given values, we get:

T=2π√(2.00/9.81)= 2.02s

θ=30°=π/6 radians

v_max=√(2×9.81×2.00×(1-cos(π/6))) = 2.29 m/s

Hence, the best choice is C with the explanation shown above.

Question 4

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

Accepted Answer

When the displacement from equilibrium is zero, the object is at its equilibrium position. At this point, the velocity of the object is at a maximum, while the acceleration is zero. Therefore, the magnitude of acceleration is at a minimum when the displacement from equilibrium is zero.

Question 5

A runaway railroad car,with mass 30 × 10⁴ kg,coasts across a level track at 2.0 m/s when it collides with a spring-loaded bumper at the end of the track.If the spring constant of the bumper is 4.0 × 10⁶ N/m,what is the maximum compression of the spring during the collision? (Assume the collision is elastic. )

Accepted Answer

The answer of A runaway railroad car,with mass 30 ×...

Question 6

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?

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 7

A mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through a point at 0.05 m displacement? The starting displacement of the mass is 0.10 m from its equilibrium position.

Accepted Answer

The period of oscillation of a mass hanging from a spring can be calculated using:

T = 2π√(m/k)

where m is the mass of the object, k is the spring constant and T is the period of oscillation.

Using the given values, we get:

T = 2π√(0.40 kg/ 160 N/m) = 0.628 s

The speed of the mass at any point in its oscillation can be calculated using:

v = ωA

where A is the amplitude (maximum displacement) of the oscillation and ω is the angular frequency, given by:

ω = 2π/T

Using the given values, we have:

ω = 2π/0.628 s = 9.98 rad/s

A = 0.10 m (given)

So the speed of the mass when it is 0.05 m from its equilibrium position can be calculated as:

v = ωA = 9.98 rad/s x 0.05 m = 0.499 m/s

Therefore, the closest answer choice is C) 1.7 m/s.

Question 8

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

Accepted Answer

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

Question 9

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

Accepted Answer

The period of a simple pendulum is proportional to the square root of its length. To quadruple the period, the length must be increased by a factor of 16 (since 4^2 = 16).

Question 10

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. )

Accepted Answer

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

Question 11

A 0.20-kg mass is oscillating on a spring over a horizontal,frictionless surface.When it is at a displacement of 2.6 cm from equilibrium,it has a kinetic energy of 2.7 J and a spring potential energy of 2.2 J.What is the maximum speed of the mass during its oscillation?

Accepted Answer

The answer of A 0.20-kg mass is oscillating on a...

Question 12

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?

Accepted Answer

The acceleration in simple harmonic motion is given by a(t) = -ω²x(t). The angular frequency ω = 2πf = 8π rad/s. At t = 3 s, x(t) = A cos(ωt) = 5 cm cos(8π * 3) = 5 cm. Thus, a(t) = -(8π)² * 0.05 m = 31.6 m/s².

Question 13

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

Accepted Answer

The frequency of a simple pendulum is independent of the mass of the bob. It depends only on the length of the pendulum and the acceleration due to gravity. Therefore, halving the mass does not change the frequency, so the factor is 1.0.

Question 14

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?

Accepted Answer

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

Question 15

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° and then released.After a time,the maximum angle of swing is only 10°.How much energy has been lost to friction?

Accepted Answer

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

Question 16

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?

Accepted Answer

The potential energy of the spring-mass system can be calculated using the formula:

U = 1/2 kx^2

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

When the object's displacement is exactly half the maximum amplitude (which is 0.080 m), x = 0.040 m.

Plugging in the given values:

U = 1/2 (30 N/m) (0.040 m)^2
U = 0.024 J

Therefore, the potential energy of the system is 0.024 J, which corresponds to choice B.

Question 17

What are the base SI units for the torsion constant 
?

Accepted Answer

The torsion constant is measured in units of force times length squared divided by angle, which in SI units is N*m^2/rad. Therefore, the answer is the only option with the correct ending identifier for this unit, which is "_TB5259_11".

Question 18

A mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.

Accepted Answer

To solve this problem, we need to use conservation of energy. At the equilibrium point, the mass has purely potential energy and no kinetic energy. As it oscillates, the potential energy is converted into kinetic energy and vice versa. At the farthest point from equilibrium, the mass has purely kinetic energy and no potential energy.

We can use the equation for the total mechanical energy of the system:

E = 1/2 kx^2 + 1/2 mv^2

where k is the spring constant, x is the displacement from equilibrium, m is the mass, and v is the speed.

At the equilibrium point, x = 0.10 m and v = 0, so the total mechanical energy is purely potential energy:

E = 1/2 kx^2 = 1/2 (160 N/m)(0.10 m)^2 = 0.80 J

At the farthest point from equilibrium, x = 0 and the potential energy is zero, so the total mechanical energy is purely kinetic energy:

E = 1/2 mv^2

We can solve for v:

v = sqrt(2E/m) = sqrt(2(0.80 J)/(0.40 kg)) = 2.0 m/s

Therefore, the answer is C.

Question 19

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?

Accepted Answer

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

Question 20

A 0.20-kg block rests on a frictionless,level surface and is attached to a horizontally aligned spring with a spring constant of 80 N/m.The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion.What is the speed of the block when it passes through the equilibrium point?

Accepted Answer

The answer of A 0.20-kg block rests on a frictionless,level...

Quiz 11: Energy in Thermal Processes Heat and Internal Energy: Part A

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

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

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

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 mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through a point at 0.05 m displacement? The starting displacement of the mass is 0.10 m from its equilibrium position.

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

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

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 0.20-kg mass is oscillating on a spring over a horizontal,frictionless surface.When it is at a displacement of 2.6 cm from equilibrium,it has a kinetic energy of 2.7 J and a spring potential energy of 2.2 J.What is the maximum speed of the mass during its oscillation?

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?

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

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?

What are the base SI units for the torsion constant ?

A mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.

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?

Quiz 11: Energy in Thermal Processes Heat and Internal Energy: Part A

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

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

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

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 mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through a point at 0.05 m displacement? The starting displacement of the mass is 0.10 m from its equilibrium position.

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

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

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 0.20-kg mass is oscillating on a spring over a horizontal,frictionless surface.When it is at a displacement of 2.6 cm from equilibrium,it has a kinetic energy of 2.7 J and a spring potential energy of 2.2 J.What is the maximum speed of the mass during its oscillation?

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?

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

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?

What are the base SI units for the torsion constant ?

A mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.

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?

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