Question 1

A mass of 0.150 kg is attached to a spring with a force constant of 3.58 N/m and undergoes simple harmonic oscillations with an amplitude of 7.50 cm. What is the total mechanical energy of the system?

Accepted Answer

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Question 2

What is the length of a simple pendulum with a period of 2.00 s?

Accepted Answer

The period of a simple pendulum is given by the formula T = 2π√(L/g). Solving for L when T = 2.00 s and g = 9.81 m/s² gives L ≈ 0.994 m.

Question 3

A pendulum that was originally erected by Foucault at the Pantheon in Paris for the Paris Exhibition in 1851 was restored in 1995. It has a 28.0 kg sphere suspended from a 67.0-m light cable. How long would it take for the bob in this pendulum to move from the position of maximum displacement back to the equilibrium point?

Accepted Answer

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.Plugging in the given values, we get:T = 2π√(67.0 m / 9.81 m/s^2) = 16.5 sThe time it takes for the bob to move from the position of maximum displacement back to the equilibrium point is half of the period, or 8.25 s.

Question 4

A 4.20-g bullet embeds itself in a 20.0-kg block, which is attached to a horizontal spring with a force constant of 950 N/m. The maximum compression of the spring is 2.10 cm. What is the speed of the bullet just before it hits the block?

Accepted Answer

The kinetic energy of the bullet-block system is converted into potential energy of the spring. Using conservation of energy: (1/2)mv^2 = (1/2)kx^2, solve for v. Here, m = 0.0042 kg + 20 kg, k = 950 N/m, x = 0.021 m.

Question 5

A mass of 1.53 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. What is the total mechanical energy of the system?

Accepted Answer

The answer of A mass of 1.53 kg is attached...

Question 6

A spring with a force constant of 1.50 N/m is attached to a mass of 120 g. The system has a damping constant of 0.0180 Ns/m. How long does it take the amplitude of the oscillations to decrease from 10.0 mm to 5.00 mm?

Accepted Answer

The amplitude of a damped harmonic oscillator decreases exponentially over time according to the formula A(t) = A_0 * e^(-bt/2m), where A_0 is the initial amplitude, b is the damping constant, m is the mass, and t is the time. Solving for t when A(t) = 5.00 mm, A_0 = 10.0 mm, b = 0.0180 Ns/m, and m = 0.120 kg gives approximately 9.24 s.

Question 7

An astronaut has landed on another planet and conducts an experiment to determine the acceleration of gravity on that planet. She uses a pendulum 0.640 m long and determines that the period of oscillation is 2.60 s. What is the acceleration of gravity on that planet?

Accepted Answer

The period of oscillation of a pendulum is given by:
T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration of gravity.
Rearranging the equation, we get:
g = (4π²L)/T²
Substituting the given values, we get:
g = (4π² x 0.640 m) / (2.60 s)² = 3.74 m/s²
Therefore, the acceleration of gravity on the planet is 3.74 m/s², which is closest to option C (3.74 m/s²).

Question 8

A meter stick with a mass of 89.5 g is being supported at the 50.0 cm mark by a pivot that is 1.00 cm above the center of mass of the meter stick. The moment of inertia of the stick about that point is 0.00746 kg∙m². What is the period of oscillation of this system?

Accepted Answer

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Question 9

You have connected a loudspeaker that is facing upwards to a variable frequency source and notice that a small screw has fallen into the loudspeaker. As you increase the frequency of the signal you notice that at a frequency of 20.0 Hz the screw no longer moves with the speaker but starts to leave the surface of the speaker intermittently. What must be the amplitude of the oscillations?

Accepted Answer

The screw leaves the surface when the acceleration of the speaker equals gravitational acceleration. Using the formula a = (2πf)^2 * A, where a = 9.8 m/s² and f = 20 Hz, solving for A gives approximately 0.621 mm.

Question 10

A pendulum that was originally erected by Foucault at the Pantheon in Paris for the Paris Exhibition in 1851 was restored in 1995. It has a 28.0 kg sphere suspended from a 67.0-m light cable. If the amplitude of the swing is 5.00 m, what is the maximum speed of the sphere?

Accepted Answer

The answer of A pendulum that was originally erected by...

Question 11

A Christmas ornament made from a thin hollow glass sphere hangs from a small hook at its surface. It is observed to oscillate with a frequency of 2.50 Hz. What is the radius of the ornament? The moment of inertia of a hollow sphere of mass M and radius R about a point on its edge is 5MR²/3.

Accepted Answer

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Question 12

How much work is required to stretch a spring 25.2 cm if its force constant is 12.3 N/m?

Accepted Answer

The work done in stretching the spring can be calculated using the formula: W = (1/2)kx^2, where k is the force constant of the spring and x is the displacement from its equilibrium position.
Given: k = 12.3 N/m and x = 25.2 cm = 0.252 m
Therefore, W = (1/2) * 12.3 * 0.252^2 = 0.391 J, which is the closest option to the calculated value. Hence, the correct answer is (A).

Question 13

A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a point on its rim is 2MR².

Accepted Answer

The period of oscillation for a hoop rotating about a fixed axis perpendicular to the plane of the hoop is given by:

T = 2π√(I/mgd)

where I is the moment of inertia of the hoop, m is its mass, g is acceleration due to gravity, and d is the distance between the axis of rotation and the center of mass of the hoop.

In this case, the hoop is supported by a nail, which means that the axis of rotation passes through its center of mass. Therefore, d = R, where R is the radius of the hoop.

Rearranging the equation for T, we get:

I = (mgR^2)/(4π^2T^2)

Substituting the given values, we get:

I = (2MR^2)/5

Setting T = 1.00 s, we can solve for R:

R = √[(5IT^2)/(2M)]

Substituting the given values, we get:

R = √[(5)(1.00 s)^2/(2)(M)] = √(5/2)R

Solving for R, we get:

R = 0.124 m

Therefore, the best choice is C.

Question 14

A mass of 1.53 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. What is the speed of the mass when it is 3.00 cm from its equilibrium position?

Accepted Answer

We can use the equation for the speed of an object undergoing simple harmonic motion at a given displacement from the equilibrium position:

v = ω√(A^2 - x^2)

where ω is the angular frequency (ω = 2πf) and x is the displacement from the equilibrium position.

Plugging in the given values, we get:

ω = 2πf = 2π(1.95 Hz) ≈ 12.24 rad/s
A = 7.50 cm = 0.075 m
x = 3.00 cm = 0.030 m

Thus, the speed of the mass at a displacement of 3.00 cm is:

v = ω√(A^2 - x^2) ≈ 0.842 m/s

Therefore, the correct answer is D.

Question 15

An astronaut has landed on an asteroid and conducts an experiment to determine the acceleration of gravity on that asteroid. She uses a pendulum that has a period of oscillation of 2.00 s on Earth and finds that on the asteroid the period is 11.3 s. What is the acceleration of gravity on that asteroid?

Accepted Answer

The period of a pendulum is proportional to the square root of the inverse of the gravitational acceleration. Using the formula T = 2π√(L/g), and comparing the periods on Earth and the asteroid, the ratio of the periods squared gives the ratio of the gravitational accelerations. Solving for the asteroid's gravity gives approximately 0.307 m/s².

Question 16

In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the Earth using a pendulum 11.0 m long, which was set up in the Paris Observatory. What was the period of oscillation of the pendulum?

Accepted Answer

The period of oscillation 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, L = 11.0 m and g = 9.81 m/s^2.Therefore, T = 2π√(11.0 m / 9.81 m/s^2) = 6.65 s.

Question 17

An underdamped spring-mass system consists of a spring with a force constant of 2.00 N/m and a mass of 550 g. It is observed that the oscillations decrease in magnitude by 10.0% every 360 s. What is the value of the damping constant for the system?

Accepted Answer

The damping constant can be calculated using the formula for the decay of amplitude in an underdamped system: A(t) = A(0) * e^(-bt/2m). Given that the amplitude decreases by 10% every 360 s, we can solve for b using the given parameters.

Question 18

A mass of 500 g is resting on a vertical spring with a force constant of 55.0 N/m. A mass of 250 g is dropped from a height of 12.0 cm onto the larger mass and sticks to it. What is the amplitude of the resulting oscillations?

Accepted Answer

First, we need to find the initial potential energy of the 250 g mass before it is dropped onto the 500 g mass:

U = mgh = (0.25 kg)(9.81 m/s^2)(0.12 m) = 0.2943 J

When the 250 g mass is dropped onto the 500 g mass, their total mass becomes 750 g. The spring is compressed, and their potential energy is converted to spring potential energy:

0.2943 J = (1/2)(0.055 kg)(x^2)

where x is the amplitude of the resulting oscillations.

Solving for x, we get:

x = 0.0597 m = 5.97 cm

Therefore, the answer is E.

Question 19

A 16.2-g bullet with an initial speed of 870 m/s embeds itself in a 40.0-kg block, which is attached to a horizontal spring with a force constant of 1010 N/m. What is the maximum compression of the spring?

Accepted Answer

The answer of A 16.2-g bullet with an initial speed...

Question 20

A spring with a force constant of 2.00 N/m is attached to a mass of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant?

Accepted Answer

The damping constant can be calculated using the formula for damped oscillations: A(t) = A_0 * e^(-bt/2m). Solving for b using the given amplitudes and time, we find b = 0.00693 Ns/m.

Quiz 13: Oscillations About Equilibrium