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AP Physics C: Mechanics : Harmonic Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

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Example Questions

Example Question #22 : Mechanics Exam

A 5.0 kg mass oscillates once. The total distance it travels is 1.5 m and it takes 4.0 s to travel that distance. What is its frequency of oscillation?

Possible Answers:

1.25Hz

30Hz

7.5Hz

0.83Hz

0.25 Hz

Correct answer:

0.25 Hz

Explanation:

Frequency is only based on the period of the oscillation; all the other given information is useless for this problem. Using $\small f=\frac{1}{t}$ , we can calculate that the frequency is 0.25 Hz.

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Example Question #2 : Understanding Principles Of Harmonic Motion

A mass oscillates on a spring with period . If the mass is doubled, what is the new period of oscillation?

Possible Answers:

$\sqrt{2}T$

$\frac{T}{\sqrt{2}}$

$\frac{T}{2}$

Correct answer:

$\sqrt{2}T$

Explanation:

The formula for the period of oscillation is

$T=2\pi\sqrt{\frac{m}{k}}$ .

When we double the mass, we get:

$T=2\pi\sqrt{\frac{2m}{k}}$

Because the new factor of 2 is under the square root sign, and also in the numerator, the new period will be increased by $\sqrt{2}$ .

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Example Question #3 : Understanding Principles Of Harmonic Motion

A mass is attached to the ceiling by a spring and hangs vertically downward. The mass is stretched downward so its amplitude is units from its equilibrium position. What is the distance the mass travels during one full oscillation?

Possible Answers:

$\frac{A}{2}$

Correct answer:

Explanation:

When the mass is at point , it hasn't traveled at all. When it reaches the spring's equilibrium point, it has traveled a distance of . The mass then continues to a point that's equal to the initial distance traveled, but on the opposite side of the equilibrium point, so the total distance traveled so far is . The mass must then travel back to the starting point to complete the oscillation, so the total distance traveled is .

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Example Question #23 : Mechanics Exam

A mass is attached to a spring, which is fixed to a wall. The mass is pulled away from the spring's equilibrium point and is then released. At what point does the mass experience its maximum kinetic energy?

Possible Answers:

Where the mass started

The mass does not have kinetic energy

At the spring's equilibrium point

Halfway through one full oscillation

Equal in all places

Correct answer:

At the spring's equilibrium point

Explanation:

The formula for determining kinetic energy is

$K= \frac{1}{2}mv^{2}$

So, kinetic energy will be greatest when the mass is moving most quickly. The force of the spring on the mass increases the mass's velocity until the spring’s equilibrium point, where the force of the spring acts against the motion of the mass, slowing it down. The mass is moving fastest at the spring's equilibrium point, so that's where its kinetic energy is greatest.

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Example Question #5 : Understanding Principles Of Harmonic Motion

A mass is attached to a spring with a spring constant of $20\frac{N}{m}$ . The mass is moved 2 m from the spring's equilibrium point. What is the total energy of the system?

Possible Answers:

320J

80J

40J

800J

120J

Correct answer:

40J

Explanation:

The total energy of the system is

$E_{total}=K+U=\frac{1}{2}mv^{2} +\frac{1}{2}kx^{2}$

but due to the face that the mass currently has no velocity, the kinetic energy term goes to zero.

$E_{total}=U=\frac{1}{2}kx^{2}$

Plugging in the given values, we can solve for the total energy of the system:

$E_{total}=\frac{1}{2}(20\frac{N}{m})(2m)^{2}$

$E_{total}=40J$

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Example Question #1 : Harmonic Motion

A mass $\small m$ is attached to a spring of force constant $\small k$ . The mass rests on a frictionless surface and oscillates horizontally, with oscillations of amplitude $\small A$ . What is the maximum velocity of this mass in terms of $\small k$ , $\small A$ , and $\small m$ ?

Possible Answers:

$\sqrt{\frac{kA^2}{m}}$

$\sqrt{\frac{kA^2}{2m}}$

$\sqrt{\frac{2kA}{m}}$

$\sqrt{\frac{kA}{2m}}$

$\sqrt{\frac{kA}{m}}$

Correct answer:

$\sqrt{\frac{kA^2}{m}}$

Explanation:

Relevant equations:

K_i+U_i=K_f+U_f

$U_{sp} = \frac{1}{2}kx^2$

$K = \frac{1}{2}mv^2$

Write expressions for the initial kinetic and potential energies, if the spring is initially stretched to the maximum amplitude before being released.

K_i = 0

$U_i = U_{sp, max}=\frac{1}{2}kA^2$

Write expressions for the final kinetic and potential energies when the spring crosses the equilibrium point.

$K_f = \frac{1}{2}mv_{max}^2$

$U_f = U_{sp,min}=\frac{1}{2}k(0)^2 = 0$

Use conservation of energy to equate the initial and final energy sums.

$0+\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2 +0$

Solve the equation to isolate $\small v_{max}$ .

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

$v_{max}^2=\frac{\frac{1}{2}kA^2}{\frac{1}{2}m}$

$v_{max}=\sqrt{\frac{kA^2}{m}}$

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Example Question #1 : Using Spring Equations

A ball is attached to a spring on a frictionless, horizontal plane. If the spring constant is $12\frac{N}{m}$ and the mass of the ball three kilograms, at what angular frequency will the system oscillate?

Possible Answers:

$3\frac{rad}{s}$

$4\frac{rad}{s}$

$2\frac{rad}{s}$

$1\frac{rad}{s}$

$5\frac{rad}{s}$

Correct answer:

$2\frac{rad}{s}$

Explanation:

The units for angular frequency, $\omega$ , are radians per second.

We need to derive the equation for angular frequency using conservation of energy.

$KE=PE_{spring}$

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Rearrange to solve for the velocity:

$v=\sqrt{\frac{kx^2}{m}}$

The velocity is also the product of angular frequency and the distance of the oscillation:

$v=\omega x$

Use this equation to derive the equation for angular frequency:

$\omega=\frac{v}{x}=\frac{\sqrt{\frac{kx^2}{m}}}{x}=\sqrt{\frac{k}{m}}$

Finally, use our given mass and spring constant to solve:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{12}{3}} = 2\ \frac{rad}{s}$

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Example Question #1 : Using Spring Equations

A 1kg ball is attached to a massless spring on a frictionless, horizontal plane. If at its equilibrium position, the ball is moving at $4\frac{m}{s}$ , how much total energy is in the system?

Possible Answers:

16J

14J

10J

12J

Correct answer:

Explanation:

At the equilibrium position, the spring does not contribute any potential energy. It is neither stretched, nor compressed.

$PE_{spring}=\frac{1}{2}kx^2\rightarrow \frac{1}{2}k(0)=0$

All of the energy in the system is kinetic energy, resulting from the given velocity:

$KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}(1kg)(4\frac{m}{s})^2=8J$

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Example Question #33 : Ap Physics C

A 500g ball is attached to a massless spring on a frictionless, horizontal plane. If at its equilibrium position, the ball is moving at $4\frac{m}{s}$ , and the spring constant is $0.125\frac{N}{m}$ , what is the maximum displacement of the ball from its equilibrium position?

Possible Answers:

0.5m

10m

Correct answer:

Explanation:

To solve this question, we will need to use conservation of energy. With no displacement, the ball has a velocity of $4\frac{m}{s}$ and zero displacement. At its maximum displacement, the velocity will be zero and all the energy will be converted to spring potential energy.

KE_i=mv_i^2

$PE_{f}=\frac{1}{2}kx_f^2$

KE_i=PE_f

mv_i^2=kx_f^2

Use our given values to solve:

$(0.5kg)(4\frac{m}{s})^2=(0.125\frac{N}{m})x^2$

$x=\sqrt{\frac{(0.5kg)(4\frac{m}{s})^2}{0.125\frac{N}{m}}}=8m$

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Example Question #31 : Mechanics Exam

A bungee-jumping company operates on a bridge 200 m above the ground. They use bungee cords that are 100 m when they are unstretched; these bungee cords have a spring constant of $25 \frac{N}{m}$ . What could the mass of their largest customer be before hitting the ground becomes an issue?

$g=9.81\frac{m}{s^2}$

Possible Answers:

127kg

90kg

135kg

150kg

100kg

Correct answer:

127kg

Explanation:

When a mass is in free fall, its potential energy (K=mgh) is increasing. In this instance, all of that energy must be counteracted by the bungee cord, which we can treat like a spring $(K=\frac{1}{2}kx^{2})$ , for the person to come to a stop. This means that we can set these two equations as equal to one another:

$mgh=\frac{1}{2}kx^{2}$

We can then rearrange this equation to solve for , the mass at which the bungee cord can be stretched as far as the ground:

$m=\frac{k\cdot x^2}{2\cdot g\cdot h}$

Substituting in the known values, we can solve for :

$m=\frac{k(total\:height - length\:of\:cord)^{2}}{2\cdot g\cdot (length\:of\:cord)}$

$m=\frac{25\frac{N}{m}(200m-100m)^2}{2\cdot 9.81\frac{m}{s^2}\cdot100m}$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Harmonic Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #22 : Mechanics Exam

Example Question #2 : Understanding Principles Of Harmonic Motion

Example Question #3 : Understanding Principles Of Harmonic Motion

Example Question #23 : Mechanics Exam

Example Question #5 : Understanding Principles Of Harmonic Motion

Example Question #1 : Harmonic Motion

Example Question #1 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #33 : Ap Physics C

Example Question #31 : Mechanics Exam

All AP Physics C: Mechanics Resources

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