AP Physics C Mechanics Quiz: Energy Of Simple Harmonic Oscillators

20 questions · exam conditions

0:00

Energy Of Simple Harmonic OscillatorsQuestion 1 of 20

The potential energy of a particle undergoing one-dimensional simple harmonic motion is given by the function $U(x) = (2.0 \text{ J/m}^2)x^2$ . If the total mechanical energy of the particle is 8.0 J, what is the amplitude of the oscillation?

0.5 m

1.0 m

2.0 m

4.0 m

← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Energy Of Simple Harmonic Oscillators

Practice Energy Of Simple Harmonic Oscillators in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Energy Of Simple Harmonic Oscillators, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

0.5 m
1.0 m
2.0 m (correct answer)
4.0 m

Explanation: The total mechanical energy

E

is equal to the maximum potential energy, which occurs at the amplitude,

x=A

. So,

E = U(A)

. We have

8.0 \text{ J} = (2.0 \text{ J/m}^2)A^2

. Solving for

A^2

gives

A^2 = 8.0/2.0 = 4.0 \text{ m}^2

. Taking the square root gives the amplitude

A = 2.0

Question 2

A physical pendulum consists of a rigid object of mass $M$ that oscillates about a pivot point a distance $d$ from its center of mass. It is released from rest at a small maximum angular displacement $\theta_{max}$ . Which of the following changes will increase the total mechanical energy of the pendulum-Earth system?

Decreasing the mass $M$ of the pendulum while keeping $d$ and $\theta_{max}$ constant.
Moving the pivot point closer to the center of mass, decreasing $d$ .
Increasing the mass $M$ of the pendulum while keeping $d$ and $\theta_{max}$ constant. (correct answer)
Performing the experiment on a planet with a smaller acceleration due to gravity.

Explanation: The total mechanical energy is determined by the maximum gravitational potential energy. The maximum height of the center of mass above its equilibrium position is

h = d(1-\cos\theta_{max})

. The total energy is

E = Mgh = Mgd(1-\cos\theta_{max})

. To increase

E

, one must increase

M

g

d

, or

\theta_{max}

. Increasing the mass

M

while other factors are constant will increase the total energy.

Question 3

A simple pendulum consists of a bob of mass $m$ attached to a string of length $L$ . The pendulum is pulled back to a maximum angle $\theta_{max}$ with the vertical and released from rest. What is the total mechanical energy of the pendulum-Earth system with respect to the lowest point of the swing?

$E = mgL$
$E = mgL(1 - \cos\theta_{max})$ (correct answer)
$E = \frac{1}{2}mgL\theta_{max}^2$
$E = mgL\sin\theta_{max}$

Explanation: The total mechanical energy is conserved. At the maximum angular displacement, the bob is momentarily at rest, so its kinetic energy is zero. The total energy is equal to the gravitational potential energy at that point. The height of the bob above the lowest point is

h = L - L\cos\theta_{max} = L(1 - \cos\theta_{max})

. Therefore, the total energy is

E = mgh = mgL(1 - \cos\theta_{max})

Question 4

A block undergoing simple harmonic motion on a frictionless surface has a total mechanical energy $E$ . At which displacement from the equilibrium position is the kinetic energy of the block equal to its potential energy?

$x = 0$
$x = \pm \frac{A}{2}$
$x = \pm \frac{A}{\sqrt{2}}$ (correct answer)
$x = \pm A$

Explanation: The total energy is

E = K + U

. We want the position where

K=U

. This means

E = U + U = 2U

. The potential energy is

U = \frac{1}{2}kx^2

and the total energy is

E = \frac{1}{2}kA^2

. Substituting these into

E = 2U

gives

\frac{1}{2}kA^2 = 2(\frac{1}{2}kx^2)

, which simplifies to

A^2 = 2x^2

. Solving for

x

gives

x = \pm \frac{A}{\sqrt{2}}

Question 5

The position of an object in simple harmonic motion is given by $x(t) = A\cos(\omega t)$ . The total energy of the system is $E$ . Which expression represents the kinetic energy of the object as a function of time, $K(t)$ ?

$E \cos(\omega t)$
$E \cos^2(\omega t)$
$E \sin(\omega t)$
$E \sin^2(\omega t)$ (correct answer)

Explanation: The velocity is

v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t)

. The kinetic energy is

K(t) = \frac{1}{2}mv(t)^2 = \frac{1}{2}m(-A\omega\sin(\omega t))^2 = \frac{1}{2}mA^2\omega^2\sin^2(\omega t)

. The total energy is

E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2

. Substituting

E

into the expression for

K(t)

gives

K(t) = E\sin^2(\omega t)

Question 6

For a frictionless spring-mass oscillator, which expression correctly represents conservation of mechanical energy during SHM?

$\tfrac12mv^2+\tfrac12kx^2=\tfrac12kA^2$ (correct answer)
$\tfrac12mv^2-\tfrac12kx^2=\tfrac12kA^2$
$mv^2+kx^2=kA^2$
$\tfrac12kx^2=\tfrac12kA^2+\tfrac12mv^2$
$\tfrac12mv^2+\tfrac12kx^2=0$

Explanation: This question tests AP Physics C: Mechanics understanding of energy conservation in simple harmonic oscillators, focusing on the mathematical expression. In SHM without friction, total mechanical energy remains constant throughout oscillation, expressed as the sum of kinetic and potential energies equaling the maximum potential energy at amplitude. Choice A correctly states ½mv² + ½kx² = ½kA², showing that K + U equals the total energy E. Choice B incorrectly uses subtraction instead of addition. Choice C omits the ½ factors. Choice D rearranges incorrectly, suggesting potential energy equals total plus kinetic. Choice E incorrectly states total energy is zero. To help students: derive the energy conservation equation from first principles, practice identifying correct forms of conservation laws, and verify equations using dimensional analysis and limiting cases.

Question 7

A frictionless spring-mass system has $k=200\,\text{N/m}$ and amplitude $A=0.10\,\text{m}$ . At what displacement magnitude $|x|$ is $K=U$ ?

$0.10\,\text{m}$
$0.071\,\text{m}$ (correct answer)
$0.050\,\text{m}$
$0.035\,\text{m}$
$0\,\text{m}$

Explanation: This question tests AP Physics C: Mechanics understanding of energy transformations in simple harmonic oscillators, specifically finding where kinetic and potential energies are equal. When K = U, each equals half the total energy since E = K + U. Setting U = E/2: ½kx² = ½(½kA²), which simplifies to x² = A²/2, giving |x| = A/√2 ≈ 0.707A. With A = 0.10 m, |x| = 0.10/√2 ≈ 0.0707 m ≈ 0.071 m. Choice B is correct because it matches this calculation. Choice C (0.050 m = A/2) is a common error from assuming linear rather than quadratic energy relationships. To help students: emphasize that energy varies as x², practice solving for positions where K and U have specific ratios, and use energy bar charts to visualize energy distribution at different positions.

Question 8

An object of mass $m$ is attached to an ideal horizontal spring with spring constant $k$ . The object is displaced from its equilibrium position by a distance $A$ and released from rest. Assuming no friction, what is the total mechanical energy of the object-spring system?

$E = \frac{1}{2}mv^2$
$E = \frac{1}{2}kA^2$ (correct answer)
$E = \frac{1}{2}kA$
$E = \frac{1}{2}mA^2$

Explanation: The total mechanical energy of a simple harmonic oscillator is constant. At the point of maximum displacement (the amplitude

A

), the object is momentarily at rest, so its kinetic energy is zero. At this point, all the energy is stored as potential energy in the spring, which is given by

U_s = \frac{1}{2}kA^2

. Therefore, the total mechanical energy is

E = \frac{1}{2}kA^2

Question 9

A block attached to a horizontal spring undergoes simple harmonic motion with amplitude $A$ . The experiment is repeated, but this time the block is released from rest at a displacement of $3A$ .

How does the new total mechanical energy $E_{new}$ compare to the original total mechanical energy $E_{orig}$ ?

$E_{new} = 3 E_{orig}$
$E_{new} = 6 E_{orig}$
$E_{new} = 9 E_{orig}$ (correct answer)
$E_{new} = \frac{1}{3} E_{orig}$

Explanation: The total mechanical energy of a mass-spring system in simple harmonic motion is given by

E = \frac{1}{2}kA^2

, where

k

is the spring constant and

A

is the amplitude. Since the energy is proportional to the square of the amplitude, tripling the amplitude (

A_{new} = 3A_{orig}

) results in the new energy being

3^2 = 9

times the original energy.

Question 10

An object with a mass of 2.0 kg is attached to a horizontal spring with a spring constant of 8.0 N/m. The object is pulled to a displacement of 0.50 m from equilibrium and released from rest.

What is the speed of the object when its displacement from equilibrium is 0.30 m?

0.60 m/s
0.80 m/s (correct answer)
1.0 m/s
1.4 m/s

Explanation: The total energy is

E = \frac{1}{2}kA^2 = \frac{1}{2}(8.0)(0.50)^2 = 1.0

J. At any position

x

, the total energy is the sum of kinetic and potential energy:

E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

. At

x=0.30

1.0 = \frac{1}{2}(2.0)v^2 + \frac{1}{2}(8.0)(0.30)^2

. This simplifies to

1.0 = v^2 + 0.36

, so

v^2 = 0.64

, and

v = 0.80

m/s.

Question 11

An object of mass $m$ is in simple harmonic motion with total energy $E$ . If the total energy is increased to $4E$ while the mass remains constant, by what factor does the maximum speed of the object change?

It increases by a factor of 2. (correct answer)
It increases by a factor of 4.
It increases by a factor of 16.
It remains the same.

Explanation: The total energy

E

is equal to the maximum kinetic energy,

E = \frac{1}{2}mv_{max}^2

. Therefore,

v_{max} = \sqrt{2E/m}

. The maximum speed is proportional to the square root of the total energy. If the energy is increased by a factor of 4, the maximum speed will increase by a factor of

\sqrt{4} = 2

Question 12

The position of an object undergoing simple harmonic motion is described by $x(t) = A\cos(\omega t)$ . Which statement correctly describes the kinetic energy $K$ and potential energy $U$ of the oscillator as functions of time?

Both $K$ and $U$ oscillate with angular frequency $\omega$ .
Both $K$ and $U$ oscillate with angular frequency $2\omega$ . (correct answer)
The kinetic energy $K$ is constant, while the potential energy $U$ oscillates.
The potential energy $U$ is constant, while the kinetic energy $K$ oscillates.

Explanation: Potential energy is

U \propto x^2 = A^2\cos^2(\omega t)

. Kinetic energy is

K \propto v^2

and

v \propto \sin(\omega t)

, so

K \propto \sin^2(\omega t)

. Using trigonometric identities,

\cos^2(\theta) = (1+\cos(2\theta))/2

and

\sin^2(\theta) = (1-\cos(2\theta))/2

. Both energy functions oscillate with an angular frequency of

2\omega

Question 13

A mass-spring system undergoes simple harmonic motion with amplitude $A$ , mass $m$ , and spring constant $k$ . The system is then modified such that the mass is changed to $2m$ , the spring constant is changed to $2k$ , and the amplitude is changed to $A/2$ .

What is the total mechanical energy of the modified system in terms of the original energy $E$ ?

$E/2$ (correct answer)
$E$
$2E$
$4E$

Explanation: The total mechanical energy of a mass-spring system is given by

E = \frac{1}{2}kA^2

. The original energy is

E_{orig} = \frac{1}{2}kA^2

. The energy of the modified system is

E_{mod} = \frac{1}{2}k_{mod}A_{mod}^2 = \frac{1}{2}(2k)(A/2)^2 = \frac{1}{2}(2k)(A^2/4) = \frac{1}{2}(\frac{1}{2}kA^2) = E_{orig}/2

. The mass does not affect the total energy for a given amplitude and spring constant.

Question 14

An object of mass $m$ is attached to a spring and undergoes simple harmonic motion with total energy $E$ . Which of the following expressions correctly gives the maximum speed, $v_{max}$ , of the object?

$v_{max} = \sqrt{E/m}$
$v_{max} = \sqrt{2E/m}$ (correct answer)
$v_{max} = E/m$
$v_{max} = 2E/m$

Explanation: The total mechanical energy

E

is constant. At the equilibrium position, the potential energy is zero and the kinetic energy is maximum. Therefore, the total energy is equal to the maximum kinetic energy:

E = K_{max} = \frac{1}{2}mv_{max}^2

. Solving for

v_{max}

gives

v_{max} = \sqrt{2E/m}

Question 15

System 1 consists of a block of mass $M$ attached to a spring of constant $k$ , oscillating with amplitude $A$ . System 2 consists of a block of mass $2M$ attached to a spring of constant $k/2$ , oscillating with amplitude $2A$ .

What is the ratio of the total mechanical energy of System 2 to that of System 1, $E_2/E_1$ ?

1/2
1
2 (correct answer)
4

Explanation: The total energy of a mass-spring oscillator is given by

E = \frac{1}{2}kA^2

. For System 1,

E_1 = \frac{1}{2}kA^2

. For System 2,

E_2 = \frac{1}{2}k_2A_2^2 = \frac{1}{2}(k/2)(2A)^2 = \frac{1}{2}(k/2)(4A^2) = 2(\frac{1}{2}kA^2) = 2E_1

. The ratio

E_2/E_1

is 2.

Question 16

A mass-spring system is oscillating with amplitude $A$ and total energy $E$ . The motion is subject to a small damping force. Which of the following correctly describes the energy of the system after a long time?

The total mechanical energy remains constant at $E$ .
The total mechanical energy gradually approaches zero. (correct answer)
The total mechanical energy increases due to the damping force.
The energy oscillates between zero and $E$ .

Explanation: A damping force is a non-conservative force that does negative work on the system. This work removes mechanical energy from the system, typically converting it into thermal energy. As a result, the amplitude of oscillation decreases, and the total mechanical energy of the system gradually decreases, approaching zero as the system comes to rest at its equilibrium position.

Question 17

A particle of mass $m$ is attached to a spring with constant $k$ and is undergoing simple harmonic motion. At a displacement $x_1$ from equilibrium, the particle has a speed $v_1$ . Which of the following expressions represents the total mechanical energy of the particle-spring system?

$\frac{1}{2}mv_1^2$
$\frac{1}{2}kx_1^2$
$\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2$ (correct answer)
$\frac{1}{2}mv_1^2 - \frac{1}{2}kx_1^2$

Explanation: The total mechanical energy

E

of a conservative system is the sum of its kinetic energy

K

and potential energy

U

. At any point in the motion,

E = K + U

. For a mass-spring system, the kinetic energy is

\frac{1}{2}mv^2

and the elastic potential energy is

\frac{1}{2}kx^2

. Thus, at the given instant, the total energy is

\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2

Question 18

Two identical masses are attached to springs with different spring constants $k_1$ and $k_2$ where $k_2 = 4k_1$ . Both systems have the same total mechanical energy. The ratio of the amplitude of oscillation for system 1 to system 2 is:

$\frac{1}{4}$
$\frac{1}{2}$
$2$ (correct answer)
$4$

Explanation: Total energy is

E = \frac{1}{2}kA^2

. Since energies are equal:

\frac{1}{2}k_1A_1^2 = \frac{1}{2}k_2A_2^2

, so

k_1A_1^2 = k_2A_2^2

. Substituting

k_2 = 4k_1

k_1A_1^2 = 4k_1A_2^2

, which gives

A_1^2 = 4A_2^2

, so

\frac{A_1}{A_2} = 2

. Choice A assumes direct proportionality. Choice B takes the square root incorrectly. Choice D assumes inverse relationship without the square root.

Question 19

A particle executes SHM with amplitude $A$ and total energy $E$ . If the amplitude is reduced to $\frac{A}{2}$ while the mass and spring constant remain unchanged, what additional work must be done by an external agent to restore the total energy to $E$ ?

$\frac{3E}{4}$ (correct answer)
$\frac{E}{2}$
$\frac{E}{4}$
$E$

Explanation: When you encounter SHM energy problems involving amplitude changes, focus on the relationship between total energy and amplitude. In simple harmonic motion, the total energy is

E = \frac{1}{2}kA^2

, where

k

is the spring constant and

A

is the amplitude. Initially, the particle has energy

E

with amplitude

A

. When the amplitude is reduced to

\frac{A}{2}

, the new energy becomes

E_{new} = \frac{1}{2}k(\frac{A}{2})^2 = \frac{1}{2}k \cdot \frac{A^2}{4} = \frac{1}{4} \cdot \frac{1}{2}kA^2 = \frac{E}{4}

. To restore the total energy back to

E

, an external agent must add energy equal to the difference:

E - \frac{E}{4} = \frac{3E}{4}

. This makes choice A correct. Choice B (

\frac{E}{2}

) incorrectly assumes the energy scales linearly with amplitude rather than quadratically. Choice C (

\frac{E}{4}

) gives the remaining energy after the amplitude reduction, not the work needed to restore the original energy. Choice D (

E

) would be the work needed if all energy were lost, ignoring that

\frac{E}{4}

remains in the system. Remember that in SHM, energy depends on the square of the amplitude (

E \propto A^2

). When solving energy problems involving amplitude changes, always calculate the energy difference between final and initial states. This quadratic relationship is crucial for understanding how energy storage changes in oscillatory systems.

Question 20

A mass attached to a vertical spring oscillates with amplitude $A$ . Taking the equilibrium position as the reference for gravitational potential energy, at what displacement from equilibrium is the total mechanical energy (including gravitational potential energy) equal to twice the elastic potential energy?

$\pm \frac{A}{\sqrt{2}}$ (correct answer)
$\pm \frac{A}{2}$
$\pm \frac{A\sqrt{3}}{2}$
$\pm A\sqrt{2}$

Explanation: Total mechanical energy

E = \frac{1}{2}kA^2

. Elastic potential energy at displacement

x

U_e = \frac{1}{2}kx^2

. When

E = 2U_e

\frac{1}{2}kA^2 = 2 \cdot \frac{1}{2}kx^2

, so

A^2 = 2x^2

, giving

x = \pm \frac{A}{\sqrt{2}}

. Note that gravitational PE is zero at equilibrium by the problem setup. Choice B uses

A^2 = 4x^2

. Choice C comes from incorrectly using energy ratios. Choice D inverts the square root.