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AP Physics C: Mechanics Practice Quiz: Representing and Analyzing SHM

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

An object is undergoing simple harmonic motion (SHM). Which statement best describes the relationship between its acceleration and its position relative to the equilibrium point?

All Questions (10)

An object is undergoing simple harmonic motion (SHM). Which statement best describes the relationship between its acceleration and its position relative to the equilibrium point?

A) The acceleration is directly proportional to the position.

B) The acceleration is constant and non-zero.

C) The acceleration is directly proportional to the negative of the position.

D) The acceleration is zero at all times.

Correct Answer: C

According to the derived equation for SHM, $a_{x}=-\omega^{2}x$. This shows that the acceleration ($a_x$) is directly proportional to the position ($x$) but in the opposite direction, as indicated by the negative sign. The term $\omega^2$ is a positive constant of proportionality.

The motion of a particle is described by a second-order differential equation. Which of the following equations is the defining characteristic of simple harmonic motion?

A) $\frac{d^{2}x}{dt^{2}}=c$

B) $\frac{d^{2}x}{dt^{2}}=kx$

C) $\frac{d^{2}x}{dt^{2}}=-\omega^{2}x$

D) $\frac{d^{2}x}{dt^{2}}=-bv$

Correct Answer: C

The provided content states that the position as a function of time for an object in SHM is a solution of the second-order differential equation $\frac{d^{2}x}{dt^{2}}=-\omega^{2}x$. This equation signifies that the acceleration is proportional to the negative of the displacement, which is the condition for SHM.

An object in simple harmonic motion is at its maximum positive displacement. At this specific point, which of the following is true about its velocity and acceleration?

A) Velocity is maximum, and acceleration is zero.

B) Velocity is zero, and acceleration is maximum in the positive direction.

C) Velocity is zero, and acceleration is maximum in the negative direction.

D) Both velocity and acceleration are zero.

Correct Answer: C

At the maximum positive displacement ($x = +A$), the object momentarily stops to change direction, so its velocity is zero. According to the equation $a_{x}=-\omega^{2}x$, the acceleration at this point is $a_{x}=-\omega^{2}A$. This is the maximum magnitude of acceleration, and it is directed in the negative direction (towards equilibrium).

The position of an object in SHM is given by the equation $x(t) = A\cos(\omega t + \phi)$. How can the object's acceleration as a function of time, $a(t)$, be derived from this equation?

A) By taking the first derivative of $x(t)$ with respect to time.

B) By taking the second derivative of $x(t)$ with respect to time.

C) By integrating $x(t)$ with respect to time.

D) By multiplying $x(t)$ by $-\omega$.

Correct Answer: B

Velocity is the first time derivative of position ($v = dx/dt$), and acceleration is the first time derivative of velocity ($a = dv/dt$). Therefore, acceleration is the second time derivative of position ($a = d^2x/dt^2$). Differentiating $x(t) = A\cos(\omega t + \phi)$ twice yields $a(t) = -A\omega^2\cos(\omega t + \phi)$.

An object's motion is described by the equation $\frac{d^{2}x}{dt^{2}}=-36x$. Based on this equation, what is the angular frequency, $\omega$, of the oscillation?

A) 36 rad/s

B) 18 rad/s

C) 6 rad/s

D) 1296 rad/s

Correct Answer: C

The general differential equation for SHM is $\frac{d^{2}x}{dt^{2}}=-\omega^{2}x$. By comparing this to the given equation, $\frac{d^{2}x}{dt^{2}}=-36x$, we can equate the coefficients of $x$ to find that $\omega^2 = 36$. Taking the square root gives the angular frequency $\omega = 6$ rad/s.

An object is executing simple harmonic motion. At which point in its motion is its speed the greatest?

A) At the points of maximum displacement.

B) At the equilibrium position ($x=0$).

C) At a point halfway between equilibrium and maximum displacement.

D) The speed is constant throughout the motion.

Correct Answer: B

The acceleration of the object is given by $a_{x}=-\omega^{2}x$. Acceleration is zero when the net force is zero, which occurs at the equilibrium position ($x=0$). At this point, the object is no longer slowing down and has not yet started to slow down in the other direction, so its speed is at a maximum.

The position of a particle in SHM is given by $x(t) = A\cos(\omega t + \phi)$. Which expression correctly represents the acceleration, $a(t)$, of the particle?

A) $a(t) = -A\omega\sin(\omega t + \phi)$

B) $a(t) = A\omega^2\cos(\omega t + \phi)$

C) $a(t) = -A\omega^2\cos(\omega t + \phi)$

D) $a(t) = A\omega\cos(\omega t + \phi)$

Correct Answer: C

Acceleration is the second derivative of position with respect to time. The first derivative (velocity) is $v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$. The second derivative (acceleration) is $a(t) = \frac{dv}{dt} = -A\omega^2\cos(\omega t + \phi)$. This is also consistent with the relationship $a_x = -\omega^2x$, since $x = A\cos(\omega t + \phi)$.

For an object exhibiting SHM, its acceleration is related to its angular frequency $\omega$ and position $x$. What is the magnitude of the acceleration when the object is at position $x = -A$, where A is the amplitude?

A) 0

B) $\omega A$

C) $\omega^2 A$

D) $-\omega^2 A$

Correct Answer: C

The acceleration is given by the equation $a_{x}=-\omega^{2}x$. Substituting $x = -A$ into this equation gives $a_{x}=-\omega^{2}(-A) = \omega^{2}A$. The question asks for the magnitude, which is the absolute value, so the magnitude is $\omega^2 A$.

The displacement of an oscillator is described by $x = 4\cos(2t)$. At time $t=\pi/4$ s, what is the acceleration of the oscillator?

A) 0 m/s²

B) -4 m/s²

C) -8 m/s²

D) -16 m/s²

Correct Answer: A

First, find the position at $t=\pi/4$ s: $x = 4\cos(2(\pi/4)) = 4\cos(\pi/2) = 4(0) = 0$ m. The acceleration is related to position by $a_{x}=-\omega^{2}x$. From the position equation, we can identify $\omega = 2$ rad/s. Since the position $x$ is 0, the acceleration is $a_{x}=-(2^2)(0) = 0$ m/s².

An object undergoes SHM with an amplitude A. At what position $x$ is the magnitude of its acceleration equal to one-fourth of its maximum possible acceleration?

A) $x = A/4$

B) $x = A/2$

C) $x = A/\sqrt{2}$

D) $x = A$

Correct Answer: A

The acceleration is given by $a_x = -\omega^2x$. The magnitude of the acceleration is $|a_x| = \omega^2|x|$. The maximum acceleration occurs at the maximum displacement, $x=\pm A$, so its magnitude is $a_{max} = \omega^2A$. We want to find the position $x$ where $|a_x| = a_{max}/4$. So, $\omega^2|x| = (\omega^2A)/4$. Canceling $\omega^2$ from both sides gives $|x| = A/4$.