---
title: "AP Physics C: Mechanics Unit 7 Practice Test"
description: "A timed AP Physics C: Mechanics Unit 7 (Oscillations) practice test in the AP format, drawn from 730 multiple-choice questions and 24 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-physics-c-mechanics/unit-test/7"
type: "unit-test"
subject: "AP Physics C: Mechanics"
unit: "Unit 7"
---

# AP Physics C: Mechanics Unit 7 Practice Test

## Overview

A timed AP Physics C: Mechanics Unit 7 (Oscillations) practice test in the AP format, drawn from 730 multiple-choice questions and 24 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

## Unit FRQs by Type

- [FRQ 1 – Mathematical Routines](/ap-physics-c-mechanics/frqs/mathematical-routines?unit=7): 8 questions
- [FRQ 2 – Translation Between Representations](/ap-physics-c-mechanics/frqs/translation-between-representations?unit=7): 1 question
- [FRQ 3 – Experimental Design](/ap-physics-c-mechanics/frqs/experimental-design?unit=7): 6 questions
- [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-c-mechanics/frqs/qualitative-quantitative-translation?unit=7): 9 questions

## Practice Questions

### Question 1

A block of mass m oscillates vertically on a spring with constant k inside an elevator accelerating upward at a. Which claim about the period T is correct based on the spring-mass model?

- (A) T remains 2π√(m/k) because the uniform inertial force only shifts the equilibrium position.
- (B) T decreases because the effective gravity g+a increases the restoring force on the mass.
- (C) T increases because the effective gravity g-a decreases the restoring force on the mass.
- (D) T changes because the amplitude of oscillation is reduced by the upward acceleration force.

**Answer: (A)** Analyzing the differential equation for a spring-mass system with a constant external force reveals that the period depends only on mass and stiffness. The upward acceleration adds a constant term to the force but does not change the restoring force coefficient k or the inertial mass m.

### Question 2

A simple pendulum of length L hangs in a cart accelerating horizontally at a. Which claim about the new period T is justified by the pendulum model?

- (A) T decreases because the effective gravitational acceleration magnitude increases to √(g²+a²).
- (B) T increases because the effective gravitational acceleration magnitude increases to √(g²+a²).
- (C) T increases because the effective gravitational acceleration magnitude decreases to √(g²-a²).
- (D) T stays constant because the period depends only on the vertical component of acceleration.

**Answer: (A)** Applying the pendulum model in a non-inertial frame requires replacing g with the effective gravitational field magnitude. The vector sum of gravity and the fictitious inertial force results in a larger effective g, which appears in the denominator of the period equation, reducing T.

### Question 3

An ideal spring of constant k is cut into two equal lengths, and a mass m is suspended from one half. Which claim describes the new period T_new compared to the original period T_old?

- (A) T_new = T_old/√(2) because the spring constant doubles when the length is halved.
- (B) T_new = T_old/2 because the spring constant quadruples when the length is halved.
- (C) T_new = T_old√(2) because the spring constant halves when the length is halved.
- (D) T_new = 2T_old because the spring constant quarters when the length is halved.

**Answer: (A)** Applying the model of spring stiffness reveals that k is inversely proportional to length; halving the length doubles the spring constant to 2k. Substituting 2k into the period equation results in the period decreasing by a factor of 1/√(2).

### Question 4

A particle of mass m moves such that its acceleration a is given by a = -Cx, where C is a positive constant. Which claim about the period T is justified by this relationship?

- (A) T = 2π/√(C) because the constant C represents the square of the angular frequency.
- (B) T = 2π√(C) because the constant C represents the inverse square of angular frequency.
- (C) T = 2π/C because the constant C represents the angular frequency of the motion directly.
- (D) T = 2πC because the constant C represents the inverse of the angular frequency directly.

**Answer: (A)** Comparing the given equation of motion to the definition of simple harmonic motion (a = -ω²x) allows for the identification of the constant C as ω². Solving for the period T = 2π/ω yields T = 2π/√(C).

### Question 5

A block of mass $$m$$ attached to a spring with constant $$k$$ oscillates with amplitude $$A$$. How does the magnitude of the block's acceleration at position $$x = A$$ compare to its acceleration at position $$x = A/2$$?

- (A) It is four times greater at $$x = A$$
- (B) It is two times greater at $$x = A$$
- (C) It is the same at both positions
- (D) It is half as large at $$x = A$$

**Answer: (B)** Comparing the restoring forces using $$F = -kx$$ and Newton's second law $$F = ma$$ shows that acceleration is directly proportional to displacement ($$a = -\frac{k}{m}x$$). Since $$A$$ is twice as large as $$A/2$$, the acceleration at $$A$$ is double the acceleration at $$A/2$$.

### Question 6

Two identical springs are attached to blocks of mass $$m$$ and $$4m$$. Both systems oscillate with the same amplitude $$A$$. How does the maximum speed $$v_1$$ of the block with mass $$m$$ compare to the maximum speed $$v_2$$ of the block with mass $$4m$$?

- (A) $$v_1$$ is one-fourth of $$v_2$$
- (B) $$v_1$$ is one-half of $$v_2$$
- (C) $$v_1$$ is twice as large as $$v_2$$
- (D) $$v_1$$ is four times as large as $$v_2$$

**Answer: (C)** Analyzing the maximum speed $$v_{max} = A\omega$$ where $$\omega = \sqrt{k/m}$$ reveals the relationship. Since mass $$m$$ is in the denominator under a square root, the system with mass $$4m$$ has half the frequency ($$\sqrt{1/4} = 1/2$$), making $$v_1$$ twice $$v_2$$.

### Question 7

A spring with constant $$k$$ and mass $$m$$ oscillates horizontally. The same system is then hung vertically. How do the period $$T$$ and equilibrium position $$x_{eq}$$ of the vertical system compare to the horizontal system?

- (A) $$T$$ is the same; $$x_{eq}$$ is different
- (B) $$T$$ is greater; $$x_{eq}$$ is different
- (C) $$T$$ is smaller; $$x_{eq}$$ is the same
- (D) $$T$$ is the same; $$x_{eq}$$ is the same

**Answer: (A)** Evaluating the period equation $$T = 2\pi\sqrt{m/k}$$ shows it depends only on mass and spring constant, not gravity or orientation. However, the vertical system has a new equilibrium position where the spring force balances gravity ($$k\Delta x = mg$$), unlike the unstretched equilibrium of the horizontal system.

### Question 8

A particle undergoes simple harmonic motion. If the amplitude of oscillation is doubled from $$A$$ to $$2A$$, how does the new period $$T_{new}$$ compare to the original period $$T_{old}$$?

- (A) $$T_{new} = 0.5 T_{old}$$
- (B) $$T_{new} = 1.0 T_{old}$$
- (C) $$T_{new} = 1.4 T_{old}$$
- (D) $$T_{new} = 2.0 T_{old}$$

**Answer: (B)** Applying the formula for the period of a simple harmonic oscillator $$T = 2\pi\sqrt{m/k}$$ reveals that period is independent of amplitude. The increased distance is exactly compensated by the higher average speed resulting from larger restoring forces.

### Question 9

A block oscillates on a horizontal spring. How does the magnitude of acceleration $$a_{eq}$$ at the equilibrium position compare to the magnitude of acceleration $$a_{turn}$$ at the turning points?

- (A) $$a_{eq}$$ is zero while $$a_{turn}$$ is maximum
- (B) $$a_{eq}$$ is maximum while $$a_{turn}$$ is zero
- (C) $$a_{eq}$$ and $$a_{turn}$$ are both zero
- (D) $$a_{eq}$$ and $$a_{turn}$$ are both maximum

**Answer: (A)** Connecting acceleration to force via Newton's second law ($$F=ma$$) and the spring force law ($$F=-kx$$) shows that at equilibrium ($$x=0$$), force and acceleration are zero. At turning points ($$x=\pm A$$), displacement is maximum, making force and acceleration maximum.

### Question 10

Two blocks with masses $$m$$ and $$3m$$ are attached to identical springs with constant $$k$$ and extended to the same amplitude $$A$$. How does the magnitude of acceleration $$a_1$$ of the lighter block compare to $$a_2$$ of the heavier block at the moment of release?

- (A) $$a_1$$ is one-third the magnitude of $$a_2$$
- (B) $$a_1$$ is equal to the magnitude of $$a_2$$
- (C) $$a_1$$ is three times the magnitude of $$a_2$$
- (D) $$a_1$$ is nine times the magnitude of $$a_2$$

**Answer: (C)** Using Newton's second law $$a = F/m$$, and recognizing that the restoring force $$F = kA$$ is identical for both blocks at release, the acceleration is inversely proportional to mass. The block with mass $$m$$ therefore has three times the acceleration of the block with mass $$3m$$.

## Review the Unit

- [7.1 Defining Simple Harmonic Motion (SHM)](/ap-physics-c-mechanics/unit-7/1-defining-simple-harmonic-motion-shm/study-guide/0XdktX7mCpAcsQF4)
- [7.2 Frequency and Period of SHM](/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw)
- [7.3 Representing and Analyzing SHM](/ap-physics-c-mechanics/unit-7/3-representing-and-analyzing-shm/study-guide/nHLSGDEgw3L81KKh)
- [7.4 Energy of Simple Harmonic Oscillators](/ap-physics-c-mechanics/unit-7/4-energy-of-simple-harmonic-oscillators/study-guide/3UvewOScTLW9RUqd)
- [7.5 Simple and Physical Pendulums](/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA)

## Other Unit Tests

- [Unit 1: Kinematics](/ap-physics-c-mechanics/unit-test/1)
- [Unit 2: Force and Translational Dynamics](/ap-physics-c-mechanics/unit-test/2)
- [Unit 3: Work, Energy, and Power](/ap-physics-c-mechanics/unit-test/3)
- [Unit 4: Linear Momentum](/ap-physics-c-mechanics/unit-test/4)
- [Unit 5: Torque and Rotational Dynamics](/ap-physics-c-mechanics/unit-test/5)
- [Unit 6: Energy and Momentum of Rotating Systems](/ap-physics-c-mechanics/unit-test/6)

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