AP Physics C: Mechanics Unit 7 practice questions
Try 10 multiple-choice questions from Unit 7 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.
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?
T remains 2π√(m/k) because the uniform inertial force only shifts the equilibrium position.
T decreases because the effective gravity g+a increases the restoring force on the mass.
T increases because the effective gravity g-a decreases the restoring force on the mass.
T changes because the amplitude of oscillation is reduced by the upward acceleration force.
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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?
T decreases because the effective gravitational acceleration magnitude increases to √(g²+a²).
T increases because the effective gravitational acceleration magnitude increases to √(g²+a²).
T increases because the effective gravitational acceleration magnitude decreases to √(g²-a²).
T stays constant because the period depends only on the vertical component of acceleration.
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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?
T_new = T_old/√(2) because the spring constant doubles when the length is halved.
T_new = T_old/2 because the spring constant quadruples when the length is halved.
T_new = T_old√(2) because the spring constant halves when the length is halved.
T_new = 2T_old because the spring constant quarters when the length is halved.
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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?
T = 2π/√(C) because the constant C represents the square of the angular frequency.
T = 2π√(C) because the constant C represents the inverse square of angular frequency.
T = 2π/C because the constant C represents the angular frequency of the motion directly.
T = 2πC because the constant C represents the inverse of the angular frequency directly.
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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 attached to a spring with constant oscillates with amplitude . How does the magnitude of the block's acceleration at position compare to its acceleration at position ?
It is four times greater at
It is two times greater at
It is the same at both positions
It is half as large at
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Answer: (B)
Comparing the restoring forces using and Newton's second law shows that acceleration is directly proportional to displacement (). Since is twice as large as , the acceleration at is double the acceleration at .
Question 6
Two identical springs are attached to blocks of mass and . Both systems oscillate with the same amplitude . How does the maximum speed of the block with mass compare to the maximum speed of the block with mass ?
is one-fourth of
is one-half of
is twice as large as
is four times as large as
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Answer: (C)
Analyzing the maximum speed where reveals the relationship. Since mass is in the denominator under a square root, the system with mass has half the frequency (), making twice .
Question 7
A spring with constant and mass oscillates horizontally. The same system is then hung vertically. How do the period and equilibrium position of the vertical system compare to the horizontal system?
is the same; is different
is greater; is different
is smaller; is the same
is the same; is the same
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Answer: (A)
Evaluating the period equation 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 (), unlike the unstretched equilibrium of the horizontal system.
Question 8
A particle undergoes simple harmonic motion. If the amplitude of oscillation is doubled from to , how does the new period compare to the original period ?
T_{new} = 0.5 T_{old}
T_{new} = 1.0 T_{old}
T_{new} = 1.4 T_{old}
T_{new} = 2.0 T_{old}
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Answer: (B)
Applying the formula for the period of a simple harmonic oscillator 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 at the equilibrium position compare to the magnitude of acceleration at the turning points?
is zero while is maximum
is maximum while is zero
and are both zero
and are both maximum
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Answer: (A)
Connecting acceleration to force via Newton's second law () and the spring force law () shows that at equilibrium (), force and acceleration are zero. At turning points (), displacement is maximum, making force and acceleration maximum.
Question 10
Two blocks with masses and are attached to identical springs with constant and extended to the same amplitude . How does the magnitude of acceleration of the lighter block compare to of the heavier block at the moment of release?
is one-third the magnitude of
is equal to the magnitude of
is three times the magnitude of
is nine times the magnitude of
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Answer: (C)
Using Newton's second law , and recognizing that the restoring force is identical for both blocks at release, the acceleration is inversely proportional to mass. The block with mass therefore has three times the acceleration of the block with mass .
What's on the AP Physics C: Mechanics Unit 7 practice test
The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:
- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)
Questions are drawn from 730 multiple-choice questions and 24 FRQs written for Unit 7, Oscillations.
Practice AP Physics C: Mechanics Unit 7 FRQs
Answer the unit's free-response questions one at a time and get a score on every part.
- FRQ 1 – Mathematical Routines (8 questions)
- FRQ 2 – Translation Between Representations (1 question)
- FRQ 3 – Experimental Design (6 questions)
- FRQ 4 – Qualitative/Quantitative Translation (9 questions)
Review Unit 7 before you start
- 7.1 Defining Simple Harmonic Motion (SHM)
- 7.2 Frequency and Period of SHM
- 7.3 Representing and Analyzing SHM
- 7.4 Energy of Simple Harmonic Oscillators
- 7.5 Simple and Physical Pendulums