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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.

  1. 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.

  2. 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.

  3. 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).

  4. 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).

  5. Question 5

    A 2.02.0 kg block oscillates on a frictionless surface attached to two parallel springs with constants 2020 N/m and 3030 N/m. What is the period of oscillation?

    0.4π0.4\pi s

    0.8π0.8\pi s

    2.0π2.0\pi s

    4.0π4.0\pi s

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    Answer: (A)

    Calculates parallel stiffness (5050 N/m) and uses the period formula T=2πm/kT = 2\pi\sqrt{m/k}.

  6. Question 6

    A mass oscillates with frequency ff when attached to a single spring. If the mass is attached to two of these springs connected in series, what is the new frequency?

    f/2f/2

    f/2f/\sqrt{2}

    f2f\sqrt{2}

    2f2f

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    Answer: (B)

    Recognizes series connection halves stiffness (k/2k/2), so frequency scales by 1/2\sqrt{1/2}.

  7. Question 7

    A block of mass mm oscillates on a spring of constant kk with period T0T_0. If a second identical spring is connected in parallel with the first spring, how does the new period of oscillation TnewT_{new} compare to T0T_0?

    Tnew=T0/2T_{new} = T_0 / \sqrt{2}

    Tnew=T0/2T_{new} = T_0 / 2

    Tnew=T02T_{new} = T_0 \sqrt{2}

    Tnew=2T0T_{new} = 2 T_0

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    Answer: (A)

    Connecting springs in parallel doubles the stiffness (keq=2kk_{eq} = 2k). Since T∝1/kT \propto 1/\sqrt{k}, the period decreases by a factor of 2\sqrt{2}.

  8. Question 8

    A block of mass mm is suspended from a spring of constant kk, resulting in a period T0T_0. If the spring is replaced by two springs, each with constant kk, connected in series, how does the new period TnewT_{new} compare to T0T_0?

    Tnew=T02T_{new} = T_0 \sqrt{2}

    Tnew=T0/2T_{new} = T_0 / \sqrt{2}

    Tnew=2T0T_{new} = 2 T_0

    Tnew=T0/2T_{new} = T_0 / 2

    Show answer

    Answer: (A)

    Springs in series reduce stiffness to k/2k/2. Since T∝1/kT \propto 1/\sqrt{k}, dividing kk by 2 increases TT by a factor of 2\sqrt{2}.

  9. Question 9

    System 1 consists of a mass mm attached to a spring of constant 2k2k. System 2 consists of a mass mm attached to two springs, each of constant 4k4k, connected in series. How does the natural frequency ω1\omega_1 of System 1 compare to ω2\omega_2 of System 2?

    ω1\omega_1 is equal to ω2\omega_2

    ω1\omega_1 is double ω2\omega_2

    ω1\omega_1 is half of ω2\omega_2

    ω1\omega_1 is 2\sqrt{2} times ω2\omega_2

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    Answer: (A)

    The equivalent stiffness of System 2 is keq=(1/4k+1/4k)−1=2kk_{eq} = (1/4k + 1/4k)^{-1} = 2k. Since both systems have stiffness 2k2k and mass mm, frequencies are equal.

  10. Question 10

    Scenario A consists of a mass mm oscillating on a spring of constant kk. Scenario B consists of a mass 4m4m oscillating on two springs (each constant kk) connected in parallel. How does the period TBT_B compare to TAT_A?

    TB=2TAT_B = \sqrt{2} T_A

    TB=TAT_B = T_A

    TB=2TAT_B = 2 T_A

    TB=4TAT_B = 4 T_A

    Show answer

    Answer: (A)

    Two identical springs in parallel have effective stiffness 2k2k, so system B has mass 4m4m and stiffness 2k2k. Since T=2πm/kT = 2\pi\sqrt{m/k}, TB=2π4m/2k=2π2m/k=2 TAT_B = 2\pi\sqrt{4m/2k} = 2\pi\sqrt{2}\sqrt{m/k} = \sqrt{2}\,T_A.

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 729 multiple-choice questions and 16 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.

Review Unit 7 before you start

Other AP Physics C: Mechanics unit tests