AP Physics 1 Oscillations — Worked Answer Explanations

Unit 7 · 9% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Oscillations practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Oscillations practice test and come back here to review, or head back to the Oscillations unit overview.

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Question 1 · Easy
A mass on a spring oscillates with period $T = 2.0 s$ . What is its frequency?
- A
  $0.50 Hz$ Correct
- B
  $2.0 Hz$
  Why not B: Confused period with frequency.
- C
  $π Hz$
  Why not C: Confused frequency with angular frequency.
- D
  $4.0 Hz$
  Why not D: Used $1/ T^{2}$ instead of $1/ T$ .
Explanation
Frequency is the reciprocal of period: $f = 1/ T = 1/2.0 = 0.50 Hz$ .

Key takeaway
$f = 1/T$ — frequency and period are reciprocals.
Question 2 · Easy
A $0.50 kg$ mass on a spring oscillates with angular frequency $ω = 4.0 rad/s$ . What is the spring constant?
- A
  $2.0 N/m$
  Why not A: Forgot to square $ω$ .
- B
  $8.0 N/m$ Correct
- C
  $16 N/m$
  Why not C: Forgot mass term.
- D
  $32 N/m$
  Why not D: Doubled the answer.
Explanation
$ω = k / m$ , so $k = m ω^{2} = 0.50 \times 16 = 8.0 N/m$ .

Key takeaway
$\omega = \sqrt{k/m}$ for a mass-spring oscillator.
Question 3 · Easy
A simple pendulum of length $L = 1.0 m$ swings near Earth's surface ( $g = 10 m/s^{2}$ ). What is its approximate period?
- A
  $1.0 s$
  Why not A: Forgot the $2 π$ factor.
- B
  $2.0 s$ Correct
- C
  $2 π s$
  Why not C: Used $T = 2 π L$ without the $g$ .
- D
  $10 s$
  Why not D: Off by a large factor.
Explanation
$T = 2 π L / g = 2 π 0.1 \approx 2 π \cdot 0.316 \approx 1.99 s$ .

Key takeaway
Pendulum period $T = 2\pi\sqrt{L/g}$ — independent of mass and amplitude (small angles).
Question 4 · Easy
A mass-spring system has period $T$ . If the mass is doubled (and the spring is unchanged), the new period is:
- A
  $T /2$
  Why not A: Inverted the relationship.
- B
  $T / 2$
  Why not B: Inverted square root.
- C
  $T 2$ Correct
- D
  $2 T$
  Why not D: Forgot the square root.
Explanation
$T = 2 π m / k$ . Doubling $m$ multiplies $T$ by $2$ .

Key takeaway
Period scales as $\sqrt{m}$ for a mass-spring system.
Question 5 · Easy
Where in its motion does a simple harmonic oscillator have its maximum acceleration?
- A
  At the equilibrium position.
  Why not A: Equilibrium has zero net force, hence zero acceleration.
- B
  At the extremes of motion (turning points).Correct
- C
  At $x = A /2$ .
  Why not C: Acceleration is half its max here.
- D
  Acceleration is constant throughout SHM.
  Why not D: Confused with uniform acceleration.
Explanation
In SHM $a = - ω^{2} x$ , so | $a$ | is maximum where | $x$ | = $A$ — at the turning points.

Key takeaway
Maximum acceleration occurs at maximum displacement; max speed at equilibrium.
Question 6 · Easy
An object undergoes SHM with amplitude $A = 0.10 m$ and angular frequency $ω = 5.0 rad/s$ . What is the maximum speed?
- A
  $0.020 m/s$
  Why not A: Divided instead of multiplied.
- B
  $0.50 m/s$ Correct
- C
  $2.5 m/s$
  Why not C: Used $ω^{2} A$ — that's max acceleration, not speed.
- D
  $5.0 m/s$
  Why not D: Forgot to multiply by amplitude.
Explanation
In SHM, the speed is maximum at equilibrium: $v_{ma x} = ω A = 5.0 \times 0.10 = 0.50 m/s$ .

Key takeaway
$v_{max} = \omega A$ (at equilibrium); $a_{max} = \omega^2 A$ (at turning points).
Question 7 · Medium
A mass on a spring undergoes simple harmonic motion with amplitude $A$ . At what displacement is its kinetic energy equal to its potential energy?
- A
  $x = 0$
  Why not A: All KE — no PE.
- B
  $x = A /2$
  Why not B: PE there is $\frac{1}{4} E_{t o t}$ , KE is $\frac{3}{4} E_{t o t}$ .
- C
  $x = A / 2$ Correct
- D
  $x = A$
  Why not D: All PE — no KE.
Explanation
Total energy $E = \frac{1}{2} k A^{2}$ . PE at $x$ : $\frac{1}{2} k x^{2}$ . Setting PE = $E /2$ : $\frac{1}{2} k x^{2} = \frac{1}{4} k A^{2}$ , so $x = A / 2$ .

Key takeaway
PE = KE in SHM at $x = A/\sqrt{2}$ (and $-A/\sqrt{2}$).
Question 8 · Medium
A pendulum on Earth has period $T_{E}$ . If you take it to a planet where gravity is one-fourth Earth's gravity, what is its new period?
- A
  $T_{E} /4$
  Why not A: Inverted relationship and forgot square root.
- B
  $T_{E} /2$
  Why not B: Inverted relationship.
- C
  $2 T_{E}$ Correct
- D
  $4 T_{E}$
  Why not D: Forgot square root.
Explanation
Pendulum period: $T = 2 π L / g$ , so $T \propto 1/ g$ . With $g \to g /4$ , $T$ scales by $4 = 2$ . New period: $T \to 2 T_{E}$ .

Key takeaway
Pendulum period is inversely proportional to $\sqrt{g}$.