AP Physics C: Mechanics Oscillations — Worked Answer Explanations

Unit 6 · 12 questions explained

1. Question 1 · Easy

   A mass–spring system has spring constant $k=400$ N/m and mass $m=1$ kg. What is the angular frequency of oscillation?

   • A
     $200$ rad/s

     Why not A: Computed $k/m=400$ and did not take the square root.
   • B
     $20$ rad/sCorrect
   • C
     $400$ rad/s

     Why not C: Used $\omega =k/m$ without the square root.
   • D
     $2$ rad/s

     Why not D: Computed $\sqrt{k/m}/10$ without justification.
   Explanation

   $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{400}{1}}=20$ rad/s.

   Key takeaway

   Angular frequency of a mass–spring system: $\omega = \sqrt{k/m}$. Period $T = 2\pi/\omega$.

2. Question 2 · Easy

   A simple pendulum of length $L=1$ m oscillates with small amplitude. What is its period? ($g=10$ m/s²)

   • A
     $\pi$ s $\approx 3.14$ s

     Why not A: Used $T=\pi \sqrt{L/g}$ (forgot the factor of 2).
   • B
     $2\pi$ s $\approx 6.28$ s

     Why not B: Used $T=2\pi$ without the $\sqrt{L/g}$ factor (i.e., set $L/g=1$ without units).
   • C
     $\approx 2$ sCorrect
   • D
     $\approx 10$ s

     Why not D: Used $T=2\pi \sqrt{g/L}$ (inverted the ratio).
   Explanation

   $T=2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{1}{10}}=\frac{2\pi }{\sqrt{10}}\approx \frac{6.283}{3.162}\approx 1.99$ s $\approx 2$ s.

   Key takeaway

   Pendulum period $T = 2\pi\sqrt{L/g}$. At $g \approx 10$ m/s², a 1 m pendulum has $T \approx 2$ s — a useful benchmark.

3. Question 3 · Easy

   A mass $m=0.5$ kg on a spring ($k=200$ N/m) oscillates with amplitude $A=0.1$ m. What is the maximum speed?

   • A
     $1$ m/s

     Why not A: Computed $kA/m$ instead of $A\omega =A\sqrt{k/m}$.
   • B
     $2$ m/sCorrect
   • C
     $\sqrt{40}\approx 6.3$ m/s

     Why not C: Forgot to multiply by $A$ after computing $\omega =20$ rad/s.
   • D
     $0.2$ m/s

     Why not D: Used $v_{max}=kA/m$ dimensionally incorrectly.
   Explanation

   $\omega =\sqrt{k/m}=\sqrt{200/0.5}=\sqrt{400}=20$ rad/s. Maximum speed occurs at equilibrium: $v_{max}=A\omega =0.1\times 20=2$ m/s.

   Key takeaway

   Maximum speed in SHM: $v_{max} = A\omega = A\sqrt{k/m}$. It occurs when the mass passes through equilibrium.

4. Question 4 · Easy

   In simple harmonic motion, a particle's position is $x(t)=0.3\cos (5t)$ m. What is the maximum acceleration?

   • A
     $1.5$ m/s²

     Why not A: Computed $A\omega$ (maximum speed) instead of $A{\omega }^2$.
   • B
     $5$ m/s²

     Why not B: Computed $A{\omega }^2/A={\omega }^2=25$ and then divided by 5.
   • C
     $7.5$ m/s²Correct
   • D
     $25$ m/s²

     Why not D: Forgot to multiply by $A$ when computing $a_{max}=A{\omega }^2$.
   Explanation

   From $x(t)=A\cos (\omega t)$: $A=0.3$ m, $\omega =5$ rad/s. $a(t)=-A{\omega }^2\cos (\omega t)$, so $|a_{max}|=A{\omega }^2=0.3\times 25=7.5$ m/s².

   Key takeaway

   Maximum acceleration in SHM: $|a_{max}| = A\omega^2$. It occurs at the amplitude positions $x = \pm A$.

5. Question 5 · Medium

   A 0.4 kg mass on a spring ($k=160$ N/m) is released from rest at $x=0.05$ m from equilibrium. Find the total mechanical energy.

   • A
     $0.1$ J

     Why not A: Computed $kA$ instead of $\frac{1}{2}k{A}^2$.
   • B
     $0.2$ JCorrect
   • C
     $0.4$ J

     Why not C: Used $k{A}^2$ (forgot the $\frac{1}{2}$).
   • D
     $0.01$ J

     Why not D: Computed $\frac{1}{2}kA$ without squaring $A$.
   Explanation

   Released from rest at $x=A=0.05$ m, all energy is potential: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(160)(0.0025)=0.2$ J. This is the conserved total mechanical energy throughout the oscillation.

   Key takeaway

   In SHM, total energy $E = \frac{1}{2}kA^2$ is constant. It oscillates between fully potential (at $\pm A$) and fully kinetic (at equilibrium).

6. Question 6 · Medium

   Show using calculus that $x(t)=A\cos (\omega t+\varphi )$ satisfies the SHM equation $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$. What does this verify?

   • A
     That the motion is exponential

     Why not A: Exponential solutions satisfy $\ddot{x}=cx$, not $\ddot{x}=-{\omega }^{2}x$.
   • B
     That $x(t)=A\cos (\omega t+\varphi )$ is the general solution to the SHM ODECorrect
   • C
     That the amplitude must equal $\omega$

     Why not C: The amplitude $A$ and frequency $\omega$ are independent parameters.
   • D
     That the phase $\varphi$ must be zero

     Why not D: The ODE is satisfied for any constant $\varphi$.
   Explanation

   $\frac{dx}{dt}=-A\omega \sin (\omega t+\varphi )$. $\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x(t)$ ✓. This confirms that $A\cos (\omega t+\varphi )$ satisfies the ODE $\ddot{x}+{\omega }^{2}x=0$ for any constants $A$ and $\varphi$. Since this is a second-order ODE, the general solution requires exactly two constants — $A$ and $\varphi$ — determined by initial conditions.

   Key takeaway

   Differentiating $A\cos(\omega t + \phi)$ twice gives $-\omega^2 x$, confirming it satisfies the SHM ODE. The two constants $A$ and $\phi$ are set by $x(0)$ and $v(0)$.

7. Question 7 · Medium

   A 0.4 kg mass on a spring ($k=160$ N/m) oscillates at $x=0.05$ m amplitude. What is the speed at $x=0.03$ m?

   • A
     $0.8$ m/sCorrect
   • B
     $1.0$ m/s

     Why not B: Used $v_{max}=A\omega =0.05\times 20=1$ m/s without accounting for position.
   • C
     $0.6$ m/s

     Why not C: Used $v=\omega \sqrt{A-x}$ (wrong: should be $A^2-x^2$).
   • D
     $0.4$ m/s

     Why not D: Computed $\omega (A-x)=20(0.02)=0.4$ m/s (wrong formula).
   Explanation

   $\omega =\sqrt{k/m}=\sqrt{160/0.4}=\sqrt{400}=20$ rad/s. Energy conservation: $\frac{1}{2}m{v}^2=\frac{1}{2}k(A^2-x^2)$, so $v=\omega \sqrt{A^2-x^2}=20\sqrt{(0.05{)}^2-(0.03{)}^2}=20\sqrt{0.0025-0.0009}=20\sqrt{0.0016}=20(0.04)=0.8$ m/s.

   Key takeaway

   In SHM, $v = \omega\sqrt{A^2 - x^2}$ (from energy conservation). Maximum at $x = 0$; zero at $x = \pm A$.

8. Question 8 · Medium

   A damped oscillator has equation $m\ddot{x}+b\dot{x}+kx=0$ with $b^2<4mk$. Which statement correctly describes the motion?

   • A
     The system oscillates at the same frequency as the undamped case.

     Why not A: Damping slightly reduces the oscillation frequency: ${\omega }_d=\sqrt{{\omega }_0^2-(b/2m{)}^2}<{\omega }_0$.
   • B
     The amplitude decays exponentially while the system oscillates at ${\omega }_d<{\omega }_0$.Correct
   • C
     The system returns to equilibrium without oscillating.

     Why not C: Non-oscillatory return describes critical or overdamping ($b^2\ge 4mk$).
   • D
     The energy grows exponentially due to damping.

     Why not D: Damping always removes energy; it never amplifies it.
   Explanation

   For underdamping ($b^2<4mk$), the solution is $x(t)=A{e}^{-(b/2m)t}\cos ({\omega }_{d}t+\varphi )$ where ${\omega }_d=\sqrt{k/m-(b/2m{)}^2}$. The amplitude envelope decays as $e^{-(b/2m)t}$, and the frequency is slightly less than the undamped ${\omega }_0=\sqrt{k/m}$.

   Key takeaway

   Underdamped SHM: oscillations with exponentially decaying amplitude at reduced frequency $\omega_d = \sqrt{\omega_0^2 - (b/2m)^2}$.

9. Question 9 · Hard

   A spring–mass system ($k$, $m$) undergoes SHM. At $t=0$, $x_0=0.06$ m and $v_0=0.8$ m/s. Given $\omega =10$ rad/s, find the amplitude.

   • A
     $0.06$ m

     Why not A: Set $A=x_0$, ignoring the initial velocity.
   • B
     $0.08$ m

     Why not B: Set $A=v_0/\omega =0.08$ m, ignoring initial displacement.
   • C
     $0.1$ mCorrect
   • D
     $0.14$ m

     Why not D: Added $x_0$ and $v_0/\omega$ linearly instead of in quadrature.
   Explanation

   Total energy: $E=\frac{1}{2}k{x}_0^2+\frac{1}{2}m{v}_0^2=\frac{1}{2}m{\omega }^2{x}_0^2+\frac{1}{2}m{v}_0^2=\frac{1}{2}m({\omega }^2{x}_0^2+v_0^2)$. Also $E=\frac{1}{2}m{\omega }^2{A}^2$. So:
   $$A=\sqrt{x_0^2+{\left(\frac{v_0}{\omega }\right)}^2}=\sqrt{(0.06{)}^2+{\left(\frac{0.8}{10}\right)}^2}=\sqrt{0.0036+0.0064}=\sqrt{0.01}=0.1\text{ m}.$$

   Key takeaway

   Amplitude from initial conditions: $A = \sqrt{x_0^2 + (v_0/\omega)^2}$. Add in quadrature (like a Pythagorean theorem), not linearly.

10. Question 10 · Hard

    A torsional pendulum has restoring torque $\tau =-\kappa \theta$ where $\kappa$ is the torsion constant and $I$ is the moment of inertia. What is the period of oscillation?

    • A
      $T=2\pi \sqrt{\frac{\kappa }{I}}$

      Why not A: Inverted $\kappa$ and $I$ — the stiffer the spring ($\kappa$ large), the shorter the period.
    • B
      $T=2\pi \sqrt{\frac{I}{\kappa }}$Correct
    • C
      $T=\frac{1}{2\pi }\sqrt{\frac{I}{\kappa }}$

      Why not C: Confused $T$ with $f=1/T$.
    • D
      $T=2\pi \sqrt{I\kappa }$

      Why not D: Multiplied $I$ and $\kappa$ instead of dividing.
    Explanation

    Newton's second law for rotation: $I\ddot{\theta }=-\kappa \theta$, giving $\ddot{\theta }=-(\kappa /I)\theta$. This is SHM with ${\omega }^2=\kappa /I$. Period: $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{\kappa }}$.

    Key takeaway

    The torsional pendulum is the rotational analog of a mass–spring: $I$ replaces $m$ and $\kappa$ replaces $k$, giving $T = 2\pi\sqrt{I/\kappa}$.

11. Question 11 · Hard

    A mass $m$ is attached to two identical springs (each constant $k$) in parallel (both springs pull on the mass simultaneously). What is the effective spring constant and resulting oscillation frequency?

    • A
      $k_{eff}=k/2$, $\omega =\sqrt{k/(2m)}$

      Why not A: This is the result for springs in series, not parallel.
    • B
      $k_{eff}=2k$, $\omega =\sqrt{2k/m}$Correct
    • C
      $k_{eff}=k$, $\omega =\sqrt{k/m}$

      Why not C: Treated two springs as one, ignoring their additive effect.
    • D
      $k_{eff}=2k$, $\omega =\sqrt{k/(2m)}$

      Why not D: Correct $k_{eff}$ but wrong $\omega$ — forgot to use $k_{eff}$ in $\omega =\sqrt{k_{eff}/m}$.
    Explanation

    For springs in parallel, both exert restoring forces on the same displacement $x$: $F=-k_{1}x-k_{2}x=-2kx$. So $k_{eff}=2k$ and $\omega =\sqrt{k_{eff}/m}=\sqrt{2k/m}$.

    For springs in series: $1/k_{eff}=1/k+1/k=2/k$, giving $k_{eff}=k/2$.

    Key takeaway

    Parallel springs: $k_{eff} = k_1 + k_2$. Series springs: $1/k_{eff} = 1/k_1 + 1/k_2$. Parallel increases stiffness; series decreases it.

12. Question 12 · Hard

    A spring–mass system ($k=50$ N/m, $m=0.5$ kg) starts at rest at $x=0.2$ m from equilibrium. Find the time at which the speed first equals half its maximum value. ($\omega =10$ rad/s)

    • A
      $t=\frac{\pi }{30}$ s

      Why not A: Solved $\cos (10t)=1/2$ correctly to get $t=\pi /30$ but this gives $v=v_{max}\sin (\pi /3)=v_{max}\frac{\sqrt{3}}{2}\ne v_{max}/2$.
    • B
      $t=\frac{\pi }{20}$ s

      Why not B: Found $\sin (10t)=1$ (maximum speed) rather than $\frac{1}{2}$.
    • C
      $t=\frac{\pi }{60}$ sCorrect
    • D
      $t=\frac{\pi }{15}$ s

      Why not D: Solved $\sin (10t)=\frac{1}{2}$ and got $10t=\pi /3$, giving $t=\pi /30$, then doubled in error.
    Explanation

    With $x(0)=A=0.2$ m and $v(0)=0$: $x(t)=A\cos (10t)$, $v(t)=-A\omega \sin (10t)$. Maximum speed $v_{max}=A\omega =2$ m/s. We want $|v|=v_{max}/2$: $|-A\omega \sin (10t)|=\frac{v_{max}}{2}\Rightarrow \sin (10t)=\frac{1}{2}\Rightarrow 10t=\frac{\pi }{6}\Rightarrow t=\frac{\pi }{60}$ s.

    Key takeaway

    With ICs $x_0 = A$, $v_0 = 0$: $x = A\cos(\omega t)$, $v = -A\omega\sin(\omega t)$. Set $|\sin(\omega t)| = \frac{1}{2}$ and use $\omega t = \pi/6$ for the first occurrence.