A College Board-style free-response question on Unit 7: Oscillations. Work through each part, then reveal the model answer to see exactly what earns each point.
Scenario: A 2.0 kg block is attached to a horizontal spring with spring constant k = 50 N/m. The block rests on a frictionless surface. The block is pulled to x = 0.20 m from equilibrium and released from rest. The block then oscillates back and forth in simple harmonic motion.
— AP Physics 1 style problem · Topics 7.2, 7.4
A
Calculate the period of the block's oscillation.
✓ Model answer (earns the points)
For a mass-spring oscillator, the period is given by:
T = 2π√(m/k)
Plugging in m = 2.0 kg and k = 50 N/m:
T = 2π√(2.0/50) = 2π√(0.04) = 2π(0.2) ≈ 1.26 s
The period is approximately 1.26 s.
Why it scores: (1) Writes the correct equation T = 2π√(m/k). (2) Plugs in values correctly. (3) Computes the numerical answer with units (seconds).
B
Use energy conservation to calculate the maximum speed of the block.
✓ Model answer (earns the points)
Setup with energy conservation: At the amplitude (x = A = 0.20 m), the block is at rest, so all the energy is elastic PE. At the equilibrium position (x = 0), the spring is unstretched, so all the energy is kinetic. By conservation of energy:
½kA² = ½mv²_max
Solve for v_max:
v_max = A·√(k/m)
v_max = (0.20)·√(50/2.0) = (0.20)·√(25) = (0.20)(5) = 1.0 m/s
The maximum speed is 1.0 m/s.
Why it scores: (1) Identifies WHERE max speed occurs (at equilibrium). (2) Writes energy conservation: ½kA² = ½mv²_max. (3) Solves correctly for v_max. (4) Reports answer with units.
C
A student claims: "If we doubled the amplitude of the oscillation (to 0.40 m), the period would also double." Is the student correct? Explain, and describe what WOULD happen to: (i) the period, (ii) the maximum speed, and (iii) the total mechanical energy.
✓ Model answer (earns the points)
The student is incorrect. The period does NOT depend on amplitude in simple harmonic motion.
(i) Period: T = 2π√(m/k). Amplitude doesn't appear in the formula. Period stays the same (~1.26 s). This is THE defining feature of SHM: equal-time oscillations regardless of amplitude.
(ii) Maximum speed: v_max = A·√(k/m). Since v_max is PROPORTIONAL to A, doubling A doubles v_max. New v_max = 2.0 m/s (double the original).
(iii) Total energy: E = ½kA². Since E is proportional to A², doubling A makes A² four times bigger. New E = 4 × original E. Original was ½(50)(0.20)² = 1.0 J; new value would be 4.0 J.
Summary: Doubling the amplitude leaves the period unchanged but doubles the maximum speed and quadruples the total energy. The student confused 'big swing' with 'slow swing' — they're independent properties.
Why it scores: (1) Clearly states the student is incorrect. (2) Explains period is independent of amplitude with the relevant equation. (3) Correctly identifies v_max ∝ A (doubles). (4) Correctly identifies E ∝ A² (quadruples). (5) Uses the right equations to justify each conclusion.
How to score points on AP Physics 1 SHM FRQs
Write the equation BEFORE plugging in numbers. Showing the symbolic equation usually earns its own point.
For "increase/decrease/stays same" questions, refer to the equation. Show what depends on what — e.g., "Since T = 2π√(m/k), amplitude doesn't appear, so T is unchanged."
For maximum speed and maximum acceleration questions, identify WHERE they occur. v_max at equilibrium; a_max at the amplitude.
Use energy conservation for max-speed problems. ½kA² = ½mv²_max is the most reliable path.
Remember: amplitude ≠ frequency. Big swing doesn't mean fast swing — only the spring constant and mass set the period.
Units, units, units. Period: seconds. Frequency: Hz. Spring constant: N/m. Energy: joules.