Question

A student does an experiment with a pendulum. In the first trial, she displaces the pendulum 5 cm. In the second trial, she displaces the same pendulum a distance of 10 cm.

In what ways would the graphs that represent these simple harmonic motion graphs be different? In what ways would they be the same?

Answer 1

Short answer: the two displacement-vs.-time graphs are the same shape and have the same period/frequency, but the 10 cm graph has twice the amplitude (peaks twice as far from zero). Other time-dependent quantities (velocity, acceleration) also keep the same frequency/shape but their amplitudes scale with the displacement amplitude; the total energy scales with amplitude squared.

Explain more precisely (small-angle pendulum assumed)
- Displacement: x(t) = A cos(ωt + φ) with ω ≈ sqrt(g/L) (independent of A for small angles). So if A doubles (5 → 10 cm) the waveform is the same sinusoid but scaled vertically by 2. The times of maxima, minima and zero crossings are unchanged.
- Period/frequency: T = 2π/ω is (to good approximation) the same for both trials, so the horizontal spacing of the waves is unchanged.
- Phase: if both are released the same way (e.g. released from rest at maximum displacement) their phase φ is the same. If released differently, the phase could differ (graph shifted left/right).
- Velocity: v(t) = −A ω sin(ωt + φ). Frequency and shape unchanged; v amplitude doubles when A doubles (v_max ∝ A).
- Acceleration: a(t) = −ω^2 x(t). Shape and frequency unchanged; a-amplitude doubles with A (a_max ∝ A).
- Energy: total mechanical energy E ∝ A^2 (so E quadruples when A doubles).

Caveat: for very large angular displacements the small-angle approximation breaks down and the period increases slightly with amplitude; for the displacements you gave (5 cm vs 10 cm) a typical lab pendulum will still be in the small-angle regime, so you can treat the period as unchanged.