Introduction

Finding the amplitude of SHM from given velocity and acceleration functions is a common physics and entrance-exam problem. This article explains, step by step, how to compute the amplitude from v(t) = −100 sin(20t + π/3) and a(t) = −2000 cos(20t + π/3) using core equations of simple harmonic motion.

Step 1: Recall SHM equations

For displacement x(t) = A sin(ωt + ϕ) in SHM:

• Velocity: v(t) = Aω cos(ωt + ϕ)
• Acceleration: a(t) = −Aω² sin(ωt + ϕ)

Here: A = amplitude, ω = angular frequency, ϕ = initial phase.

The given functions are v(t) = −100 sin(20t + π/3) and a(t) = −2000 cos(20t + π/3).

Step 2: Match phases and identify ω

Different SHM forms differ only by a phase shift between sine and cosine, but both must share the same angular frequency ω and phase argument.

The arguments in the question are both 20t + π/3, so ω = 20 rad s⁻¹.

In standard form, v(t) is proportional to cos(ωt + ϕ) while a(t) is proportional to −sin(ωt + ϕ). The question instead gives velocity with sin and acceleration with cos; this is equivalent to a phase shift of π/2, but amplitude and ω remain unchanged. Hence, using magnitudes is sufficient.

Step 3: Use the acceleration equation

From SHM theory, the maximum magnitude of acceleration is |a_max| = Aω².

From the given acceleration: a(t) = −2000 cos(20t + π/3). The maximum magnitude is |a_max| = 2000 m s⁻².

Set this equal to Aω²: A × 20² = 2000 → A × 400 = 2000 → A = 2000/400 = 5 m. So the amplitude of oscillation is 5 m.

Step 4: Cross-check with velocity

Maximum speed in SHM is |v_max| = Aω.

With A = 5 m and ω = 20 rad s⁻¹, |v_max| = 5 × 20 = 100 m s⁻¹.

The given velocity function has amplitude 100, so this matches perfectly, confirming the result.