Skip to main content

SHM amplitude from two speeds

Short answer

In SHM, v² = ω²(a² – x²). Substituting v1, x1 and v2, x2 and dividing eliminates ω, giving v1²/v2² = (a² – x1²)/(a² – x2²). Solving yields a = √((v2²x1² – v1²x2²)/(v2² – v1²)).

A particle is moving with simple harmonic motion in a straight line. If it has a speed v1 when the displacement is x1 and speed v2 when the displacement is x2, then show that the amplitude of the motion is a = [(v2²x1² − v1²x2²)/(v2² − v1²)]^(1/2).

For SHM, x = a sin(ωt + φ) and v = aω cos(ωt + φ). Eliminating t:

v² = ω² (a² − x²)

Proof

  1. At displacement x1 the speed is v1, and at x2 the speed is v2:
v1² = ω² (a² − x1²)       ... (1)
v2² = ω² (a² − x2²)       ... (2)
  1. Divide (1) by (2) to eliminate ω:
v1² / v2² = (a² − x1²) / (a² − x2²)
v1² (a² − x2²) = v2² (a² − x1²)
  1. Expand and collect terms in a²:
v1² a² − v1² x2² = v2² a² − v2² x1²
a² (v1² − v2²) = v1² x2² − v2² x1²
a² (v2² − v1²) = v2² x1² − v1² x2²
  1. Therefore
a = [ (v2² x1² − v1² x2²) / (v2² − v1²) ]^(1/2)

which is the required result.

Also, from (1) − (2): ω² = (v1² − v2²)/(x2² − x1²), so the period T = 2π/ω can be found from the same measurements.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Physics chapter 1 questions →