Question

A particle performing linear S.H.M. of period 2p seconds about the mean position O is observed to have a speed of \(b\sqrt{3}\) m / s , when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the time required by the particle, to travel a further distance b.

Answer 1

Given : T = 2π s, v = \(b\sqrt{3}\)   m/s at x=b

ω = 2π/T = 2π/2π =1 rad/s

v = ω\(\sqrt{A^2-x^2}\)

∴ at x = b, \(b\sqrt{3}\)  = (1)\(\sqrt{A^2-x^2}\)

∴ 3b² = A² – b² ,   ∴ A = 2b

∴ Assuming the particle starts from the mean position, its displacement is given by

x = A sin ωt = 2b sint

If the particle is at x = b at t = t₁,

b = 2b sin t₁  ∴ t₁= sin^-1(1/2) = π/6 s

Also, with period T = 2π s, on travelling a further distance b the particle will reach the positive extremity at time t₂ = π/2 s.

 The time taken to travel a further distance b from  x = b is t₂ – t₁ = π/2 − π/6 = π/3 s