Question

A particle of mass 10 g executes linear S.H.M.
of amplitude 5 cm with a period of 2 s. Find its
PE and KE, 1/6 s after it has crossed the mean
position.​

Answer 1

KE = = 9.25 * 10^-5 J

PE =   = 3.085 * 10^-5 J

Explanation:

m = 10g = 10^-2 kg

T = 2s

ω = 2π/T = 2π/2 = π rad/s

A = 5cm = 5 * 10^-2m

KE = 1/2 m.A².ω².cos²ωt

At t = 1/6s, KE = 1/2*10^-2 * (5 * 10^-2)² (π²) Cos²π/6

   = 25 * 10^-6/2 * π² * (√3/2)²

  = 9.25 * 10^-5 J

At t = 1/6s, PE = 1/2 m.A².ω². sin²ωt

  = 1/2*10^-2 * (5 * 10^-2)² (π²) Sin²π/6

  = 25 * 10^-6/2 * π² * (1/2)²

  = 3.085 * 10^-5 J