Question

A particle executes simple harmonic acceleration with an amplitude α in the period of oscillation is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
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Answer 1

Acceleration in SHM

It is velocity per unit time. ... The differential equation of linear S.H.M. is d2x/dt2 + (k/m)x = 0 where d2x/dt2 is the acceleration of the particle, x is the displacement of the particle, m is the mass of the particle and k is the force constant.

Answer 2

Question : The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is?

AnswEr

Let the displacement equation of the particle executing SHM is

x = asin ωt

As the particle travels half of the amplitude from the equilibrium position, so

x = a/2

Therefore, a/2 asin ωt

Sin ωt = 1/2 = sin π/6

ωt = π/6 or t = π/6ω

t = π/6 [ 2π/T ] [ as ω = 2π/T ]

t = T/12

Hence the particle travels half of the amplitude from the equilibrium in T/12 second !