Question

Calculate the period of a particle performing linear S.H.M. with maximum speed of 0.08 m/s and maximum acceleration of 0.32 m/s².

Answer 1

Given: v_max = 0.08 m/s = 8 × 10⁻² m/s

a_max = 0.32 m/s² = 32 × 10⁻² m/s²

To find: Period (T) = ?

Formula:

v_max = ωA  ...(i)

a_max = ω²A  ...(ii)

Dividing equation (i) by equation (ii),

`(v_max)/(a_max) = (ωA)/(ω^2A)`

`(8 xx 10^-2)/(32 xx 10^-2) = 1/ω`

`8/32 = T/(2pi)`

`(2pi)/4` = T

T = `pi/2`

T = `3.142/2`

∴ T = 1.571 sec.