Question

A body performs SHM on a path 0.12 m long. Its velocity at the centre of the path is 0.12 m/s. Find the period of SHM. Also find the magnitude of the velocity of the body at √3 × 10^-2 m from the centre of the path.

Answer 1

The path length of the SHM is the range 2 A, and the velocity at the centre of the path, i.e., at the equilibrium position, is the maximum velocity v_max.

Data : 2A = 0.12 m, v_max = 0.12 m/s,

x = ± √3 x 10^-2 m

The magnitude of the velocity is

Answer 2

A body describe simple harmonic motion in a path of 12m long. its velocity at the centre of the line is 0.12 m/s
velocity at the centre of line , means velocity at mean position

Here A is amplitude .

Amplitude is half of path length

So, A = 0.12/2 = 0.06 m