Question

A block of mass 2 kg is attached to the spring of spring constant 50 Nm–1. The block is pulled to a distance of 5 cm from is equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. The displacement of the block at any time t is

• A. x = 0.05 sin 5t m
• B. x = 0.05 cos 5t m
• C. x = 0.5 sin 5t m
• D. x = 5 sin 5t m

Answer 1

Answer: (A)

x = 0.05 sin 5t m

Answer 2

Here, m = 2 kg k = 50 $Nm^{- 1}$

A = 5 cm = 0.05 m

The block executes SHM. Its angular frequency is given by

$\omega = \sqrt{\frac{k}{m} = \sqrt{\frac{50Nm^{- 1}}{2kg}}} = 5rads^{- 1}$

Since the time is noted from the equilibrium position its displacement at any time t is given by

$x = A\sin\omega t = 0.05\sin 5tm$