Question

A 300 g block connected to a light spring with a force constant of k = 3 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 3 cm from equilibrium and released from rest. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.)

Answer 1

Time period, T = 1.98 seconds

Step-by-step explanation:

It is given that,

Mass of the block, m = 300 g = 0.3 kg

Force constant of the spring, k = 3 N/m

Displacement in the block, x = 3 cm

Let T is the period of the motion of the block. The time period of the block is given by :

`T=2\pi \sqrt{(m)/(k)}`

`T=2\pi \sqrt{(0.3\ kg)/(3\ N/m)}`

T = 1.98 seconds

So, the period of the motion of the block is 1.98 seconds. Hence, this is the required solution.