Question

A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that oscillates in simple harmonic motion. Calculate the period of motion.

Answer 1

The period of motion is 0.5 second.

Step-by-step explanation:

Given;

extension of the spring, x = 3.8 cm = 0.038 m

mass of the object, m = 13 g = 0.013 kg

Determine the force constant of the spring, k;

F = kx

k = F / x

k = mg / x

k = (0.013 x 9.8) / 0.038

k = 3.353 N/m

When the object is replaced with a block of mass 20 g, the period of motion is calculated as;

`T = 2\pi\sqrt{(m)/(k) } \\\\T = 2\pi\sqrt{(0.02)/(3.353) } \\\\T = 0.5 \ second`

Therefore, the period of motion is 0.5 second.