Question

A spring with a mass of 0.5 kg and a spring constant of 40 N/m is compressed 0.35 m. When released, the spring-mass system will oscillate horizontally on a frictionless surface. What is the period of oscillation?

Answer 1

The period of oscillation of the spring T = 0.702 sec

Step-by-step explanation:

Mass = 0.5 kg

Spring constant k = 40
`(N)/(m)`

Compression of spring = 0.35 m

Period of oscillation

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

Put all the values in above equation

`T = 2 \pi \sqrt{(0.5)/(40) }`

T = 0.702 sec

This is the period of oscillation of the spring.