Question

A particle of mass 100 g undergoes a simple harmonic motion. The restoring force is provided by a spring with a spring constant of 40 N·m−1. What is the period of oscillation?

- $10 \pi$
- $0.5 \pi$
- $0.1 \pi$
- $1 \pi$

Final Answer

(c)

Solution Video

#### Sign up to view this solution video!

View sample solution## Calculator Screenshots

Video Transcript

This is College Physics Answers with Shaun Dychko.
The period of oscillation for a simple harmonic oscillator is two pi times the square root of the mass times the spring constant. So that's two pi times the square root of 100 grams written as 100 times ten to the minus three kilograms, divided by the spring constant of 40 Newtons per meter. This gives 0.05 times two pi which is 0.1 pi. This makes answer C