Question

A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?

Answer 1

315.5 N/m

Step-by-step explanation:

m = 500 g = 0.5 kg

T = 0.25 second

Total energy, E = 4 J

Let K be the spring constant.

The formula for the time period is given by

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

`0.25 = 2* 3.14* \sqrt{(0.5)/(K)}`

`0.0398=\sqrt{(0.5)/(K)}`

`1.585* 10^(-3)={(0.5)/(K)}`

K = 315.5 N/m