Question

When a mass m is attached to the spring of force constant k then spring stretched by length l . if the mass oscillates with amplitude l, what will be maximum potential energy stroded i the spring?

Answer 1

iska answer kya Hoga.

Answer 2

Answer:1/2Mgl

Explanation:

Mg=kl

Pe(max) =1/2kl^2

1/2×k×l×l

1/2*mg*l