Question

A linear harmonic oscillator of force ocnstant `2xx10^(6)N//m` and amplitude 0.01 m has a total mechanical energy of 160J. What is the maximum K.E. and minimum P.E.?

Answer 1

Here, maximum K.E. `=` max. PE `=(1)/(2)kr^(2)`
`=(1)/(2)xx(2xx10^(6))xx(0.01)^(2)=100J`
In SHM, P.E. is minimum at mean position and K.E. is maximum at mean position. But the total meachanical energy in SHM is constant. Therefore, at mean position.,
total mechanical energy `=` max. K.E. `+` minimum P.E.
`160=100+` minimum P.E.
So, minimum P.E. `=160-100=60J`