A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its.

Options

A. maximum potential energy is 100 J
B. maximum kinetic energy is 160 J
C. maximum potential energy is 160 J
D. maximum potential energy is zero

Answer

C

Explanation

The energy of the osciallator
`E=(1)/(2)KA^(2)=(1)/(2)xx2xx10^(6)xx(10^(-2))^(2)=10^(2)=100 J`
(i) At the mean position x=0. Velocity is maximum and K.E. is maximum.
`therefore K.E. =100 J`

But the total energy =160 J
`thererfore P.E. =160-100=60 J`
This means that the oscillator must have some initial potential energy. Because in normal circums"tan"ce its P.E. at x=0 is 0.
(ii) At the extreme position x=A.
Its K.E. =0, E=160 J and hence P.E. =160 J= max. P.E.
(c ) is the correct option