1.

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period would beA. `(pi)/(10)s`B. `(pi)/(20)`sC. `(pi)/(50)s`D. `(pi)/(100)s`

Answer» Correct Answer - D
`(1)/(2)mv_(max)^2=9-5=4`
or `mv_(max)^2=8`
or `2v_(max)^2=8`
or `v_(max)=2(m)/(s)=Aomega`
`omega=(v_(max))/(A)=(2)/(0.01)=200`
or `(2pi)/(T)=200`
`T=(2pi)/(200)=(pi)/(100)`


Discussion

No Comment Found

Related InterviewSolutions