The potential energy of a simple harmonic oscillator of mass 2 kg at its mean position is 5 J. If its total energy is 9 J and amplitude is 1 cm, then its time period is

Options

A. `(pi)/(100)s`
B. `(pi)/(50) s`
C. `(pi)/(20) s`
D. `(pi)/(10)s`

Answer

A

Explanation

Given, total energy = 9J PE at mean position = 5J So, maximum KE=9J-5J =4J
Now, in SHM
Maximum (at mean) KE= Maximum PE (at extremes)
`therefore " "(1)/(2) ka^(2) =4J`
`rArr" "k=(8)/(a^(2))=(8)/(10^(-4))=8xx10^(4) J//m^(2)`
Now, time period
`T= 2pi sqrt((m)/(k))=2pi xx sqrt((2)/(8xx10^(4)))rArr T =(pi)/(100) s`