Question

A certain simple harmonic vibrator of mass 0.1 kg has a total energy of 10 J. Its displacement from the mean position is 1 cm when it has equal kinetic and potential energies. The amplitude A and frequency n of vibration of the vibrator are
A. `A=sqrt2cm`,`n=(500)/(pi)Hz`
B. `A=sqrt2cm`,`n=(1000)/(pi)Hz`
C. `A=(1)/(sqrt2)cm`,`n=(500)/(pi)Hz`
D. `A=(1)/(sqrt2)cm`,`n=(1000)/(pi)`Hz

Answer 1

Correct Answer - A
For a displacement x, the kinetic and potential energies are `KE=(1)/(2)m(A^2-x^2)omega^2` and `PE=(1)/(2)mx^2omega^2`
Each of these `=(10)/(2)=5J` when `x=1cm`
Hence `(1)/(2)mx^2omega^2=(1)/(2)xx0.1xx(1xx10^-2)^2omega^2=5`
This gives `omega^2=(10)/(0.1xx10^-4)=10^6`
Giving `omega=10^3=1000(rad)/(s)`
Hence `T=(2pi)/(omega)=(pi)/(500)s`
And frequency `f=(500)/(pi)s`
Also, `KE=5=(1)/(2)xx0.1[A^2-(1xx10^-2)^2](10^6)`
`A^2=2xx10^-4impliesA=sqrt2xx10^-2m=sqrt2cm`