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

Options

A. `A=sqrt(2)cm,upsilon=(500)/(pi)Hz`
B. `A=sqrt(2)cm,upsilon=(1000)/(pi)Hz`
C. `A=(1)/sqrt(2)cm,upsilon=(500)/(pi)Hz`
D. `A=(1)/sqrt(2)cm,upsilon=(1000)/(pi)Hz`

Answer

D

Explanation

For a displacement x, the kinetic and potential energies are
`K=(1)/(2)m(A^(2)-x^(2))omega^(2)andU=(1)/(2)mx^(2)omega^(2)`
Each of these `=(10)/(2)=5J" when "x=1cm`
`therefore (1)/(2)mx^(2)omega^(2)=(1)/(2)xx0.1xx(1xx10^(-2))^(2)omega^(2)=5`
or `omega^(2)=(10)/(0.1xx10^(-4))=10^(6)oromega=10^(3)=1000`
`therefore` Frequency, `upsilon=(omega)/(2pi)=(1000)/(2pi)s^(-1)=(500)/(pi)Hz`
`K=5=(1)/(2)xx0.1[A^(2)-(1xx10^(-2))^(2)](10^(6))orA^(2)-10^(-4)=10^(-4)`
or `A^(2)=2xx10^(-4)orA=sqrt(2)xx10^(-2)m=sqrt(2)cm`