A particle executes SHM on a straight line path. The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm`, the numerical value of magnitude of acceleration is equal to the mumerical value of velocity. Find the frequency of SHM (in `Hz`).

Options

A. `(1)/(pi)`
B. `(sqrt2)/(pi)`
C. `(sqrt3)/(pi)`
D. `(1)/(2pisqrt3)`

Answer

C

Explanation

at P, `a_(p)=omega^(2)`
`u_(p)=omegasqrt(a^(2)-x^(2))`, `a_(p)=u_(p)` `implies omega^(2)x=omegasqrt(A^(2)-x^(2))`
`omega =(1)/(x)sqrt(A^(2)-x^(2))` `implies f=(1)/(2pi)xx(1)/(1)sqrt4-1)`
`f=sqrt((3)/(2pi)`