A particle executies linear simple harmonic motion with an amplitude `3cm` .When the particle is at `2cm` from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds is

Options

A. `(sqrt(5))/(pi)`
B. `(sqrt(5))/(2pi)`
C. `(4pi)/(sqrt(5))`
D. `(2pi)/(sqrt(3))`

Answer

C

Explanation

Here, `r=3cm, x=2cm,`
velcity, `v=omegasqrt(r^(2)-x^(2))`
Acceleration `A=omega^(2)x`
As per question, `omegasqrt(r^(2)-x^(2))=omega^(2)x`
or `r^(2)-x^(2)=omega^(2)x^(2)=(4pi^(2))/(T^(2))x^(2)`
or`3^(2)-2^(2)=(4pi^())/(T^(2))xx2^(2)`
or `5=(16pi^(2))/(T^(2)) or T=(4pi)/(sqrt(5))`