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

A particle executes linear simple harmonic motion with an amplitude of

1.	A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.
Answer» SOLUTION :`a=omega^(2)x,a=v,A=3cm,x=2cm` `v=omegasqrt(A^(2)-x^(2))` `omega^(2)x=omegasqrt(A^(2)-x^(2))` `((2pi)/(T))2=sqrt(3^(2)-2^(2)),(4PI)/(T)=sqrt(5)` `T=(4pi)/(sqrt(5))impliesT=5.6sec`