A particle executes SHM on a straight line path. The amplitude of oscialltion is 3 cm. Magnitude of its acceleration is eqal to that of its velocity when its displacement from the mean position is 1 cm. Find the time period of S.H.M.
Hint `: A= 3cm`
When`x=1 cm` , magnitude of velocity `=`magnitude of acceleration
i.e., `omega(A^(2) - x^(2))((1)/(2)) = omega^(2) x`
Find `omega`
`T = ( 2pi)(omega)`

Explanation

`A= 3 cm`,
When `x=1 cm omega ( A^(2)- x^(2)) ^(1//2) =omega^(2) x`
or `(A^(2) - x^(2)) = omega^(2) x^(2)`
`3^(2) -1^(2) = omegat^(2)xx 1^(2)`
or `omega = 2 sqrt(2)`
Time period, `T = ( 2pi)/(omega)= ( 2xx 3.14 )/( 2sqrt(2)) = ( 3.14)/(2)xx sqrt(2)`
`=2.23s`