A particle exectues a linear S.H.M. of amplitude 2 cm. When it is at 1cm from the mean position, the magnitudes of its velocity and acceleration are equal. What is its maximum velocity ? `[sqrt(3)=1.732]`

Options

A. `2.5 cm//s`
B. `1.4 cm//s`
C. `3.464 cm//s`
D. `4.322 cm//s`

Answer

C

Explanation

When x=1 cm, |v|=|Acceleration|
`therefore omega sqrt(A^(2)-x^(2))=omega^92)x`
`sqrt((A^(2)-x^(2)))=omega x`
`therefore A^(2)-x^(2)=omega^(2)x^(2) but A=2cm and x=1 cm`
`therefore 2^(2)-1^(2)=omega^(2)xx1`
`therefore omega^(2)=3 or omega =sqrt(3) rad//s`
and Maxim velocity `=omega A=sqrt(3)xx2`
`=2xx1.732=3.464 cm//s`