1.

A particle executing SHM has a maximum speed of 30 cm"/"s and a maximum acceleration of 60 cm"/"s^(2). The period of oscillation is………..

Answer»

`pi s`
`(pi)/(2)s`
`2pi s`
`(pi)/(t) s`

Solution :Maximum speed in SHM, `v_("max")= A omega`
`therefore 30 = A omega """…….."(1)`
and maximum acceleration, `a_("max")= A omega^(2)`
`therefore 60 = A omega^(2)"""…….."(2)`
Taking RATIO of equation (2) and (1) `(60)/(30)= omega`
`therefore 2= (2pi)/(T)""therefore T= pi s`.


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