1.

A simple pendulum executing S.H.M. has period T and amplitude A. Its speed, when at a distance (A)/(4) is :

Answer»

`(piAsqrt(15))/(2T)`
`(piA sqrt(15))/(T)`
`(piA)/(2T)`
`(2piA)/(T)`.

Solution :`v=omega sqrt(A^(2)-y^(2))=(2pi)/(T)sqrt(A^(2)-(A^(2))/(16))`
`=(2pi)/(T)sqrt((15)/(16)A^(2))=(PI sqrt(15)A)/(2T)`
Hence correct choice is(a).


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