1.

A simplependulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x=(a)/(2) will be :

Answer»

`(PIA)/(T)`
`(3pi^(2)a)/(T)`
`(pi a sqrt(3))/(T)`
`(pi a sqrt(3))/(2T)`.

Solution :Speed `v=omega sqrt(a^(2)-x^(2)),x=(a)/(2)`
`:.""v=omega sqrt(a^(2)-(a^(2))/(4))=omega sqrt((3a^(2))/(4))`
`=(2PI)/(T)(a sqrt(3))/(2)=(pi a sqrt(3))/(T)`
Correctchoice is ( c ).


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