1.

The bob of simple pendulumof length L is released at time t = 0 from a position of small angular displacementtheta. Its linear displacement at time t is given by :

Answer»

`X=thetao sin 2pi SQRT((L)/(g))xxt`
`x=L thetao cos 2pi sqrt((g)/(L))xxt`
`x=L thetao sin sqrt((g)/(L))xx t`
`x=L thetao cos sqrt((g)/(L))xxt`.

Solution :Here displacement is given by x = A `cos omegat` when A is LINEAR amplitude.
Now `""theta=(A)/(L)` (when `theta` is small)
or `""A=L theta`
Also `""omega=(2pi)/(T)=(2pi)/(2pi sqrt((L)/(g)))=sqrt((g)/(L))`
`:.""x=L theta cos sqrt((g)/(L)).t.`
Correct CHOICE is (d).


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