Question

A simple pendulum simple harmonic motion about `x = 0` with an amplitude a and time period `T` speed of the pendulum at `x = a//2` will be
A. `pi A sqrt(3) / T`
B. `pi A/T`
C. `pi A sqrt(3) / 2T`
D. `3pi^(2) A/T`

Answer 1

Correct Answer - A
`v=omegasqrt(A^(2)-x^(2))=omegasqrt(A^(2)-(A^(2))/(4))`
`=omegasqrt((4A^(2)-A^(2))/(2))=(Aomega)/(2)sqrt(3)`
`=(sqrt(3))/(2)A(2pi)/(T)=piAsqrt(3)//T`