A pendulum of length `l=1m` is released from `theta_(0)=60^(@)` . The rate of change of speed of the bob at `theta=30^(@)` is.

Options

A. `5sqrt(3)m//s^(2)`
B. `5 m//s^(2)`
C. `10 m//s^(2)`
D. `2.5 m//s^(2)`

Answer

A

Explanation

Vertical height on both sides is same.
`sinalpha=(h)/(l_(1))`
`implies " "l_(1)=(h)/(sinalpha)`, mass of left part `=(m)/(L)xx(h)/(sinalpha)=m_(1)`
`l_(2)=(h)/(sinbeta)`,mass of right part `=(m)/(L)xx(h)/(sinbeta)=m_(2)`
acceleration `a=(m_(2)gsinbeta-m_(1)gsinalpha)/(m)=0`