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A pendulum of length l=1m is released fr...

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.

A

`5sqrt(3)m//s^(2)`

B

`5m//s^(2)`

C

`10m//s^(2)`

D

`2.5m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Rate of change of speed
`(dv)/(dt)` =tangential acceleration
`=(tang ential fo rce)/(mass)`
`=(mgsin30^(@))/(m)`
`=10((1)/(2))m//s^(2)=5m//s^(2)` .
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