A pendulum bob has a speed 3m/s while passing thorugh its lowest position. What is its speed when it makes an angle of `60^0` with the vertical? The length of the pendulum is 0.5m Take `g=10 m/s^2`.

Options

A. `(1//3)ms^(-1)`
B. `(1//2)ms^(-1)`
C. `2ms^(-1)`
D. `3ms^(-1)`

Answer

C

Explanation

K.E. of the bob at lowest position
`=(1)/(2)mxx(3)^(2)=(9)/(2)mJ`
Let the bob velocity be `upsilon` at the point `A`, where `/_OSA=60^(@)`, figure, Using law of conservation fo energy, we have
`(9)/(2)m =(1)/(2)m upsilon^(2)+mgl (1-costheta)`
or `9=upsilon^(2)+2xx10xx0.5(1-1//2)`
On solving, `upsilon=2m//s`