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Solve it, - Oscillations and Waves - JEE Main

A pendulum of length 1 m is released from \theta = 60\degree. the rate of chage of speed of bob at \theta = 30\degree is (g = 10 ms^{-2})

  • Option 1)

    10\; ms^{-2}

  • Option 2)

    7.5\; ms^{-2}

  • Option 3)

    5\; ms^{-2}

  • Option 4)

    5\sqrt{3}\; ms^{-2}

 
Answers (1)
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\omega^{2} for pendulum is  ^{g}/_{l} 

When \theta is half the amplitude of 60o, acceleration  = \omega^{2}\theta\Rightarrow a = -10\times \frac{\pi}{6} = -5 ms^{-2}

 

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

 


Option 1)

10\; ms^{-2}

This is incorrect.

Option 2)

7.5\; ms^{-2}

This is incorrect.

Option 3)

5\; ms^{-2}

This is correct.

Option 4)

5\sqrt{3}\; ms^{-2}

This is incorrect.

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