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Physics Question on Oscillations

A pendulum of length 1 m is released from θ\theta = 60∘^{\circ}. The rate of change of speed of the bob at θ\theta= 30∘^{\circ} is (g=10 ms−1)(g=10 \, ms^{-1})

A

10 ms−210\, ms^{-2}

B

7.5 ms−27.5\, ms^{-2}

C

5 ms−25\, ms^{-2}

D

53 ms−25\sqrt 3\, ms^{-2}

Answer

5 ms−25\, ms^{-2}

Explanation

Solution

For a simple pendulum time period
         T=2πlg\, \, \, \, \, \, \, \, \, T=2\pi\sqrt{\frac{l}{g}}
or         2πT=gl \, \, \, \, \, \, \, \, \frac{2\pi}{T} =\sqrt{\frac{g}{l}}
∴         ω=gl\therefore \, \, \, \, \, \, \, \, \, \omega=\sqrt{\frac{g}{l}}
          ω2=gl\, \, \, \, \, \, \, \, \, \, \omega^2 =\frac{g}{l}
Amplitude, when angular displacement is 60∘^{\circ}
          =2πl360∘×60∘=2πl6\, \, \, \, \, \, \, \, \, \, =\frac{2\pi l}{360^{\circ}} \times 60^{\circ} =\frac{2\pi l}{6}
Therefore, displacement when angular displacement is 30∘^{\circ}
             =12(2πl6)\, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{1}{2}\bigg(\frac{2\pi l}{6}\bigg)
            y=πl6           \, \, \, \, \, \, \, \, \, \, \, \, y=\frac{\pi l}{6} \, \, \, \, \, \, \, \, \, \, \,... (ii)
Acceleration (a) = -ω2y\omega^2 y
Using Eqs. (i) and (ii), we get
          a=−gl×πl6\, \, \, \, \, \, \, \, \, \, a=-\frac{g}{l} \times \frac{\pi l}{6}
           =−10×3.146\, \, \, \, \, \, \, \, \, \, \, =-\frac{10 \times 3.14}{6}
           =−5.2ms−2=−5ms−2\, \, \, \, \, \, \, \, \, \, \, =-5.2 ms^{-2} =-5ms^{-2}