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प्रश्न
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
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उत्तर
\[\frac{18}{100} = 0 . 18 m = 0 . 2 m\]
Let I be the moment of inertia and \[\omega\] be the angular speed.
Using the energy equation, we can write:
\[mgl(1 - \cos \theta) + \frac{1}{2}I \omega^2 = \text { constant }\]
\[mg\left( 0 . 20 \right) \left( 1 - \cos \theta \right) + \frac{1}{2}I \omega^2 = C . . . \left( 1 \right)\]
\[\text { Moment of inertia about the point of suspension A is given by, } \]
\[I = \frac{2}{3}m r^2 + m l^2 \]
\[\text { Substituting the value of l in the above equation, we get: }\]
\[I = \frac{2}{3}m \left( 0 . 02 \right)^2 + m \left( 0 . 2 \right)^2 \]
\[ = \frac{2}{3}m\left( 0 . 0004 \right) + m\left( 0 . 04 \right)\]
\[ = m\left[ \frac{0 . 0008}{3} + 0 . 04 \right]\]
\[ = m\left( \frac{0 . 1208}{3} \right)\]
On substituting the value of I in equation (1) and differentiating it, we get:
\[\frac{d}{dt}\left[ mg \left( 0 . 2 \right) \left( 1 - \cos \theta \right) + \frac{1}{2}\frac{0 . 1208}{3}m \omega^2 \right] = \frac{d}{dt}\left( c \right)\]
\[ \Rightarrow mg\left( 0 . 2 \right)\sin\theta\frac{d\theta}{dt} + \frac{1}{2}\left( \frac{0 . 1208}{3} \right)m \times 2\omega\frac{d\omega}{dt} = 0 \]
\[ \Rightarrow 2\sin \theta = \frac{0 . 1208}{3}\alpha \left[ \text { because }, g = 10 m/ s^2 \right]\]
\[ \Rightarrow \frac{\alpha}{\theta} = \frac{6}{0 . 1208}\]
\[ \Rightarrow \omega^2 = 49 . 66\]
\[ \Rightarrow \omega = 7 . 04\]
\[\text { Thus, time period }\left( T \right) \text { will be: }\]
\[T = \frac{2\pi}{\omega} = 0 . 89 s\]
For a simple pendulum, time period (T) is given by,
% change in the value of time period = \[\frac{0 . 89 - 0 . 86}{0 . 89} \times 100 = 0 . 3\]
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