First slide

Simple harmonic motion

Question

A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations , If two masses each of mass ‘m’ are attached at distance $\frac{L}{2}$ from centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is closed to

Moderate

A

0.17
B

0.37
C

0.57
D

0.77

Solution

I L l
$\begin{array}{rcl} m o m e n t o f i n e r t i a o f r o d i s, I & = & \frac{M {(2 L)}^{2}}{12}, h e r e M i s m a s s o f r o d, 2 L i s l e n g t h o f r o d \\ m o m e n t o f i n e r t i a o f r o d w i t h m a s s e s m a t t a c h e d i s, I^{1} & = & I + 2 [m {(\frac{L}{2})}^{2}] - - - (1) \\ n o w, I & = & \frac{M L^{2}}{3} - - - (2) \\ o n s u b s t i t u t i n g e q u a t i o n 2 i n 1, \\ I^{1} & = & L^{2} (\frac{M}{3} + \frac{m}{2}) \\ I^{1} & = & L^{2} [\frac{2 M + 3 m}{6}] \\ f r e q u e n c y o f t o r s i o n a l o s c i l l a t i o n s i s f & = & \frac{1}{2 π} \sqrt{\frac{K}{I}} - - - (3) \\ \Rightarrow & f^{1} = \frac{1}{2 π} \sqrt{\frac{k}{I^{1}}} - - - (4) \\ S o b y q u e s t i o n, \\ \frac{f - f^{1}}{f} & = & 1 - \frac{f^{1}}{f} = \frac{20}{100} = \frac{1}{5} \\ o n s u b s t i t u t i n g e q u a t i o n s 3 a n d 4 \\ 1 - \frac{f^{1}}{f} & = & 1 - \sqrt{\frac{I}{I^{1}}} \\ \Rightarrow & \frac{1}{5} = 1 - \sqrt{\frac{I}{I^{1}}} \\ \Rightarrow & 1 - \frac{1}{5} = \sqrt{\frac{I}{I^{1}}} \\ \Rightarrow & {(\frac{4}{5})}^{2} = \frac{I}{I^{1}} \\ s u b s t i t u t e t h e v a l u e s o f I, I^{1} \\ \Rightarrow & \frac{\frac{M L^{2}}{3}}{L^{2} [\frac{2 M + 3 m}{6}]} = \frac{16}{25} \\ \frac{2 M}{2 M + 3 m} & = & \frac{16}{25} \\ 25 M & = & 16 M + 24 m \\ 9 M & = & 24 m \\ ∴ \frac{m}{M} & = & \frac{3}{8} = 0.37 \end{array}$

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