Simple Harmonic Motion Ques 2
- A rod of mass ’ $M$ ’ and length ’ $2 L$ ’ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of ’ $m$ ’ are attached at distance ’ $L / 2$ ’ from its centre on both sides, it reduces the oscillation frequency by $20 %$. The value of ratio $m / M$ is close to
(2019 Main, 9 Jan II)
(a) 0.57
(b) 0.37
(c) 0.77
(d) 0.17
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Answer:
Correct Answer: 2.( b )
Solution:
5 We know that in case of torsonal oscillation frequency
$ \mathrm{v}=\frac{k}{\sqrt{I}} $
where, $I$ is moment of inertia and $k$ is torsional constant.
$\therefore$ According to question, $\mathrm{v}_1=\frac{k}{\sqrt{\frac{M(2 L)^2}{12}}}$
(As, MOI of a bar is $I=\frac{M L^2}{12}$ )
or $ \mathrm{v}_1=\frac{k}{\sqrt{\frac{M L^2}{3}}} $
When two masses are attached at ends of rod. Then its moment of inertia is
$ \frac{M(2 L)^2}{12}+2 m\left(\frac{L}{2}\right)^2 $
So, new frequency of oscillations is,
$ \begin{aligned} & \mathrm{v}_2=\frac{k}{\sqrt{\frac{M(2 L)^2}{12}+2 m\left(\frac{L}{2}\right)^2}} \\ & \mathrm{v}_2=\frac{k}{\sqrt{\frac{M L^2}{3}+\frac{m L^2}{2}}} \end{aligned} $
As, $ v_2=80 % \text { of } v_1=0.8 v_1 $
So, $ \frac{k}{\sqrt{\frac{M L^2}{3}+\frac{m L^2}{2}}}=\frac{0.8 \times k}{\sqrt{\frac{M L^2}{3}}} $
After solving it, we get,
$ \frac{m}{M}=0.37 $
BETA
