Simple Harmonic Motion Ques 42
- Two light identical springs of spring constant $k$ are attached horizontally at the two ends of an uniform horizontal $\operatorname{rod} A B$ of length $l$ and mass $m$. The rod is pivoted at its centre ’ $O$ ’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure.
The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is (2019 Main, 12 Jan I)
(a) $\frac{1}{2 \pi} \sqrt{\frac{2 k}{m}}$
(b) $\frac{1}{2 \pi} \sqrt{\frac{3 k}{m}}$
(c) $\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}}$
(d) $\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
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Answer:
Correct Answer: 42.(c)
Solution:
Formula:
Compound Pendulum / Physical Pendulum:
- When a system oscillates, the magnitude of restoring torque of system is given by
$ \tau=C \theta $ $\quad$ …….(i)
where, $C=$ constant that depends on system.
$ \text { Also, } \quad \tau=I \alpha $ $\quad$ …….(ii)
where, $I=$ moment of inertia
and $\alpha=$ angular acceleration
From Eqs. (i) and (ii),
$ \alpha=\frac{C}{I} \cdot \theta $ $\quad$ …….(iii)
and time period of oscillation of system will be
$ T=2 \pi \sqrt{\frac{I}{C}} $
In given case, magnitude of torque is
$\tau=$ Force $\times$ perpendicular distance
$ \tau=2 k x \times \frac{l}{2} \cos \theta $
For small deflection,
$ \tau=(\frac{k l^{2}}{2}) \theta $ $\quad$ …….(iv)
$\because$ For small deflections, $\sin \theta=\frac{x}{(l / 2)} \approx \theta$
$\Rightarrow \quad x=\frac{l \theta}{2}$
Also,
$ \cos \theta \approx 1 $
comparing Eqs. (iv) and (i), we get
$ \begin{aligned} C & =\frac{k l^{2}}{2} \Rightarrow \alpha=\frac{\left(k l^{2} / 2\right)}{(\frac{1}{12} m l^{2})} \cdot \theta \\ \Rightarrow \quad \alpha & =\frac{6 k}{m} \cdot \theta \end{aligned} $
Hence, time period of oscillation is
$ T=2 \pi \sqrt{\frac{m}{6 k}} $
Frequency of oscillation is given by
$ f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}} $
BETA
