Simple Harmonic Motion 4 Question 3
3. 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: 3. (c)
Solution:
- When a system oscillates, the magnitude of restoring torque of system is given by
$$ \tau=C \theta $$
where, $C=$ constant that depends on system.
$$ \text { Also, } \quad \tau=I \alpha $$
where, $I=$ moment of inertia
and $\alpha=$ angular acceleration
From Eqs. (i) and (ii),
$$ \alpha=\frac{C}{I} \cdot \theta $$
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,
$$ \tau=\frac{k l^{2}}{2} \theta $$
$\because$ 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
$$ f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}} $$
