Question

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m.$ The rod is pivoted at its centre ${}^{\prime }{O}^{\prime }$ 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

• A. $\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{3k}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Answer 1

The correct option is A $\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$

Given,

$A$ uniform horizontal rod $AB$

Length $=l$

Mass $=m$

Torque of spring force,

$\tau =r\times F\Rightarrow \tau =-2kx\left(\frac{1}{2}\right)\cos \theta$

$\Rightarrow \tau =\left(\frac{k{l}^2}{2}\right)\theta =-C\theta$

Compare equation we get,

$C=\left(\frac{k{l}^2}{2}\right)andl=\left(\frac{m{l}^2}{12}\right)$

The frequency of oscillation,

$f=\frac{1}{2\pi }\sqrt{\frac{C}{l}}$

$\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{\frac{k{l}^2}{2}}{\frac{m{l}^2}{12}}}$

$\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$

Final Answer: $\left(a\right)$