Question

Two light identical spring of spring constant k are attached horizontally at the two ends of a uniform horizontally rod AB of length $\ell$ and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two spring 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. $\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{3k}{m}}$

Answer 1

Answer: (A)

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

$\tau =-2Kx\frac{\ell }{2}\cos \theta$

$\Rightarrow \tau =\frac{Kill{\ell }^2}{2}\theta =-C\theta$
$f=\frac{1}{2\pi }\sqrt{\frac{c}{1}}=\frac{1}{2\pi }\sqrt{\frac{\frac{K{\ell }^2}{2}}{\frac{M{\ell }^2}{12}}}$
$f=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$