Q: Figure below shows a uniform rod of length 2*l* and mass *m*. At one end it is connected to a frictionless axis at *A*. The other end is connected to a spring of force constant *k*. The rod is displaced by a small angle θ from the end connected to the spring and released. What is the frequency of oscillations of the rod ?

(a) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{3k}{m}} $

(b) $ \displaystyle \frac{1}{2\pi} \sqrt{\frac{k}{3 m}} $

(c) $ \displaystyle \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

(d) $ \displaystyle \frac{1}{2\pi} \sqrt{\frac{4 k}{3 m}} $

Ans: (a)

Sol: Resting force in spring when stretched through length x = k x= k(2lθ) ; where θ is the small angle through which the rod is rotated

Restoring Torque $ \displaystyle \tau = – kx \times 2l $

$ \displaystyle \tau = – k(2l\theta) \times 2l $

$ \displaystyle \tau = – 4k l^2 \theta $

As $ \displaystyle \tau = I \alpha $

$ \displaystyle I \alpha = – 4k l^2 \theta $

$ \displaystyle (\frac{m(2l)^2}{3})\alpha = – 4k l^2 \theta $

$ \displaystyle \alpha = – \frac{3 k}{m} \theta $

$ \displaystyle \omega^2 = \frac{3 k}{m} $

$ \displaystyle \omega = \sqrt{\frac{3 k}{m}} $

Frequency , $ \displaystyle \nu = \frac{2\pi}{\omega} $