Question

Find the frequency of oscillation of a uniform rod of length L and mass M which is pivoted at the centre and its both ends are attached to two springs (which are fixed to the walls) of equal spring constant k and it is also pushed through a small angle $\theta $ in one direction and released

• A. $\frac{1}{2 \pi } \sqrt{\frac{\text{k}}{\text{M}}}$
• B. $\frac{1}{2 \pi } \sqrt{\frac{2 \text{k}}{\text{M}}}$
• C. $\frac{1}{2 \pi } \sqrt{\frac{2 \text{Y} \text{k}}{\text{M}}}$
• D. $\frac{1}{2 \pi } \sqrt{\frac{6 \text{k}}{\text{M}}}$

Answer 1

Answer: (D)

Consider the situation below

From the above diagram
$\theta = \frac{\text{x}}{ l / 2}$
$\Rightarrow \text{x} = \frac{\text{L}}{2} \theta $
In the above diagram the rod is displaced through an angle $\theta $ as in above diagram
Restoring torque $= \left(- 2 \text{kx}\right) \times \frac{\text{L}}{2} = - \text{kxL}$
$\Rightarrow \alpha =$ angular acceleration $= \frac{- \text{k} \times \frac{\text{L}}{2} \theta \times \text{L}}{\text{l}}$
$= \frac{- \left(\text{kL}\right)^{2} / 2}{\left(\text{ML}\right)^{2} / 1 2} = - \left(\frac{6 \text{k}}{\text{M}}\right) \theta $
$\Rightarrow \text{f} = \frac{1}{2 \pi } \sqrt{\frac{\text{kl}}{\theta }} = \frac{1}{2 \pi } \sqrt{\frac{6 \text{k}}{\text{M}}}$