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 $=\frac{-\text{k}\times \frac{\text{L}}{2}\theta \times \text{L}}{\text{l}}$
$=\frac{-{\left(\text{kL}\right)}^2/2}{{\left(\text{ML}\right)}^2/12}=-\left(\frac{6\text{k}}{\text{M}}\right)\theta$
$\Rightarrow \text{f}=\frac{1}{2\pi }\sqrt{\frac{\text{kl}}{\theta }}=\frac{1}{2\pi }\sqrt{\frac{6\text{k}}{\text{M}}}$