Question

A rod of mass m and length l is connected by two spring of spring constants $k_1$ and $k_2$, so that it is horizontal at equilibrium. What is the natural frequency of the system?

• A. $\frac{1}{2\pi }\sqrt{\frac{k_1{b}^2+k_2{l}^2}{m{l}^2}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{2k_1{b}^2+k_2{l}^2}{m{l}^2}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{k_1{b}^2+k_2{l}^2}{2m{l}^2}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}}$

Answer 1

Answer: (D)

$\frac{1}{2\pi }\sqrt{\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}}$

Let rod is displaced by an angle $\theta$, taking torque about edge of rod.

$k_{1}b\theta \times bcos\theta +k_{2}l\theta \times cos\theta =\frac{I{d}^2\theta }{d{t}^2}$

Here, $I=\frac{m{l}^2}{3}$ and $\theta$ is small $cos\theta =1$

$\therefore \frac{m{l}^2\alpha }{3}+(k_1{b}^2+k_2{l}^2)\theta =0$

$\Rightarrow \alpha =\frac{-3(k_1{b}^2+k_2{l}^2)\theta }{m{l}^2}$

$\therefore \omega =\sqrt{\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}}$ and frequency

$v=\frac{1}{2\pi }\sqrt{\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}}$