Question

The frequency of small oscillations of thin uniform vertical rod of mass \({}'m'\) and length \({}'l'\) hinged at point O with the help of two springs with spring constants \(k_{1}\) and \(k_{2}\), is

• A. \(\frac{1}{2\pi }\sqrt{\frac{(k_{1}+k_{2})}{m}+\frac{3g}{2l}}\)
• B. \(\frac{1}{2\pi }\sqrt{\frac{3(k_{1}+k_{2})}{m}+\frac{3g}{2l}}\)
• C. \(2\pi \sqrt{\frac{(k_{1}+k_{2})}{m}+\frac{g}{2l}}\)
• D. \(2\pi \sqrt{\frac{(k_{1}+k_{2})}{3m}+\frac{2g}{3l}}\)

Hint:

Find the restoring torque from the springs and gravity about the hinge, and divide by the moment of inertia of the rod about the hinge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For small angular displacement \(\theta\) the motion is angular SHM, with \(\omega^{2}=\frac{\text{restoring torque per unit angle}}{I}\).

Step 2: Calculation
The lower end moves by \(l\theta\), so the two springs together give a torque \((k_{1}+k_{2})l^{2}\theta\). Gravity acting at the centre gives \(mg\frac{l}{2}\theta\). The moment of inertia about the hinge is \(I=\frac{ml^{2}}{3}\). So \(\omega^{2}=\frac{(k_{1}+k_{2})l^{2}+mg\,l/2}{ml^{2}/3}=\frac{3(k_{1}+k_{2})}{m}+\frac{3g}{2l}\).

Step 3: Compare the options
The frequency is \(f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{3(k_{1}+k_{2})}{m}+\frac{3g}{2l}}\). Options C and D start with \(2\pi\) instead of \(\frac{1}{2\pi}\), which is the period form, not the frequency.

Final Answer:
The frequency is \(\frac{1}{2\pi}\sqrt{\frac{3(k_{1}+k_{2})}{m}+\frac{3g}{2l}}\). \[ \boxed{\text{(B) }\dfrac{1}{2\pi}\sqrt{\dfrac{3(k_{1}+k_{2})}{m}+\dfrac{3g}{2l}}} \]