JEE Advanced · Physics · 15. Oscillations
As shown in the figures, a uniform rod \(O O^{\prime}\) of length \(l\) is hinged at the point \(O\) and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end \(\left(O^{\prime}\right)\) of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is \(f_1\). On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is \(f_2\). Ignoring gravity and assuming motion only in the plane of the diagram, the value of \(\frac{f_1}{f_2}\) is:
- A \(2\)
- B \(\sqrt{2}\)
- C \(\sqrt{\frac{5}{2}}\)
- D \(\sqrt{\frac{2}{5}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{5}{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\mathrm{ML}^2}{3} \ddot{\theta}+\mathrm{K} \cdot \frac{\ell}{2} \theta \cdot \frac{\ell}{2}+\mathrm{K} \cdot \ell \cdot \theta \cdot \ell=0 \\ & \ddot{\theta}+\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}} \theta=0 \\ & \ddot{\theta}=-\left(\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}}\right) \theta \\ & \omega_1=\sqrt{\frac{15 \mathrm{~K}}{4 \mathrm{M}}}\end{aligned}\)
\(\frac{1}{3} \mathrm{ML}^2 \ddot{\theta}+2 \mathrm{~K} \frac{\mathrm{~L}}{2} \theta \cdot \frac{\mathrm{~L}}{2}=0\)
\(\begin{aligned} & \ddot{\theta}+\frac{3}{2} \frac{K}{M} \theta=0 \\ & \ddot{\theta}=-\frac{3}{2} \frac{K}{M} \theta \\ & \omega_2=\sqrt{\frac{3}{2} \frac{K}{M}} \\ & \frac{\omega_1}{\omega_2}=\sqrt{\frac{15}{4} \times \frac{2}{3}}=\sqrt{\frac{5}{2}}\end{aligned}\)
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