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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Oscillations

JEE Advanced 2025 · Paper 2 · Physics · Oscillations

As shown in the figures, a uniform rod O O^ of length l is hinged...

As shown in the figures, a uniform rod OO′O O^{\prime} of length ll is hinged at the point OO 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 (O′)\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 f1f_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 f2f_2. Ignoring gravity and assuming motion only in the plane of the diagram, the value of f1f2\frac{f_1}{f_2} is: As shown in the figures, a uniform rod O O^ of length l is hinged at the point O and held in place vertically between

Options

  1. A22
  2. B2\sqrt{2}
  3. C52\sqrt{\frac{5}{2}}
  4. D25\sqrt{\frac{2}{5}}

Correct answer

C. 52\sqrt{\frac{5}{2}}

Step-by-step solution

As shown in the figures, a uniform rod O O^ of length l is hinged at the point O and held in place vertically between
ML23θ¨+K⋅ℓ2θ⋅ℓ2+K⋅ℓ⋅θ⋅ℓ=0θ¨+154kMθ=0θ¨=−(154kM)θω1=15 K4M\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}As shown in the figures, a uniform rod O O^ of length l is hinged at the point O and held in place vertically between
13ML2θ¨+2 K L2θ⋅ L2=0\frac{1}{3} \mathrm{ML}^2 \ddot{\theta}+2 \mathrm{~K} \frac{\mathrm{~L}}{2} \theta \cdot \frac{\mathrm{~L}}{2}=0θ¨+32KMθ=0θ¨=−32KMθω2=32KMω1ω2=154×23=52\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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