JEE Advanced 2025 — Paper 2 — PHY-I solution

JEE Advanced2025Paper 2PHY-I

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 two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end ( $O^{'}$ ) 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
$2$
C
$\frac{5}{2}$
D
$\frac{2}{5}$

Solution

Let $M$ be the rod's mass and $K$ the spring constant. The moment of inertia about $O$ is $I_{O} = \frac{M ℓ ^{2}}{3}$ . For a small angular displacement $θ$ , each spring exerts a restoring torque equal to (force) $\times$ (lever arm).

Case 1 (Fig. 1): spring at midpoint contributes restoring torque $K (ℓ /2 \cdot θ) \cdot (ℓ /2)$ and spring at the top contributes $K (ℓ θ) \cdot ℓ$ . Equation of motion:

$\frac{M ℓ ^{2}}{3} \ddot{θ} + K \frac{ℓ ^{2}}{4} θ + K ℓ^{2} θ = 0 \Rightarrow \ddot{θ} + \frac{15 K}{4 M} θ = 0$

So $ω_{1} = \frac{15 K}{4 M}$ .

Case 2 (Fig. 2): both springs act at the midpoint, giving an effective $2 K$ at lever arm $ℓ /2$ :

$\frac{M ℓ ^{2}}{3} \ddot{θ} + 2 K \frac{ℓ ^{2}}{4} θ = 0 \Rightarrow \ddot{θ} + \frac{3 K}{2 M} θ = 0$

So $ω_{2} = \frac{3 K}{2 M}$ .

Therefore

$\frac{f _{1}}{f _{2}} = \frac{ω _{1}}{ω _{2}} = \frac{15 K /4 M}{3 K /2 M} = \frac{15}{4} \cdot \frac{2}{3} = \frac{5}{2}$

$∴$ Option (C) is correct.

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