JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Elasticity & SHM

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) $\frac{5}{2}$ ✓ Correct
(D) $\frac{2}{5}$

Answer: (C)

Solution

For Fig. 1, the restoring torque about the hinge is due to springs at

ℓ /2

and

ℓ

. The equation of motion is:

\frac{M ℓ ^{2}}{3} \ddot{θ} + k \frac{ℓ}{2} θ \cdot \frac{ℓ}{2} + k ℓ θ \cdot ℓ = 0

\ddot{θ} + \frac{15 k}{4 M} θ = 0 ⟹ ω_{1} = \frac{15 k}{4 M}

For Fig. 2, both springs at

ℓ /2

\frac{M ℓ ^{2}}{3} \ddot{θ} + 2 k \frac{ℓ}{2} θ \cdot \frac{ℓ}{2} = 0

\ddot{θ} + \frac{3 k}{2 M} θ = 0 ⟹ ω_{2} = \frac{3 k}{2 M}

Therefore:

\frac{ω _{1}}{ω _{2}} = \frac{15}{4} \times \frac{2}{3} = \frac{5}{2}

The answer is (C).

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Elasticity & SHM

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