As shown in the figures, a uniform rod OO' 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 f₁. 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 f₂. Ignoring gravity and assuming motion only in the plane of the diagram, the value of \fracf_1f_2 is:

Simple Harmonic Motion PYQ — JEE Advanced 2025 · Shift 2 · Q35 with Solution

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Correct answer: C

Physical model

A uniform rod of length $l$ is hinged at $O$ and undergoes a small angular displacement $\theta$ in the plane.

For small oscillations, each spring provides a restoring force due to the horizontal displacement of its point of attachment.

Since gravity is ignored, the restoring torque comes only from the springs.

Moment of inertia of the rod about hinge $O$

For a uniform rod of length $l$ about one end,

I_O = \frac{1}{3} m l^2

where $m$ is the mass of the rod.

Small displacement of a point at distance $r$ from hinge

If the rod rotates by a small angle $\theta$ , a point at distance $r$ from the hinge moves horizontally by

x \approx r\theta

So the spring attached there is stretched/compressed by amount $r\theta$ .

Hence spring force magnitude is

F = kx = kr\theta

This force is horizontal, so for small $\theta$ , its torque about the hinge is approximately

\tau = Fr = kr^2\theta

restoring in nature, i.e.

\tau = -kr^2\theta

Thus, a spring attached at distance $r$ contributes torsional constant

C = kr^2

Case 1: Fig. 1

One spring is attached at midpoint:

r_1 = \frac{l}{2}

Another spring is attached at top end:

r_2 = l

So total torsional constant is

C_1 = k\left(\frac{l}{2}\right)^2 + k(l)^2

C_1 = k\frac{l^2}{4} + kl^2 = \frac{5}{4}kl^2

Angular frequency is

\omega_1 = \sqrt{\frac{C_1}{I_O}}

\omega_1 = \sqrt{\frac{\frac{5}{4}kl^2}{\frac{1}{3}ml^2}}

\omega_1 = \sqrt{\frac{15k}{4m}}

Case 2: Fig. 2

Both springs are attached at midpoint, so each has

r = \frac{l}{2}

Hence total torsional constant is

C_2 = 2k\left(\frac{l}{2}\right)^2

C_2 = 2k\frac{l^2}{4} = \frac{1}{2}kl^2

Thus,

\omega_2 = \sqrt{\frac{C_2}{I_O}}

\omega_2 = \sqrt{\frac{\frac{1}{2}kl^2}{\frac{1}{3}ml^2}}

\omega_2 = \sqrt{\frac{3k}{2m}}

Ratio of frequencies

Since frequency $f = \dfrac{\omega}{2\pi}$ ,

\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2}

So,

\frac{f_1}{f_2} = \sqrt{\frac{\frac{15k}{4m}}{\frac{3k}{2m}}}

\frac{f_1}{f_2} = \sqrt{\frac{15}{4}\cdot\frac{2}{3}}

\frac{f_1}{f_2} = \sqrt{\frac{5}{2}}

Option check

A: $2$ ❌
B: $\sqrt{2}$ ❌
C: $\sqrt{\frac{5}{2}}$ ✅
D: $\sqrt{\frac{2}{5}}$ ❌

Therefore, the correct answer is

\boxed{\sqrt{\frac{5}{2}}}

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View written solutionFree

Correct answer: C

Physical model

A uniform rod of length $l$ is hinged at $O$ and undergoes a small angular displacement $\theta$ in the plane.

For small oscillations, each spring provides a restoring force due to the horizontal displacement of its point of attachment.

Since gravity is ignored, the restoring torque comes only from the springs.

Moment of inertia of the rod about hinge $O$

For a uniform rod of length $l$ about one end,

I_O = \frac{1}{3} m l^2

where $m$ is the mass of the rod.

Small displacement of a point at distance $r$ from hinge

If the rod rotates by a small angle $\theta$ , a point at distance $r$ from the hinge moves horizontally by

x \approx r\theta

So the spring attached there is stretched/compressed by amount $r\theta$ .

Hence spring force magnitude is

F = kx = kr\theta

This force is horizontal, so for small $\theta$ , its torque about the hinge is approximately

\tau = Fr = kr^2\theta

restoring in nature, i.e.

\tau = -kr^2\theta

Thus, a spring attached at distance $r$ contributes torsional constant

C = kr^2

Case 1: Fig. 1

One spring is attached at midpoint:

r_1 = \frac{l}{2}

Another spring is attached at top end:

r_2 = l

So total torsional constant is

C_1 = k\left(\frac{l}{2}\right)^2 + k(l)^2

C_1 = k\frac{l^2}{4} + kl^2 = \frac{5}{4}kl^2

Angular frequency is

\omega_1 = \sqrt{\frac{C_1}{I_O}}

\omega_1 = \sqrt{\frac{\frac{5}{4}kl^2}{\frac{1}{3}ml^2}}

\omega_1 = \sqrt{\frac{15k}{4m}}

Case 2: Fig. 2

Both springs are attached at midpoint, so each has

r = \frac{l}{2}

Hence total torsional constant is

C_2 = 2k\left(\frac{l}{2}\right)^2

C_2 = 2k\frac{l^2}{4} = \frac{1}{2}kl^2

Thus,

\omega_2 = \sqrt{\frac{C_2}{I_O}}

\omega_2 = \sqrt{\frac{\frac{1}{2}kl^2}{\frac{1}{3}ml^2}}

\omega_2 = \sqrt{\frac{3k}{2m}}

Ratio of frequencies

Since frequency $f = \dfrac{\omega}{2\pi}$ ,

\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2}

So,

\frac{f_1}{f_2} = \sqrt{\frac{\frac{15k}{4m}}{\frac{3k}{2m}}}

\frac{f_1}{f_2} = \sqrt{\frac{15}{4}\cdot\frac{2}{3}}

\frac{f_1}{f_2} = \sqrt{\frac{5}{2}}

Option check

A: $2$ ❌
B: $\sqrt{2}$ ❌
C: $\sqrt{\frac{5}{2}}$ ✅
D: $\sqrt{\frac{2}{5}}$ ❌

Therefore, the correct answer is

\boxed{\sqrt{\frac{5}{2}}}

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