As shown in the figures, a uniform rod OO' of length is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. [2025]

Question

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 $f_{1}$ . 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: [2025]

Smart Achievers · Accepted Answer

(3)

$τ_{net} = K (θ ℓ) ℓ + K (θ . \frac{ℓ}{2}) \frac{ℓ}{2}$

$τ_{net} = (K θ \frac{ℓ}{2}) \frac{ℓ}{2} \times 2$

$= \frac{5 K ℓ^{2}}{4} θ = ℓ α = \frac{K ℓ^{2}}{2} θ = ℓ α$

As $α = ω^{2} θ$

So, $\frac{ω_{1}^{2}}{ω_{2}^{2}} = \frac{5 K ℓ^{2}}{\frac{4 K ℓ^{2}}{2}} = \frac{5}{2}$ or, $\frac{ω_{1}}{ω_{2}} = \sqrt{\frac{5}{2}}$

Solution

1 $2$

2 $\sqrt{2}$

3 $\sqrt{\frac{5}{2}}$

4 $\sqrt{\frac{2}{5}}$

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Solution

1 2

2 2

3 52

4 25

Ans. (3) τnet=K(θℓ)ℓ+K(θ.ℓ2)ℓ2 τnet=(Kθℓ2)ℓ2×2 =5Kℓ24θ=ℓα=Kℓ22θ=ℓα As α=ω2θ So, ω12ω22=5Kℓ24Kℓ22=52 or, ω1ω2=52

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1 $2$

2 $\sqrt{2}$

3 $\sqrt{\frac{5}{2}}$

4 $\sqrt{\frac{2}{5}}$