Two identical simple pendulums each of length ‘l’ are connected by a weightless spring as shown: In equilibrium, the pendulums are vertical and the spring is horizontal and unreformed. The time period of small oscillations of the linked pendulums, when they are deflected from their equilibrium positions through equal displacements in the same vertical plane in the opposite direction and released, is.....

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2 π \sqrt{\frac{2 l}{g} + \frac{m}{k}}

2 π \sqrt{\frac{l}{(g + \frac{k l}{m})}}

2 π \sqrt{\frac{l}{(g + \frac{2 k l}{m})}}

2 π \sqrt{\frac{l}{g}}

answer is B.

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Detailed Solution

Restoring torque on particle $= τ = force \times ⊥^{r}$ Distance

$θ = \frac{x}{l} \Rightarrow x = l θ$

$τ = (l \sin θ) mg \to (1)$

Force on spring $= 2F = 2K x = 2K θ l$

Torque on spring $= 2 F \times l \cos θ = 1 \to (2)$

Restoring Torque on particle is $' θ'$

$\sin θ = θ; \cos θ = 1$

$τ = - mg l θ + 2 K l^{2} θ$

$\Rightarrow τ = - m l θ (g + \frac{2K l}{m})$

We know that $τ = I α$

$\Rightarrow I α = - m l θ (g + \frac{2 K l}{m})$

$\Rightarrow m l^{2} α = - m l θ (g + \frac{2 K l}{m})$

$α = - \frac{(g + \frac{2 K l}{m})}{l} θ$

$∴ α = - ω^{2} θ \Rightarrow ω^{2} = \frac{g + \frac{2 K l}{m}}{l}$

$ω = \sqrt{\frac{(g + \frac{2 K l}{m})}{l}}$

$\Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{l}{(g + \frac{2 K l}{m})}}$

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