Two identical simple pendulums, each of length L are connected by a weightless spring as shown in the figure. The force constant of the spring is k. In equilibrium, the pendulums are vertical and the spring is horizontal and undeformed. They are deflected from their equilibrium positions through small and equal displacements in the same vertical plane(i) in the same directions and (ii) in opposite direction and released, In case (i), time period is $T_{1}$ and in case (ii) time period is $T_{2}$ . Then,

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$T_{2} = 2 π \sqrt{\frac{2 m L}{m g + 2 k L}}$

$T_{1} = 2 π \sqrt{\frac{L}{g}}$

$T_{1} = 2 π \sqrt{\frac{2 L}{g}}$

$T_{2} = 2 π \sqrt{\frac{m L}{m g + 2 k L}}$

answer is C, D.

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Detailed Solution

(i)In this case, spring exerts no restoring torque and restoring torque is only due to the weight of the bob. So, time period is as usual $T_{1} = 2 π \sqrt{\frac{L}{g}}$

(ii)In this case, restoring torque on one pendulum is

$τ = - [m g L \sin θ + k (2 L \sin θ) L] = - (m g L + 2 k L^{2}) θ ∴ l α = - (m g L + 2 k L^{2}) θ S o, α = - \frac{(m g L + 2 k L^{2}) θ}{m L^{2}} = - (\frac{g}{L} + \frac{2 k}{m}) θ o r α = - (\frac{m g + 2 k L}{m L}) θ, c o m p a r i n g w i t h α = - ω^{2} θ ω = \sqrt{\frac{m g + 2 k L}{m L}} \Rightarrow T_{2} = \frac{2 π}{ω} = 2 π \sqrt{\frac{m L}{m g + 2 k L}}$