Most Important Selected Qs for JEE AdvancedPhysicsOscillations
Two identical simple pendulums each of length L and mass m are connected by a weightless spring as shown in figure. The force constant of the spring is k. In equilibrium the pendulums are vertical and the spring is horizontal and un-deformed. The frequency of small oscillations of the linked pendulums, when they are deflected from their equilibrikum positions through equal displacements in the same vertical plane in
Options
- Af₁= 1 2 g 2 L
- Bf₁= 1 2 g L
- Cf₂= 1 2 g L + 2 k m
- Df₂= 1 2 g 2 L + k m
Correct answer
C. f₂= 1 2 g L + 2 k m
Step-by-step solution
(a) When both the pendulums are displaced in the same direction by same amount, the spring will neither compress or stretch, so the restoring torque on each pendulum about the point of suspension will be due to its own weight only. i.e. =-m g L =-m g L [ as for small , = ] But by definition =l =M L^2 d^2 d t^2 [ . as .l=mL^2 ]