SHMHard
Question
A system of two identical rods (L - shaped) of mass m and length 2L are resting on a peg P as shown in the figure. If the system is displaced in its plane by a small angle, find the period of oscillations
Options
A.
B.
C.
D.
Solution
Center of mass 2 m of a system is at a distance from peg P is 
and moment of inertia of the system is 
τ = r × F = - mg sin θ
⇒ Iα = - mg sin (in θ = θ for small θ)
⇒ α = - mgθ ⇒ α = - 
= - ω2θ
⇒ ω =
and moment of inertia of the system is τ = r × F = - mg sin θ
⇒ Iα = - mg sin
(in θ = θ for small θ)⇒
α = - mgθ ⇒ α = - = - ω2θ⇒ ω =
Create a free account to view solution
View Solution FreeMore SHM Questions
A spring is vibration with frequency under same mass. If it is cut into two equal pieces and same mass. is suspended the...The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/...The time taken by a particle performing S.H.M to pass from point A to B where its velocities are same is 2 seconds . Aft...In the above question, the value of maximum velocity of the particle will be :...The length of a simple pendulum is increased four times of its initial value, its time period with respect to its previo...