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PhysicsSHMAngular SHMEasy2 minQB
PhysicsEasysingle choice

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constant k. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is (the springs are fixed to the walls)

Options:

Answer:
D
Solution:

Consider the situation below



From the above diagram

θ = x l⁡ / 2

⇒     x = L 2 θ

In the above diagram the rod is displaced through an angle θ as in above diagram

Restoring torque = - 2 kx × L 2 = - kxL
angular acceleration,
⇒ a= −k× L 2 θ×L Ι = −6k M θ

=− ω 2 θ

ω= 6K M

f= 1 2π 6K M

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Angular SHM
⏱ 2mℹ️ Source: QB

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