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PhysicsEMI/ACDamped SHM and forced oscillationsHard2 minQB

PhysicsHardnumerical

Passage / Comprehension

A uniform rod of mass m and length $ℓ$ is free to rotate about an axis passing through its end. Neglect gravity. There is a spring of constant k at $\frac{ℓ}{2}$ from axis and damper of $F = - cv$ at the free end. If rod is displaced slightly, it executes damped oscillations. Take $\frac{k}{m} = \frac{25}{3} \frac{c ^{2}}{m ^{2}}, \frac{c}{m} = 10/ sec$ .

What is the angular frequency of these oscillations.

Answer:

20.00

Solution:

Spring at $ℓ /2$ : restoring torque $- kx^{2} θ$ with $x = ℓ /2$ ; damper at free end: torque $- c ℓ^{2} \dot{θ}$ .

$\frac{m ℓ ^{2}}{3} \ddot{θ} + c ℓ^{2} \dot{θ} + \frac{k ℓ ^{2}}{4} θ = 0$ .

$ω_{0}^{2} = \frac{3 k}{4 m} = \frac{3}{4} \times \frac{25}{3} \frac{c ^{2}}{m ^{2}} = 625/ s^{2}$ using $\frac{c}{m} = 10$ .

Damping coefficient $γ = \frac{c ℓ ^{2}}{2 I} = \frac{3 c}{2 m} = 15/ s$ .

$ω = ω_{0}^{2} - γ^{2} = 625 - 225 = 20 rad / s$ .

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Damped SHM and forced oscillations

⏱ 2mℹ️ Source: QB

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