Simple Harmonic Motion 4 Question 3

3. Two light identical springs of spring constant k are attached horizontally at the two ends of an uniform horizontal rodAB of length l and mass m. The rod is pivoted at its centre ’ O ’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure.

The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is (2019 Main, 12 Jan I)

(a) 12π2km

(b) 12π3km

(c) 12π6km

(d) 12πkm

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Answer:

Correct Answer: 3. (c)

Solution:

  1. When a system oscillates, the magnitude of restoring torque of system is given by

τ=Cθ

where, C= constant that depends on system.

 Also, τ=Iα

where, I= moment of inertia

and α= angular acceleration

From Eqs. (i) and (ii),

α=CIθ

and time period of oscillation of system will be

T=2πIC

In given case, magnitude of torque is

τ= Force × perpendicular distance

τ=2kx×l2cosθ

For small deflection,

τ=kl22θ

For small deflections, sinθ=x(l/2)θ

x=lθ2

Also,

cosθ1

comparing Eqs. (iv) and (i), we get

C=kl22α=(kl2/2)112ml2θα=6kmθ

Hence, time period of oscillation is

T=2πm6k

Frequency of oscillation is given by

f=1T=12π6km



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