Oscillations

JEE Main level, single correct, medium

74 of 194

Question 74 of 194

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176rad/s176 \mathrm{rad} / \mathrm{s}. The frequency of this simple harmonic oscillator is ____\_\_\_\_ Hz. [\left[\right. take π=227]\left.\pi=\frac{22}{7}\right]

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Answer: (A) 14

In a simple harmonic oscillator, if the displacement is x=Asin⁡(ωt+ϕ)x = A \sin(\omega t + \phi), the velocity is v=Aωcos⁡(ωt+ϕ)v = A\omega \cos(\omega t + \phi).
The kinetic energy is given by K=12mv2=12mA2ω2cos⁡2(ωt+ϕ)K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi).
Using the identity cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = \frac{1 + \cos 2\theta}{2}, we get K=14mA2ω2[1+cos⁡(2ωt+2ϕ)]K = \frac{1}{4}m A^2 \omega^2 [1 + \cos(2\omega t + 2\phi)].
The angular frequency of the kinetic energy oscillation is ωKE=2ω\omega_{KE} = 2\omega, where ω\omega is the angular frequency of the SHM.
Given ωKE=176\omega_{KE} = 176 rad/s, we have 2ω=1762\omega = 176, which gives ω=88\omega = 88 rad/s.
The frequency of the simple harmonic oscillator is f=ω2πf = \frac{\omega}{2\pi}.
Substituting the values: f=882×227=88×744=2×7=14f = \frac{88}{2 \times \frac{22}{7}} = \frac{88 \times 7}{44} = 2 \times 7 = 14 Hz.

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