Test Index

Oscillations Part 2

Section: Physics

Question : 14 of 81

Marks: +1, -0

176\,\text{rad/s}

\_\_\_\_

\pi=\;\frac{22}{7}

Solution:

Concept:

In simple harmonic motion, the kinetic energy oscillates at twice the frequency of the oscillator itself.

Explanation:

Let the angular frequency of the oscillator be

\omega

Kinetic energy is given by

KE = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)

Using

\sin^2\theta = \frac{1 - \cos 2\theta}{2}

, we get

KE = \frac{1}{4} m \omega^2 A^2 \left[1 - \cos(2\omega t + 2\phi)\right]

Thus, the angular frequency of the KE oscillation is

2\omega

Given

2\omega = 176\ \text{rad/s}

, so

\omega = 88\ \text{rad/s}

Frequency of the oscillator

f = \frac{\omega}{2\pi} = \frac{88}{2 \times \frac{22}{7}} = \frac{88 \times 7}{44} = 2 \times 7 = 14\ \text{Hz}

Answer:

Option B:

14

Hz.

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