Test Index

Oscillations Part 1

Section: Physics

Question : 15 of 100

Marks: +1, -0

\;\frac{\pi}{\omega}

Solution:

Time period,

T=\;\frac{2\pi}{\omega'}

Given, time period,

T=\;\frac{\pi}{\omega}

\;\frac{2\pi}{\omega'} = \;\frac{\pi}{\omega} \Rightarrow \omega' = 2\omega

Options (a) and (b) are incorrect.

Option (c),

\sin^{2} \omega t = \;\frac{1}{2}(2 \sin^{2} \omega t) = \;\frac{1}{2}(1-\cos 2\omega t)

Hence, the angular frequency is

2\omega

Option (d),

3 \cos \left(\;\frac{\pi}{4} - 2\omega t\right)

Angular frequency of

\text{SHM}

2\omega

Option (d) is the correct answer.

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