Oscillations

JEE Main level, single correct, medium

187 of 194

Question 187 of 194

The function of time representing a simple harmonic motion with a period of πω\dfrac{\pi }{\omega } is :

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Answer: (D) 3cos⁡(π4−2ωt)3\cos \left(\dfrac{\pi }{4}-2\omega t\right)

Time period T=2πω′T=\dfrac{2\pi }{\omega '}
πω=2πω′\dfrac{\pi }{\omega }=\dfrac{2\pi }{\omega '}
ω′=2ω→\omega '=2\omega \rightarrow Angular frequency of SHM
Option (3)(3)
sin⁡2ωt=12(2sin⁡2ωt)=12(1−cos⁡2ωt){\sin }^{2}\omega t=\dfrac{1}{2}\left(2{\sin }^{2}\omega t\right)=\dfrac{1}{2}\left(1-\cos 2\omega t\right)
Angular frequency of (12−12cos⁡2ωt)\left(\dfrac{1}{2}-\dfrac{1}{2}\cos 2\omega t\right) is 2ω2\omega
Option (4)(4)
Angular frequency of SHM
3cos⁡(π4−2ωt)3\cos \left(\dfrac{\pi }{4}-2\omega t\right) is 2ω2\omega.
So option (3)&(4) both have angular frequency 2ω2\omega but option (4)(4) is direct answer.

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