Oscillations

JEE Main level, single correct, medium

21 of 194

Question 21 of 194

For a periodic motion represented by the equation y=sinωt+cosωty=sin\omega t+cos\omega t the amplitude of the motion is

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Answer: (D) 2\sqrt{2}

Given the equation of motion is
y=sinωt+cosωt...(1)y=sin\omega t+cos\omega t...\left(1\right)
Simplify equation (1) to obtain the standard form:
y=2(sin⁡ωtcos⁡π4+cos⁡ωtsin⁡π4)=2sin⁡(ωt+π4)...(2)\begin{matrix}y & = & \sqrt{2}\left(\sin \omega t\cos \dfrac{\pi }{4}+\cos \omega t\sin \dfrac{\pi }{4}\right) \\ & = & \sqrt{2}\sin \left(\omega t+\dfrac{\pi }{4}\right)...\left(2\right)\end{matrix}
Comparing equation (2) with standard equation of SHM y=Asin⁡(ωt+ϕ)y=A\sin \left(\omega t+\phi \right), we get amplitude of the given oscillation as 2\sqrt{2}.

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