Oscillations

JEE Main level, numerical answer, medium

37 of 194

Question 37 of 194

A particle executes simple harmonic motion represented by displacement function as x(t)=Asin⁡(ωt+ϕ)x(t)=A\sin (\omega t+\phi ). If the position and velocity of the particle at t=0st=0s are 2cm2cm and 2ωcms−12\omega cm{s}^{-1} respectively, then its amplitude is x2cmx\sqrt{2}cm where the value of xx is

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Answer: 2

x(t)=Asin⁡(ωt+ϕ)x(t)=A\sin (\omega t+\phi )
v(t)=Aωcos⁡(ωt+ϕ)v(t)=A\omega \cos (\omega t+\phi )
2=Asinϕ2=Asin\phi ...(1)...\left(1\right)
2ω=Aωcos⁡ϕ2\omega =A\omega \cos \phi ...(2)...\left(2\right)
From (1)(1) and (2)(2)
tan⁡ϕ=1\tan \phi =1
ϕ=45∘\phi =45^{\circ}
Putting the value of ϕ\phi in equation (1)(1)
2=A{12}2=A\left\{\dfrac{1}{\sqrt{2}}\right\}
A=22A=2\sqrt{2}
x=2x=2

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