A particle executes simple harmonic motion… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Waves & Oscillations

2021integermedium

A particle executes simple harmonic motion represented by displacement function as $x(t)=A\mathrm{sin}(\omega t+\phi )$ . If the position and velocity of the particle at $t=0s$ are $2\mathrm{cm}$ and $2\omega \mathrm{cm}{s}^{-1}$ respectively, then its amplitude is $x\sqrt{2}\mathrm{cm}$ where the value of $x$ is

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Held on 27 Jul 2021 · Verified 6 Jul 2026.

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Solution

$x(t)=A\mathrm{sin}(\omega t+\phi )$

$v(t)=A\omega \mathrm{cos}(\omega t+\phi )$

$2=\mathrm{Asin}\phi$ $...(1)$

$2\omega =A\omega \mathrm{cos}\phi$ $...(2)$

From $(1)$ and $(2)$

$\mathrm{tan}\phi =1$

$\phi =45^{\circ}$

Putting the value of $\phi$ in equation $(1)$

$2=A{\frac{1}{\sqrt{2}}}$

$A=2\sqrt{2}$

$x=2$

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