A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is x = A/2 and it moves along positive x-axis. The displacement of particle in time t is x = Asin (ω t + δ), then the value of δ will be (a) π/2 (b) π/3 (c) π/4 (d) π/6

Answer: (d). At t = 0: A/2 = Asin δ ⇒ sin δ = 1/2 Moving along +x means v = Aωcos δ > 0, so δ = π/6 (not 5π/6). So the answer is (d) π/6.

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the… | JEE Main 2023 (10 Apr, Shift 1)

Q36JEE Main 2023 · 10 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $x = \frac{A}{2}$ and it moves along positive x-axis. The displacement of particle in time t is $x = A\sin(\omega t + \delta)$ , then the value of $\delta$ will be

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Answer: (d)

At $t = 0$ : $\frac{A}{2} = A\sin\delta \Rightarrow \sin\delta = \frac{1}{2}$
Moving along $+x$ means $v = A\omega\cos\delta > 0$ , so $\delta = \frac{\pi}{6}$ (not $\frac{5\pi}{6}$ ).
So the answer is (d) $\frac{\pi}{6}$ .

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