Y = A sin( ω t + φ 0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is Y = A/2 and it is moving along negative x-direction. Then the initial phase angle φ 0 will be: (a) 5π/6 (b) π/3 (c) 2π/3 (d) π/6

Answer: (a). At t = 0: A/2 = Asin φ₀ ⇒ sin φ₀ = 1/2, so φ₀ = π/6 or 5π/6. Velocity v = Aωcos φ₀ must be negative, so cos φ₀ < 0. cos 5π/6 < 0, so φ₀ = 5π/6. So the answer is (a) 5π/6.

Y = A sin( ω t + φ 0) is the time-displacement equation of a SHM. At… | JEE Main 2021 (25 Feb, Shift 2)

Q46JEE Main 2021 · 25 FEB · SHIFT 2Easy

Periodic Motion & SHM Basics

Y = A sin( $\omega$ t + $\phi$ 0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $Y = \frac{A}{2}$ and it is moving along negative x-direction. Then the initial phase angle $\phi$ 0 will be:

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Answer: (a)

At $t = 0$ : $\frac{A}{2} = A\sin\phi_0 \Rightarrow \sin\phi_0 = \frac{1}{2}$ , so $\phi_0 = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .
Velocity $v = A\omega\cos\phi_0$ must be negative, so $\cos\phi_0 < 0$ .
$\cos\frac{5\pi}{6} < 0$ , so $\phi_0 = \frac{5\pi}{6}$ .
So the answer is (a) $\frac{5\pi}{6}$ .

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