Two simple harmonic motions are represented by the equations y₁ = 0.1sin (100π t + π/3) and y₂ = 0.1cos π t. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is (a) π/3 (b) -π/6 (c) π/6 (d) -π/3

Answer: (b). Take y₂ = 0.1cos 100π t (same ω as y₁). v₁ ∝ cos (100π t + π/3), v₂ ∝ -sin 100π t = cos (100π t + π/2) Δφ = π/3 - π/2 = -π/6 So the answer is (b) -π/6.

Two simple harmonic motions are represented by the equations y₁ =… | JEE Main 2005 (AIEEE)

Q79JEE Main 2005 · AIEEEMedium

Periodic Motion & SHM Basics

Two simple harmonic motions are represented by the equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos \pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

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Answer: (b)

Take $y_2 = 0.1\cos 100\pi t$ (same $\omega$ as $y_1$ ).
$v_1 \propto \cos\left(100\pi t + \frac{\pi}{3}\right)$ , $v_2 \propto -\sin 100\pi t = \cos\left(100\pi t + \frac{\pi}{2}\right)$
$\Delta\phi = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$
So the answer is (b) $-\frac{\pi}{6}$ .

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