Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x -axis. Their mean position is separated by distance X₀ (X₀ > A). If the maximum separation between them is (X₀ + A), the phase difference between their motion is: (a) π/3 (b) π/4 (c) π/6 (d) π/2

Answer: (a). Take x₁ = Asin ω t and x₂ = X₀ + Asin (ω t + φ). Separation = X₀ + A[sin (ω t+φ) - sin ω t] = X₀ + 2Asin φ/2cos (ω t + φ/2) Maximum separation = X₀ + 2Asin φ/2 = X₀ + A ⇒ sin φ/2 = 1/2 φ/2 = π/6 ⇒ φ = π/3 So the answer is (a) π/3.

Two particles are executing simple harmonic motion of the same… | JEE Main 2011 (AIEEE)

Q70JEE Main 2011 · AIEEEDifficult

Periodic Motion & SHM Basics

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$ -axis. Their mean position is separated by distance $X_0\ (X_0 > A)$ . If the maximum separation between them is $(X_0 + A),$ the phase difference between their motion is:

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Answer: (a)

Take $x_1 = A\sin\omega t$ and $x_2 = X_0 + A\sin(\omega t + \phi)$ .
Separation $= X_0 + A[\sin(\omega t+\phi) - \sin\omega t] = X_0 + 2A\sin\frac{\phi}{2}\cos\left(\omega t + \frac{\phi}{2}\right)$
Maximum separation $= X_0 + 2A\sin\frac{\phi}{2} = X_0 + A \Rightarrow \sin\frac{\phi}{2} = \frac{1}{2}$
$\frac{\phi}{2} = \frac{\pi}{6} \Rightarrow \phi = \frac{\pi}{3}$
So the answer is (a) $\frac{\pi}{3}$ .

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Periodic Motion & SHM Basics question from JEE Main 2011 (AIEEE)

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