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 :

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Basics of Simple Harmonic Motion

Superposition of Simple Harmonic Motion

Question

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 :

$\frac{π}{6}$

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{3}$

x₁ = A sin(ωt + φ₁)

x₂ = A sin(ωt + f₂)

$x_{1} - x_{2} = A [2 \sin [ωt + \frac{ϕ_{1} + ϕ_{2}}{2}] \sin [\frac{ϕ_{1} - ϕ_{2}}{2}]]$

$A = 2 Asin (\frac{ϕ_{1} - ϕ_{2}}{2})$

$\frac{ϕ_{1} - ϕ_{2}}{2} = \frac{π}{6}$

$ϕ_{1} = \frac{π}{3}$

Two particles are executing simple harmonic motion (SHM) with the same amplitude A and angular frequency ω along the x-axis. Their mean positions are separated by a distance X₀, where X₀ > A. The maximum separation between them is given as (X₀ + A), and we need to find the phase difference between their motions.

Let the equations of motion for the two particles be:

Particle 1: $x_{1} = A \sin (ω t + ϕ_{1})$

Particle 2: $x_{2} = X ₀ + A \sin (ω t + ϕ_{2})$

The separation between them at any time t is: $S = x_{2} - x_{1} = X ₀ + A [\sin (ω t + ϕ_{2}) - \sin (ω t + ϕ_{1})]$

Using the trigonometric identity for the difference of sines:

$\sin (α) - \sin (β) = 2 \cos (\frac{α + β}{2}) \sin (\frac{α - β}{2})$

Let α = ωt + ϕ₂ and β = ωt + ϕ₁. Then:

$S = X ₀ + 2 A \cos (ω t + \frac{ϕ_{2} + ϕ_{1}}{2}) \sin (\frac{ϕ_{2} - ϕ_{1}}{2})$

The maximum value of S occurs when the cosine term is at its maximum absolute value of 1. Therefore:

$S_{m a x} = X ₀ + 2 A | \sin (\frac{ϕ_{2} - ϕ_{1}}{2}) |$

Given that S_max = X₀ + A, we equate:

$X ₀ + A = X ₀ + 2 A | \sin (\frac{ϕ_{2} - ϕ_{1}}{2}) |$

Subtracting X₀ from both sides:

$A = 2 A | \sin (\frac{ϕ_{2} - ϕ_{1}}{2}) |$

Dividing both sides by A (A ≠ 0):

$1 = 2 | \sin (\frac{ϕ_{2} - ϕ_{1}}{2}) |$

Therefore:

$| \sin (\frac{ϕ_{2} - ϕ_{1}}{2}) | = \frac{1}{2}$

This implies:

$\frac{ϕ_{2} - ϕ_{1}}{2} = \frac{π}{6}, \frac{5 π}{6}, \dots$ (and other angles with sine = 1/2)

So the phase difference Δϕ = |ϕ₂ - ϕ₁| = π/3 or 5π/3, etc. The smallest positive phase difference is π/3.

Thus, the phase difference between their motions is π/3.

Formulae

General equation of SHM: $x = A \sin (ω t + ϕ)$

Trigonometric identity: $\sin (α) - \sin (β) = 2 \cos (\frac{α + β}{2}) \sin (\frac{α - β}{2})$

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