Two particles are executing simple harmonic motion of the same amplitude A and frequency w along the

Q: Why do we add $X_0$ to $x_2$?

Because $x_1$ and $x_2$ are displacements from their *own* mean positions. To find the actual distance between them, we need their coordinates on a common axis.

Q: Why is the maximum of $x_2 - x_1$ equal to the length of the difference phasor?

The value $x_2 - x_1$ is the projection of the difference vector on the x-axis. The projection is maximum when the vector is parallel to the x-axis, and its value is exactly the magnitude of the vector.

Q: Can we solve this using trigonometry?

Yes, by expanding $\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)$ into $2 \sin(\frac{\Delta \phi}{2}) \cos(\omega t + \text{avg})$. The amplitude of this new SHM is $2A \sin(\frac{\Delta \phi}{2})$. Equating this to $A$ gives the same result.

Q: What if the maximum separation was $X_0 + \sqrt{2}A$?

Then the difference phasor would have length $\sqrt{2}A$. A triangle with sides $A, A, \sqrt{2}A$ is a right-angled triangle, so the phase difference would be $\pi/2$.

Q: Does the frequency $\omega$ affect the maximum separation?

No, as long as both particles have the *same* frequency, the phase difference remains constant, and the maximum separation depends only on the amplitudes and the phase difference.

Question

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

Options:

(A) $\frac{\pi}{3}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{2}$

Sigma Percentile · Accepted Answer

The problem of finding the phase difference between two oscillating particles might seem like a daunting trigonometric nightmare at first glance. But what if I told you that we can solve it in seconds using pure geometry? Let's dive into the magic of phasors!

### Analyzing the Setup

Imagine two particles executing Simple Harmonic Motion (SHM) along the x-axis. They both have the same amplitude $A$ and the same angular frequency $\omega$. However, they are oscillating about different mean positions.

Let's say the first particle oscillates about the origin, so its mean position is at $x = 0$. The second particle oscillates about a mean position located at a distance $X_0$ from the origin.

If we denote their displacements from their *respective* mean positions as $x_1$ and $x_2$, we can write their absolute coordinates on the x-axis as:
$y_1 = x_1$
$y_2 = X_0 + x_2$

### The Master Equation for Separation

The question gives us a crucial piece of information: the maximum separation between the two particles is $X_0 + A$.

To use this, we first need an expression for the separation distance $d$ at any given instant $t$. The separation is simply the difference between their absolute positions:
$d = y_2 - y_1$
$d = (X_0 + x_2) - x_1$
$d = X_0 + (x_2 - x_1)$

For the separation $d$ to be maximum, the relative displacement term $(x_2 - x_1)$ must be at its maximum value. 
$d_{	ext{max}} = X_0 + (x_2 - x_1)_{	ext{max}}$

We are given that $d_{	ext{max}} = X_0 + A$. Comparing the two equations, we arrive at a beautiful conclusion:
$(x_2 - x_1)_{	ext{max}} = A$

### The Phasor Trick

Now, we could write $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$, subtract them, and use trigonometric identities to find the maximum amplitude of the resulting wave. But that's the long way around!

Instead, let's use **phasors**. In a phasor diagram, an SHM is represented by a vector of length $A$ rotating in a circle. The actual displacement is just the horizontal projection of this vector.

Since we have two particles, we draw two phasors, $\vec{A}_1$ and $\vec{A}_2$, both of length $A$. The angle between them is their phase difference, $\Delta \phi$.

The relative displacement $x_2 - x_1$ is simply the horizontal projection of the vector difference $\vec{A}_2 - \vec{A}_1$.

### The Equilateral Triangle

What is the maximum possible value of the projection of a vector? It's simply the magnitude (length) of the vector itself!

Therefore, the maximum value of $(x_2 - x_1)$ is exactly the magnitude of the difference vector $|\vec{A}_2 - \vec{A}_1|$.

From our earlier deduction, we know this maximum value is $A$. 
$|\vec{A}_2 - \vec{A}_1| = A$

Now, look at the geometry of our phasor diagram. We have a triangle formed by the origin and the tips of the two phasors. 
- The length of the first phasor is $A$.
- The length of the second phasor is $A$.
- The length of the difference vector connecting their tips is also $A$.

All three sides of this triangle are equal! This means it is an **equilateral triangle**.

In an equilateral triangle, all internal angles are $60^\circ$. Therefore, the angle between the two phasors—which is exactly the phase difference $\Delta \phi$—must be $60^\circ$.

Converting this to radians, we get:
$\Delta \phi = \frac{\pi}{3}$

And just like that, by visualizing the problem geometrically, we bypassed all the complex algebra and arrived at the correct answer instantly. Always look for the phasor shortcut in SHM problems!

Select Answer:

Visualized Solution

$y_{1} = x_{1}, y_{2} = X_{0} + x_{2}$

$d = y_{2} - y_{1} = X_{0} + (x_{2} - x_{1})$

$d_{max} = X_{0} + (x_{2} - x_{1})_{max} = X_{0} + A$

$A_{rel} = A_{2} - A_{1}$

$∣ A_{2} - A_{1} ∣ = A$

$Δ ϕ = \frac{π}{3}$

$Phasors > Trigonometry$

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Master Equation for Separation

The Phasor Trick

The Equilateral Triangle

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​=x1​,y2​=X0​+x2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=y2​−y1​=X0​+(x2​−x1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dmax​=X0​+(x2​−x1​)max​=X0​+A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Arel​=A2​−A1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A2​−A1​∣=A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δϕ=3π​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phasors>Trigonometry