Two particles are in SHM with the same amplitude and frequency along the same line and about the same point. If the maximum separation between them is `sqrt(3)` times their amplitude, the phase difference between them is `(2pi)/n`. Find value of n

Answer

To solve the problem step by step, let's break down the information and derive the required value of \( n \).

### Step 1: Understand the problem
We have two particles in Simple Harmonic Motion (SHM) with the same amplitude \( A \) and frequency. The maximum separation between them is given as \( \sqrt{3} \) times their amplitude.

### Step 2: Write the equations of motion
The displacement of the first particle can be expressed as:
\[ x_1 = A \sin(\omega t) \]

The displacement of the second particle, which has a phase difference \( \phi \), can be expressed as:
\[ x_2 = A \sin(\omega t + \phi) \]

### Step 3: Find the separation between the two particles
The separation \( S \) between the two particles is given by:
\[ S = x_2 - x_1 = A \sin(\omega t + \phi) - A \sin(\omega t) \]

Using the sine subtraction formula, we can rewrite this as:
\[ S = A \left( \sin(\omega t + \phi) - \sin(\omega t) \right) \]

### Step 4: Apply the sine difference identity
Using the identity \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \):
\[ S = A \cdot 2 \cos\left(\frac{(\omega t + \phi) + \omega t}{2}\right) \sin\left(\frac{(\omega t + \phi) - \omega t}{2}\right) \]
\[ S = 2A \cos\left(\omega t + \frac{\phi}{2}\right) \sin\left(\frac{\phi}{2}\right) \]

### Step 5: Maximum separation
The maximum separation occurs when \( \cos\left(\omega t + \frac{\phi}{2}\right) = 1 \):
\[ S_{\text{max}} = 2A \sin\left(\frac{\phi}{2}\right) \]

According to the problem, this maximum separation is given as:
\[ S_{\text{max}} = \sqrt{3} A \]

### Step 6: Set the equations equal
Setting the two expressions for maximum separation equal gives:
\[ 2A \sin\left(\frac{\phi}{2}\right) = \sqrt{3} A \]

Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 2 \sin\left(\frac{\phi}{2}\right) = \sqrt{3} \]

### Step 7: Solve for \( \sin\left(\frac{\phi}{2}\right) \)
Dividing both sides by 2:
\[ \sin\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \]

### Step 8: Find \( \frac{\phi}{2} \)
The value of \( \frac{\phi}{2} \) that satisfies this equation is:
\[ \frac{\phi}{2} = \frac{\pi}{3} \]

### Step 9: Find \( \phi \)
Multiplying both sides by 2:
\[ \phi = \frac{2\pi}{3} \]

### Step 10: Relate \( \phi \) to \( n \)
According to the problem, the phase difference \( \phi \) is also given as:
\[ \phi = \frac{2\pi}{n} \]

Setting the two expressions for \( \phi \) equal:
\[ \frac{2\pi}{3} = \frac{2\pi}{n} \]

### Step 11: Solve for \( n \)
By equating the two expressions, we can solve for \( n \):
\[ n = 3 \]

### Final Answer
Thus, the value of \( n \) is \( 3 \).

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