Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is one third their amplitude. What is the phase difference between them?

Options

A. 0
B. `(2pi)/(3)`
C. `pi`
D. `(pi)/(6)`

Answer

To find the phase difference between two particles executing simple harmonic motion (SHM) with the given conditions, we can follow these steps:

### Step 1: Define the equations of motion for both particles
Let the displacement of the first particle be given by:
\[ x_1 = A \sin(\omega t) \]
For the second particle, with a phase difference \( \theta \), the displacement can be expressed as:
\[ x_2 = A \sin(\omega t + \theta) \]

### Step 2: Set the condition for when they pass each other
The particles pass each other when their displacements are one-third of their amplitude. Therefore, we have:
\[ x_1 = \frac{A}{3} \]
This implies:
\[ \frac{A}{3} = A \sin(\omega t) \]
From this, we can simplify to find:
\[ \sin(\omega t) = \frac{1}{3} \]

### Step 3: Calculate \( \cos(\omega t) \)
Using the Pythagorean identity, we can find \( \cos(\omega t) \):
\[ \sin^2(\omega t) + \cos^2(\omega t) = 1 \]
Substituting \( \sin(\omega t) = \frac{1}{3} \):
\[ \left(\frac{1}{3}\right)^2 + \cos^2(\omega t) = 1 \]
\[ \frac{1}{9} + \cos^2(\omega t) = 1 \]
\[ \cos^2(\omega t) = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus, we have:
\[ \cos(\omega t) = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} \]

### Step 4: Set the condition for the second particle
For the second particle, at the same time \( t \), we have:
\[ x_2 = \frac{A}{3} \]
This gives us:
\[ \frac{A}{3} = A \sin(\omega t + \theta) \]
Simplifying, we find:
\[ \sin(\omega t + \theta) = \frac{1}{3} \]

### Step 5: Use the sine addition formula
Using the sine addition formula:
\[ \sin(\omega t + \theta) = \sin(\omega t) \cos(\theta) + \cos(\omega t) \sin(\theta) \]
Substituting the known values:
\[ \frac{1}{3} = \left(\frac{1}{3}\right) \cos(\theta) + \left(\frac{\sqrt{8}}{3}\right) \sin(\theta) \]
Multiplying through by 3 to eliminate the denominator:
\[ 1 = \cos(\theta) + \sqrt{8} \sin(\theta) \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ \cos(\theta) + \sqrt{8} \sin(\theta) = 1 \]
Now, we can isolate the terms:
\[ \cos(\theta) = 1 - \sqrt{8} \sin(\theta) \]

### Step 7: Square both sides
Squaring both sides leads to:
\[ \cos^2(\theta) = (1 - \sqrt{8} \sin(\theta))^2 \]
Expanding the right-hand side:
\[ \cos^2(\theta) = 1 - 2\sqrt{8} \sin(\theta) + 8 \sin^2(\theta) \]
Using the identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \):
\[ \cos^2(\theta) = 1 - 2\sqrt{8} \sin(\theta) + 8(1 - \cos^2(\theta)) \]
This simplifies to:
\[ \cos^2(\theta) + 8\cos^2(\theta) = 1 - 2\sqrt{8} \sin(\theta) + 8 \]
\[ 9\cos^2(\theta) = 9 - 2\sqrt{8} \sin(\theta) \]

### Step 8: Solve the quadratic equation
Rearranging gives:
\[ 9\cos^2(\theta) + 2\sqrt{8} \sin(\theta) - 9 = 0 \]
This is a quadratic equation in terms of \( \cos(\theta) \).

### Step 9: Find solutions for \( \cos(\theta) \)
Solving this quadratic equation will yield values for \( \cos(\theta) \). The possible values are \( \cos(\theta) = 1 \) or \( \cos(\theta) = -\frac{7}{9} \). The first case is not possible since it would imply \( \theta = 0 \).

Thus, we take:
\[ \cos(\theta) = -\frac{7}{9} \]
Calculating \( \theta \) gives:
\[ \theta = \cos^{-1}\left(-\frac{7}{9}\right) \]
This results in an approximate phase difference of \( 141.4^\circ \) or \( \frac{2\pi}{3} \) radians.

### Final Answer
The phase difference \( \theta \) between the two particles is approximately \( \frac{2\pi}{3} \) radians.

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