Two waves are propagating to the point P along a straight line produced by two sources A and B of simple harmonic and of equal frequency. The amplitude of every wave at P is `a` and the phase of A is ahead by `pi//3` than that of B and the distance AP is greater than BP by `50 cm`. Then the resultant amplitude at the point P will be if the wavelength `1` meter

Options

A. (a)2a
B. (b)`asqrt3`
C. (c)`asqrt2`
D. (d)a

Answer

To solve the problem step by step, we will follow the principles of wave interference and superposition.

### Step 1: Understand the Given Information
We have two waves from sources A and B, both with the same amplitude \( a \) and frequency. The phase of wave A is ahead of wave B by \( \frac{\pi}{3} \). The distance from A to point P (AP) is greater than the distance from B to P (BP) by 50 cm, and the wavelength \( \lambda \) is 1 meter.

### Step 2: Convert Distance to Meters
Given that the distance difference \( \Delta x = AP - BP = 50 \, \text{cm} = 0.5 \, \text{m} \).

### Step 3: Calculate the Phase Difference
The phase difference \( \Delta \phi \) due to the path difference can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the values:
\[ \Delta \phi = \frac{2\pi}{1} \times 0.5 = \pi \]

### Step 4: Total Phase Difference
The total phase difference between the two waves at point P is given by:
\[ \text{Total Phase Difference} = \Delta \phi + \text{Phase of A} - \text{Phase of B} \]
Since the phase of A is \( \frac{\pi}{3} \) ahead of B, we have:
\[ \text{Total Phase Difference} = \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 5: Calculate the Resultant Amplitude
The resultant amplitude \( A_r \) of two waves can be calculated using the formula:
\[ A_r = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos(\text{Total Phase Difference})} \]
Substituting the known values:
\[ A_r = \sqrt{A^2 + A^2 + 2A^2 \cos\left(\frac{2\pi}{3}\right)} \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ A_r = \sqrt{A^2 + A^2 - A^2} = \sqrt{A^2} = A \]

### Final Result
The resultant amplitude at point P is:
\[ A_r = A \]