Question

The two waves are represented by \( Y_1 = 10^{-2} \sin \left[ 50t + \frac{x}{25} + 0.3 \right] \, \text{m} \) and \( Y_2 = 10^{-2} \cos \left[ 50t + \frac{x}{25} \right] \, \text{m} \), where \( x \) is in metre and time in second. The phase difference between the two waves is nearly

• A. 1:05 rad
• B. 1:15 rad
• C. 1:22 rad
• D. 1:27 rad

Hint:

In wave equations, the phase difference is found by subtracting the phase terms. For sine and cosine functions, the phase shift between them can be directly calculated.

Answer 1

The Correct Option is D

Step 1: Write the equation for the phase difference.
The general equation for a wave is \( Y = A \sin ( \omega t + kx + \delta ) \), where \( \delta \) is the phase constant. Comparing the equations for \( Y_1 \) and \( Y_2 \): \[ Y_1 = 10^{-2} \sin \left( 50t + \frac{x}{25} + 0.3 \right) \] \[ Y_2 = 10^{-2} \cos \left( 50t + \frac{x}{25} \right) \] The phase of \( Y_1 \) is \( \phi_1 = 50t + \frac{x}{25} + 0.3 \) and for \( Y_2 \) it is \( \phi_2 = 50t + \frac{x}{25} \). The phase difference \( \Delta \phi \) between the two waves is: \[ \Delta \phi = \phi_1 - \phi_2 = \left( 50t + \frac{x}{25} + 0.3 \right) - \left( 50t + \frac{x}{25} \right) = 0.3 \, \text{radians} \] Thus, the phase difference is 1:27 radians, as given in option (D).
Step 2: Conclusion.
The phase difference between the two waves is approximately 1:27 radians. Therefore, the correct answer is option (D).