AIPMT PRELIMS 2004 Physics Phase Difference MCQ Question

To find the phase difference between the two waves given by

• $y_1 = 10^{-6} \sin\left(100t + \frac{x}{50} + 0.5\right)$
• $y_2 = 10^{-6} \cos\left(100t + \frac{x}{50}\right)$,

we start by expressing both waves in a similar form, focusing on their arguments.

Step 1: Identify the Phase of Each Wave

1. For $y_1$: The wave $y_1$ is in the sine form, and its phase is: $\phi_1 = 100t + \frac{x}{50} + 0.5$

2. For $y_2$: The wave $y_2$ is in the cosine form, and its phase is: $\phi_2 = 100t + \frac{x}{50}$

Step 2: Calculate the Phase Difference

The phase difference $\Delta \phi$ between two waves is given by the difference of their phases: $\Delta \phi = \phi_1 - \phi_2$

Substituting the values of $\phi_1$ and $\phi_2$: $\Delta \phi = \left(100t + \frac{x}{50} + 0.5\right) - \left(100t + \frac{x}{50}\right)$

Simplifying this expression: $\Delta \phi = 0.5$

Step 3: Convert Phase Difference to Radians

The phase difference calculated above is in radians. Since $0.5$ is already a value in radians, we can directly consider it as the phase difference.

Step 4: Consider the Options

Now, we need to compare with the given options:

• A) 0.207 radians
• B) 0.5 radians
• C) 1.5 radians
• D) 1.07 radians

The calculated phase difference of $0.5$ radians directly matches option B, but it is stated that the correct answer is D (1.07 radians). This indicates a need for further analysis.

Step 5: Re-evaluate the Phase Difference

Since the question states that the correct answer is D (1.07 radians), let’s check the phase of the cosine function more closely. The cosine function can be rewritten in terms of sine:

$$\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$$

Thus, we can rewrite $y_2$ as:

$$y_2 = 10^{-6} \sin\left(100t + \frac{x}{50} + \frac{\pi}{2}\right)$$

Now, the phase for $y_2$ becomes: $\phi_2' = 100t + \frac{x}{50} + \frac{\pi}{2}$

Now, recalculating the phase difference: $\Delta \phi = \phi_1 - \phi_2' = \left(100t + \frac{x}{50} + 0.5\right) - \left(100t + \frac{x}{50} + \frac{\pi}{2}\right)$

This simplifies to: $\Delta \phi = 0.5 - \frac{\pi}{2}$

Given $\frac{\pi}{2} \approx 1.57$, we have: $\Delta \phi = 0.5 - 1.57 \approx -1.07$

Since phase differences can be expressed in positive terms (taking the absolute value), we find: $|\Delta \phi| = 1.07 \text{ radians}$

Conclusion

The correct answer is D (1.07 radians), as derived from the sine and cosine relationship. The other options are incorrect because they do not reflect the accurate calculation of the phase difference when accounting for the transformation from cosine to sine.

Summary

• The phase difference $\Delta \phi$ between the two waves is approximately $1.07$ radians.
• Option B is misleading, as it only considers the sine form without accounting for the phase shift introduced by the cosine function. Hence, the correct answer is D (1.07 radians).