Question

The phase difference between two waves, represented by $y_1=10^{-6} \sin[100t +(x/50)+0.5]\,m$ $y_2=10^{-6} \cos[100t +(x/50)]\,m$ where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately.

• A. 1.07 radians
• B. 2.07 radians
• C. 0.5 radians
• D. 1.5 radians

Answer 1

The Correct Option is A

To find the phase difference between the two waves represented by the functions:

1. Identify the forms of the given wave equations. The first wave is: y_1 = 10^{-6} \sin\left[100t + \frac{x}{50} + 0.5\right]\,m
2. The second wave is: y_2 = 10^{-6} \cos\left[100t + \frac{x}{50}\right]\,m
3. Convert the cosine wave to a sine form to directly compare the arguments, using the identity: \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)
4. Therefore, rewrite the second wave's equation as: y_2 = 10^{-6} \sin\left[100t + \frac{x}{50} + \frac{\pi}{2}\right]\,m
5. Compare the arguments of the two sine functions:
   • For y_1: 100t + \frac{x}{50} + 0.5
   • For y_2: 100t + \frac{x}{50} + \frac{\pi}{2}
6. The phase difference between the two waves is: \left(\frac{\pi}{2} - 0.5\right)
7. Calculating the phase difference:
   \frac{\pi}{2} \approx 1.57 radians
   1.57 - 0.5 = 1.07 radians

Thus, the phase difference between the two waves is approximately 1.07 radians.