Uncategorized: Practice Problem & Solution
The phase difference between two waves, represented by: $y_1 = 10^{-6} \sin{100t + (x/50) + 0.5} \text{m}$, $y_2 = 10^{-6} \cos{100t + (x/50)} \text{m}$ where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately: (2004)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Write $y_2$ as a sine function: $y_2 = 10^{-6} \sin(100t + x/50 + \pi/2)$. The phase difference is $\Delta\phi = \frac{\pi}{2} - 0.5 = 1.57 - 0.5 = 1.07 \text{ radians}$.
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