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Chapter 10: Problem 1438

Two waves are represented by $\mathrm{y}_{1}=\mathrm{A} \sin \omega \mathrm{t}\( and \)\mathrm{y}_{2}=\mathrm{A}\( cos \)\omega \mathrm{t}$. The phase of the first wave, \(\mathrm{w}\). r. t. to the second wave is (A) more by radian (B) less by \(\pi\) radian (C) more by \(\pi / 2\) (D) less by \(\pi / 2\)

Short Answer

Expert verified
The phase of the first wave relative to the second wave is less by \(\frac{\pi}{2}\) radians. Hence, the correct answer is (D).

Step by step solution

01

Recall the relationship between sine and cosine functions

The sine and cosine functions are related by the following identity: \[\sin(x) = \cos\left( x - \frac{\pi}{2} \right)\] This means that the sine function is simply a shifted version of the cosine function by a phase of \(\frac{\pi}{2}\).
02

Compare the given waves

We have the two waves, \(y_1 = A\sin(\omega t)\) and \(y_2 = A\cos(\omega t)\). We can rewrite \(y_1\) using the identity mentioned in Step 1: \[y_1 = A\sin(\omega t) = A\cos\left(\omega t - \frac{\pi}{2}\right)\]
03

Determine the phase difference

Comparing the rewritten equation of \(y_1\) with that of \(y_2\), we can see that the first wave has a phase that is less than the second wave by \(\frac{\pi}{2}\) radians. Therefore, the correct answer is (D) less by \(\frac{\pi}{2}\)

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