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Chapter 10: Problem 1475

Two sound waves are represented by $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\( and \)\mathrm{y}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$. The phase difference between the waves in water is \(\ldots \ldots \ldots\) (A) \((\pi / 2)\) (B) \((\pi / 4)\) (C) \(\pi\) (D) \((3 \pi / 4)\)

Short Answer

Expert verified
The phase difference between the two sound waves in water is \(\boxed{\Delta \phi = \frac{\pi}{2}}\). The correct answer is (A) \((\pi / 2)\).

Step by step solution

01

Identify the wave equations

The given waves are represented by the equations: 1. \(y_1 = a \sin(\omega t - kx)\) 2. \(y_2 = a \cos(\omega t - kx)\)
02

Find the phase of each wave

The phase of a wave is the term inside the sine or cosine function, in our case it is: 1. \(\phi_1 = \omega t - kx\) 2. \(\phi_2 = \omega t - kx\) Notice that the phase for both waves is the same.
03

Find the phase difference

The phase difference between two waves can be represented as \(\Delta \phi = \phi_2 - \phi_1\). Here: \(\Delta \phi = (\omega t - kx) - (\omega t - kx)\) Since the phases of both waves are the same, their difference is: \(\Delta \phi = 0\)
04

Relating the sine and cosine functions with phase difference

We know that: \(\sin(x) = \cos(\frac{\pi}{2} - x)\) So we can rewrite \(y_2\) in terms of the sine function: \(y_2 = a \cos(\omega t - kx) = a \sin(\frac{\pi}{2} - (\omega t - kx))\)
05

Determine the phase difference

Now we can see that the phase difference between the two waves is: \(\Delta \phi = \frac{\pi}{2} - (\omega t - kx) - (\omega t - kx)\) Since \(\omega t - kx\) cancels out: \(\Delta \phi = \frac{\pi}{2}\) So the phase difference between the two waves in water is: \(\boxed{\Delta \phi = \frac{\pi}{2}}\) The correct answer is (A) \((\pi / 2)\).

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