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Chapter 10: Problem 1515

When two sound waves travel in the same direction in a medium, the displacement of a particle located at \(\mathrm{x}\) at time \(\mathrm{t}\) is given by \(\mathrm{y}_{1}=0.05 \cos (0.50 \mathrm{px}-100 \mathrm{pt}) \&\) \(\mathrm{y}_{2}=0.05 \cos (0.46 \mathrm{px}-92 \mathrm{pt})\), where \(\mathrm{y}_{1}, \mathrm{y}_{2}\) and \(\mathrm{x}\) are in meter and \(t\) is in seconds. What is the speed of sound in the medium ? (A) \(332 \mathrm{~m} / \mathrm{s}\) (B) \(100 \mathrm{~m} / \mathrm{s}\) (C) \(92 \mathrm{~m} / \mathrm{s}\) (D) \(200 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The speed of sound in the medium is \(92 \:\text{m/s}\) (option C).

Step by step solution

01

Write down the given equations

We are given the equations: \(y_1 = 0.05 \cos(0.50x - 100t)\) \(y_2 = 0.05 \cos(0.46x - 92t)\)
02

Identify the phase velocity terms

The term inside the cosine brackets represents the phase of the wave. So, we identify the phase velocities for both waves: For \(y_1\), the phase is \(0.50x - 100t\) For \(y_2\), the phase is \(0.46x - 92t\)
03

Extract the coefficients of x and t and find the ratio

From the phases of both waves, extract the coefficients of x and t: \(Y_1 = 0.05 \cos(0.50x - 100t)\) has coefficients 0.50 for x and -100 for t. \(Y_2 = 0.05 \cos(0.46x - 92t)\) has coefficients 0.46 for x and -92 for t. Now, we find the ratio of coefficients for both waves: \(\frac{0.50}{-100} = \frac{0.46}{-92}\)
04

Solve for the speed of sound

Now, we solve the equation found in step 3, to get the speed of sound, v: \(\frac{0.50}{-100} = \frac{0.46}{-92}\) Cross-multiplying, we get: \(-100 \times 0.46 = -92 \times 0.50\) Divide both sides by -1 and simplify: \(100 \times 0.46 = 92 \times 0.50\) Divide both sides by 0.5: \(100 \times 0.92 = 92\) Since both waves are traveling in the same medium, their average speed is the speed of sound in the medium: \(v = 0.92 \times 100 = 92 \:\text{m/s}\) Therefore, the speed of sound in the medium is 92 m/s (option C).

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