Question

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

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

Step-by-step solution

Write down the given equations

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

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 \(0.46x - 92t\)

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: \(\frac{0.50}{-100} = \frac{0.46}{-92}\)

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).

Key concepts

Phase Velocity

Phase velocity refers to the speed at which a wave phase propagates through space. This is a crucial concept when studying waves, as it helps us understand how quickly the crests or troughs of a wave move from one point to another in a medium. For a mathematical representation, phase velocity \( v_p \) is given by the formula \( v_p = \frac{\omega}{k} \), where \( \omega \) is the angular frequency and \( k \) is the wave number. These parameters describe how the wave oscillates and propagates. In the context of sound waves, as described in the original exercise, these phase velocities can be found directly from the wave equations. For instance, a sound wave represented by \( y_1 = 0.05 \cos(0.50x - 100t) \) has a phase velocity where \( \omega = 100 \) rad/s and \( k = 0.50 \) rad/m, leading to a phase velocity \( v_{p1} = \frac{100}{0.50} = 200 \, \text{m/s} \). Such calculations help in analyzing how sound travels through different media.

Wave Equation

The wave equation is a fundamental equation used to describe the behavior of waves in various mediums. It is expressed in its simplest form as:\[\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}\]This equation states that the acceleration of wave displacement \( y \) at any point is proportional to the curvature of the wave at that point, scaled by the square of the wave speed \( v \). In practical terms, this equation helps us understand how waves like sound waves, water ripples, and electromagnetic waves propagate.In our exercise, the wave equations \( y_1 = 0.05 \cos(0.50x - 100t) \) and \( y_2 = 0.05 \cos(0.46x - 92t) \) showcase two waves traveling in the same direction. By comparing the coefficients of \( x \) and \( t \), we can derive important properties such as phase velocity and speed of sound in the medium.

Coefficient Comparison

Coefficient comparison is a technique used to analyze equations or expressions by comparing the numerical factors influencing each variable. In the context of wave equations, this involves examining the coefficients multiplying \( x \) and \( t \) in the phases of the given waves.For the specific sound waves in the problem, their equations \( y_1 = 0.05 \cos(0.50x - 100t) \) and \( y_2 = 0.05 \cos(0.46x - 92t) \) have given coefficients:

• For \( y_1 \): Coefficient of \( x \) is 0.50, and coefficient of \( t \) is -100.
• For \( y_2 \): Coefficient of \( x \) is 0.46, and coefficient of \( t \) is -92.

By setting the ratios \( \frac{0.50}{-100} \) and \( \frac{0.46}{-92} \) equal, we deduce properties about how these waves relate, specifically that they travel at nearly equivalent speeds in the same medium. This procedure of comparing coefficients provides insights necessary for solving complex wave problems.

Sound Wave Interference

Sound wave interference occurs when two or more sound waves overlap and combine, forming a new wave pattern. This principle is based on the superposition of waves, where the resultant wave displacement is the sum of the individual displacements at any point in the medium.When sound waves like \( y_1 = 0.05 \cos(0.50x - 100t) \) and \( y_2 = 0.05 \cos(0.46x - 92t) \) travel through the same medium, they might interfere constructively, reinforcing each other, or destructively, cancelling each other out. This interference can lead to patterns of nodes and antinodes, affecting the overall sound heard, such as beats in musical acoustics.By understanding interference, we gain insights into wave behavior in acoustics and solve problems related to sound intensities and wave interactions effectively.