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Prove that sound waves propagate with a speed given by Equation \(17.1 .\) Proceed as follows. In Figure \(17.3,\) consider a thin cylindrical layer of air in the cylinder, with face area \(A\) and thickness \(\Delta x\). Draw a free-body diagram of this thin layer. Show that \(\Sigma F_{x}=m a_{x}\) implies that \(-[\partial(\Delta P) / \partial x] A \Delta x=\) \(\rho A \Delta x\left(\partial^{2} s / \partial t^{2}\right) .\) By substituting \(\Delta P=-B(\partial s / \partial x),\) obtain the wave equation for sound, \((B / \rho)\left(\partial^{2} s / \partial x^{2}\right)=\left(\partial^{2} s / \partial t^{2}\right) .\) To a mathematical physicist, this equation demonstrates the existence of sound waves and determines their speed. As a physics student, you must take another step or two. Substitute into the wave equation the trial solution \(s(x, l)=\) \(s_{\max } \cos (k x-\omega t) .\) Show that this function satisfies the wave equation provided that \(\omega / k=\sqrt{B / \rho} .\) This result reveals that sound waves exist provided that they move with the speed \(v=f \lambda=(2 \pi f)(\lambda / 2 \pi)=\omega / k=\sqrt{B / \rho}\)

Step by step solution

01

Understanding given

According to the given, a cylindrical layer of air is being considered with face area \(A\) and thickness \(\Delta x\). We have to first draw a free-body diagram of this layer.

02

Applying Newton's second law

According to Newton's second law, the net force on an object is equal to its mass times its acceleration. Mathematically, it can be written as \(\Sigma F_{x}=m a_{x}\). From this, we can conclude that \(-[\partial(\Delta P) / \partial x] A \Delta x=\rho A \Delta x\left(\partial^{2} s / \partial t^{2}\right)\). Here, \(\Delta P\) denotes the change in pressure, \(\rho\) is the density and \(s\) is displacement.

03

Obtain the wave equation for sound

Next, we substitute the given \(\Delta P=-B(\partial s / \partial x)\) into the obtained equation. Here, \(B\) is the bulk modulus. After the substitution, we get \((B / \rho)\left(\partial^{2} s / \partial x^{2}\right)=\left(\partial^{2} s / \partial t^{2}\right)\). This equation is the wave equation for sound.

04

Substitute the trial solution

Now we substitute the trial solution \(s(x, l)=s_{\max } \cos (k x-\omega t)\) into the wave equation and show that this function satisfies it. Here, \(s_{\max }\) is the maximum displacement, \(k\) is the wavelength, and \(\omega\) is the angular frequency.

05

Show the existence and speed of sound waves

Lastly, we need to prove that sound waves exist and they move with the speed \(v=f \lambda=(2 \pi f)(\lambda / 2 \pi)=\omega / k=\sqrt{B / \rho}\). By substituting the given values and simplifying, we can conclude that \(\omega / k=\sqrt{B / \rho}\) and thus prove the existence and speed of sound waves.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law

Newton's Second Law is a fundamental principle in physics. It states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it's stated as \( \Sigma F = ma \). In the context of sound waves, we analyze a thin cylindrical layer of air within a cylinder. By applying Newton's Second Law to this layer, we find the relationship between pressure change, density, and acceleration.

Consider the net force in the x-direction on the layer: \(-\frac{\partial(\Delta P)}{\partial x} A \Delta x = \rho A \Delta x \frac{\partial^2 s}{\partial t^2} \). This equation tells us how changes in pressure create acceleration in the particles of the medium, leading to the propagation of sound waves.

Wave Equation

The wave equation describes the behavior of waves and is crucial in understanding sound propagation. By substituting \( \Delta P = -B \frac{\partial s}{\partial x} \) into our force equation, we derive the standard wave equation for sound:

\[ \frac{B}{\rho} \frac{\partial^2 s}{\partial x^2} = \frac{\partial^2 s}{\partial t^2} \]

Here, \( B \) is the bulk modulus, a measure of the medium's resistance to compression, and \( \rho \) is its density. This equation shows how the mechanical properties of the medium affect the wave's speed and behavior. It highlights the interplay between space and time derivatives of displacement, fundamentally linking wave mechanics to sound propagation.

Trial Solution Method

The trial solution method helps us test if a particular form of a solution satisfies the wave equation. In this context, we use a cosine function \( s(x, t) = s_{\max} \cos(kx - \omega t) \) as our trial solution for sound waves.

By substituting this trial solution into the wave equation, we verify its validity provided the condition \( \omega / k = \sqrt{B / \rho} \) holds.

• \( s_{\max} \): Maximum displacement
• \( k \): Wave number, related to wavelength by \( k = 2\pi/\lambda \)
• \( \omega \): Angular frequency, related to the period by \( \omega = 2\pi f \)

This trial solution shows how wave parameters must align with medium properties for sound waves to propagate effectively.

Sound Wave Speed

Sound wave speed refers to how fast disturbances travel through a medium. From our derived wave equation and trial solution, we find that the speed of sound \( v \) is given by \( v = \sqrt{B/\rho} \).

This result emerges as we demonstrated that \( \omega / k = \sqrt{B/\rho} \). Here,

• \( v = f \lambda \)
• \( v = \omega / k \)
• \( \omega = \sqrt{B / \rho} k \)

Consequently, the speed depends on the medium's bulk modulus and density. High bulk modulus or low density leads to faster propagation. Understanding sound speed is vital in applications like acoustics and audio technology, where sound propagation is key.