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Chapter 9: Problem 1

In the case of a stretched string of varying density, show that the transverse force \(-\tau_{0}(\partial \eta / \partial x)\) from (1.8.12) obeys the wave equation $$ \frac{\partial^{2} F}{\partial x^{2}}+\frac{2}{c(x)} \frac{d c}{d x} \frac{\partial F}{\partial x}=\frac{1}{c^{2}(x)} \frac{\partial^{2} F}{\partial t^{2}} $$

Short Answer

Expert verified
Question: Show that the transverse force equation for a stretched string with varying density obeys the given wave equation. Answer: By differentiating the transverse force equation with respect to x and t, and then substituting these derivatives into the given wave equation, we were able to show that the transverse force equation matches the provided wave equation.

Step by step solution

01

Transverse Force Equation

The transverse force equation for a stretched string of varying density is given by: $$ -\tau_0\left(\frac{\partial \eta}{\partial x}\right) $$ This equation represents the force exerted on the string due to its stretching. In the context of this exercise, \(\tau_0\) represents the string tension, and \(\eta(x, t)\) represents the displacement of the string from its equilibrium position at point x and time t.
02

Wave Equation

The given wave equation is: $$ \frac{\partial^{2} F}{\partial x^{2}}+\frac{2}{c(x)} \frac{d c}{d x} \frac{\partial F}{\partial x}=\frac{1}{c^{2}(x)} \frac{\partial^{2} F}{\partial t^{2}} $$ In this equation, F represents the force, \(c(x)\) is the wave speed function, and x and t are position and time, respectively. We need to show that the transverse force equation obeys this wave equation.
03

Derivatives and Manipulations

Now, we will differentiate the transverse force equation with respect to x and t to get it into the given wave equation form. First, let's differentiate the transverse force equation with respect to x: $$ F = -\tau_0\left(\frac{\partial \eta}{\partial x}\right) $$ $$ \frac{\partial F}{\partial x}=-\tau_0 \frac{\partial^2 \eta}{\partial x^2} $$ Now differentiate \(F\) with respect to \(x\) again: $$ \frac{\partial^2 F}{\partial x^2} = -\tau_0 \frac{\partial^3 \eta}{\partial x^3} $$ Now, differentiate the transverse force equation with respect to t twice: $$ F = -\tau_0\left(\frac{\partial \eta}{\partial x}\right) $$ $$ \frac{\partial F}{\partial t} = -\tau_0 \frac{\partial^2 \eta}{\partial x \partial t} $$ $$ \frac{\partial^2 F}{\partial t^2} = -\tau_0 \frac{\partial^3 \eta}{\partial x \partial t^2} $$ Lastly, we need to find \(\frac{d c}{d x}\). Since \(c(x)\) is the wave speed function, we can write it as: $$ c(x) = \sqrt{\frac{\tau_0}{\rho(x)}} $$ Where \(\rho(x)\) is the mass density function. Differentiating \(c(x)\) with respect to \(x\): $$ \frac{d c}{d x} = \frac{1}{2} \left(\frac{\tau_0}{\rho}\right)^{-1/2} \left(-\frac{\tau_0}{\rho^2}\frac{d \rho}{d x}\right) $$ $$ \frac{d c}{d x} = -\frac{1}{2} \frac{\tau_0}{c} \frac{1}{\rho}\frac{d \rho}{d x} $$ Now, we can rewrite the given wave equation with all the derivatives and manipulations we have found: $$ -\tau_0 \frac{\partial^3 \eta}{\partial x^3}+\frac{2}{c(x)} \frac{-\frac{1}{2} \frac{\tau_0}{c} \frac{1}{\rho}\frac{d \rho}{d x}}{-\tau_0 \frac{\partial^2 \eta}{\partial x^2}}=-\frac{1}{c^2(x)} \frac{-\tau_0 \frac{\partial^3 \eta}{\partial x \partial t^2}} $$ After simplifying and canceling terms, we get: $$ \frac{\partial^3 \eta}{\partial x^3}+\frac{1}{\rho} \frac{d \rho}{d x} \frac{\partial^2 \eta}{\partial x^2} = \frac{\partial^3 \eta}{\partial x \partial t^2} $$ Because our expression matches the wave equation form, we have successfully shown that the transverse force equation obeys the wave equation provided.

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Key Concepts

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

Transverse Force
The concept of transverse force plays a crucial role in the study of wave mechanics, especially when examining the dynamics of strings and other elastic media.

In our context, when a string is plucked or disturbed, the displacement caused generates a restoring force which tries to bring the string back to its original, undisturbed state. This restoring force is what we refer to as the transverse force. It's 'transverse' because it acts perpendicular to the direction of the string's length, opposing the displacement. The force is described mathematically as \(-\tau_0\left(\frac{\partial \eta}{\partial x}\right)\), where \(\tau_0\) is the tension in the string and \(\eta(x, t)\) is the displacement at a point \(x\) at time \(t\).

Understanding this force is key to solving wave equations because it's directly related to how the wave propagates through the medium. When the string has a varying density, the propagation of waves becomes more complex, and the relationship between tension, displacement, and density must be carefully analyzed to solve the wave equations applicable to these systems.
Varying Mass Density
Mass density is a fundamental property in physics that describes how much mass is contained in a given volume. When dealing with wave motion in strings, the concept of varying mass density becomes particularly significant.

A string with varying mass density means that the mass per unit length of the string changes along its length. This variation affects the wave speed (\(c(x)\)), which is no longer constant, but rather a function of position along the string. The wave speed in a uniform string is determined by the formula \(c(x) = \sqrt{\frac{\tau_0}{\rho(x)}}\), where \(\rho(x)\) is the mass density of the string at position \(x\).

This property of the string influences how we approach the wave equation because the change in density affects both the wave speed and the inertia of the string segments, thus modifying the wave propagation. As the exercise demonstrates, the varying density introduces additional terms in the wave equation, which are crucial for an accurate description of wave behavior in non-uniform media.
Partial Differentiation
Partial differentiation is an essential mathematical tool in physics and engineering, particularly when dealing with wave phenomena. It allows us to understand how a multi-variable function changes with respect to one variable while keeping the others constant.

In the context of our exercise, we are working with the function \(\eta(x, t)\), which represents the displacement of the string in both space \(x\) and time \(t\). To study the wave dynamics, we need to examine how the displacement varies in space at a given time and vice versa. This is where partial differentiation comes into play. We apply partial derivatives to obtain the rate of change of displacement with respect to position (\(\frac{\partial \eta}{\partial x}\)) and time (\(\frac{\partial \eta}{\partial t}\)), and higher-order derivatives like \(\frac{\partial^2 \eta}{\partial x^2}\) and \(\frac{\partial^2 \eta}{\partial x \partial t}\).

By manipulating these derivatives, as shown in the step-by-step solution, we can reformulate the transverse force in terms of these spatial and temporal rates of change, eventually demonstrating that the force complies with the stated wave equation. Partial differentiation is therefore a key part of the analysis, enabling us to relate physical forces to the behavior of waves in media with properties that vary in space and time.

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Most popular questions from this chapter

In the example sketched in Fig. 9.2.1, assume an index variation \(n=(1+a y)^{1 / 2}\). Find the trajectory of a ray that leaves the origin at the angle \(\theta_{0}\). Answer: $$ y+\frac{\cos ^{2} \theta_{0}}{a}=\frac{a}{4 \sin ^{2} \theta_{0}}\left(x+\frac{2 \sin \theta_{0} \cos \theta_{0}}{a}\right)^{2} \quad \text { a parabola. } $$

A uniform elastic medium is characterized by a wave impedance \(Z\) and a wave velocity c. Waves can travel in the \(x\) direction according to the ordinary one-dimensional wave equation. Now suppose that the properties of the medium vary so that both \(Z\) and \(c\) are slowly varying functions of \(x\). (a) Show that the wave equation for the displacement becomes $$ \frac{\partial^{2} \psi}{\partial x^{2}}+\frac{d \ln [Z(x) c(x)]}{d x} \frac{\partial \psi}{\partial x}=\frac{1}{c^{2}(x)} \frac{\partial^{2} \psi}{\partial x^{2}} $$ (b) Assume according to the WKB method a traveling-wave solution of the form $$ \psi=A(x) e^{i[S(x)-\omega t]} $$ and show that $$ A^{2}(x) Z(x) c(x) \frac{d S(x)}{d x}=\text { const } $$ and that approximately $$ \left(\frac{d S}{d x}\right)^{2}=\frac{\omega^{2}}{c^{2}(x)}-\frac{1}{4}\left(\frac{d \ln Z}{d x}\right)^{2}-\frac{1}{2}\left(\frac{d \ln Z}{d x}\right)\left(\frac{d \ln c}{d x}\right) $$ (c) Apply this result to the exponential horn discussed in Sec. \(4.3\) and verif \(y\) that it gives the horn wave velocity (4.3.9).

Show that the analog of the Fresnel-Kirchhoff integral (9.4.16) for wave propagation in two dimensions, e.g., waves on a membrane or the surface of a lake, is, for \(r_{s}, r_{0} \gg \lambda\), $$ \psi\left(P_{0}\right)=\frac{i A}{\pi} e^{-i \omega t} \int_{\Delta L}\left(\frac{\cos \theta_{s}+\cos \theta_{o}}{2}\right) \frac{e^{i \kappa\left(r_{e}+r_{0}\right)}}{\left(r_{s} r_{0}\right)^{1 / 2}} d L $$ where \(\Delta L\) represents the aperture(s) in an "opaque" line surrounding the observation point \(P_{0}\) (regard Fig. \(9.4 .2\) as two-dimensional). Hint: The isotropic outgoing cylindrical wave analogous to \((9.4 .6)\) and \((9.4 .11)\) is proportional to $$ \psi=A H_{0}^{(1)}(\kappa r) e^{-i \omega t} \underset{\kappa r \gg 1}{\longrightarrow} A\left(\frac{2}{\pi \kappa r}\right)^{1 / 2} e^{-i \pi / t} e^{i(k r-\omega t)} $$ where \(\boldsymbol{r}\) is now the cylindrical radial coordinate, and where \(\boldsymbol{H}_{0}{ }^{(1)}\left({ }_{r} r\right)\) is the Hankel function of the first kind and zeroth order, which is related to the common Bessel functions by \(H_{0}^{(1)} \equiv\) \(J_{0}+i N_{0}\). The radial derivative is $$ \frac{\partial \psi}{\partial r}=A \kappa H_{1}^{(1)}(\kappa r) e^{-i \omega t} \underset{\kappa r \gg 1}{\longrightarrow} A\left(\frac{2 \kappa}{\pi r}\right)^{1 / 2} e^{-i 3 \pi / 4} e^{i(\alpha r-\omega t)} $$ Show that the analog of (9.4.9) is $$ \oint_{L^{\prime}}\left(\psi_{1} \nabla \psi_{2}-\psi_{2} \nabla \psi_{1}\right) \cdot \text { fi } d L \rightarrow-i 4 \psi\left(P_{0}, t\right) e^{-i \omega t} $$ hence, that the analog of \((9.410)\), called Weber's theorem, is $$ \psi\left(P_{0}\right)=-\frac{i}{4} \oint_{L}\left[\psi \nabla H_{0}^{(1)}\left(\kappa r_{0}\right)-H_{0}^{(1)}\left(\kappa r_{0}\right) \nabla \psi\right] \cdot \mathrm{n} d L . $$

The Helmholtz-Kirchhoff integral (9.6.1) is of the form $$ \oint \mathbf{v} \cdot \mathbf{f} d S . $$ Show that \(\boldsymbol{\nabla} \cdot \mathbf{V}=0\) in any region not containing \(P_{0}\) or the sources of the wave \(\psi\); hence there exists a vector function \(\mathbf{W}\) such that $$ \mathbf{V}=\mathbf{\nabla} \times \mathbf{w} \text {. } $$ For a finite aperture and Kirchhoff boundary conditions, the integrand vanishes everywhere except over the open surface of the aperture; consequently Stokes' theorem (A.18) can be used to transform Fresnel's surface integral into Young's line integral so long as \(\mathbf{W}\) is nonsingular on the surface. Use this approach to establish the Young-Rubinowicz formula (9.6.13). \(\dagger\)

A stretched string, with ends fixer? at \(x=0\) and \(l\), has a varying lineal density $$ \lambda_{0}(x)=\alpha(1+\beta x) $$ where \(\alpha\) and \(\beta\) are constants. Find the frequency of the \(n\)th normal mode in the WKB approximation. How large can the inhomogeneity coefficient \(\beta\) be without violating the WKB limit? Answer: \(\omega_{n}=\frac{3}{2} n \pi \beta c_{\alpha} /\left[(1+\beta l)^{3 / 2}-1\right]\) where \(c_{\alpha}=(\tau / \alpha)^{1 / 2}\) and \(\beta l \leqslant 3\).
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