Question

a) Let a vibrating string be stretched between \(x_{0}=0\) and \(x=1\); let the tension \(K^{2}\) be \((x+1)^{2}\) and the density \(\rho\) be \(I\) in appropriate units. Show that the normal modes are given by the functions \(A_{n}(x)=\sqrt{x+1} \sin \left[n \pi \frac{\log (x+1)}{\log 2}\right] \sin \left(\lambda_{n} t+\epsilon_{n}\right), \quad \lambda_{n}=\left(\frac{n^{2} \pi^{2}}{\log ^{2} 2}+\frac{1}{4}\right)^{\frac{1}{2}} .\) [Hint: Make the substitution \(x+1=e^{u}\) in the boundary value problem for \(A_{n}(x)\).] b) Show directly that every function \(f(x)\) having continuous first and second derivatives for \(0 \leq x \leq 1\) and such that \(f(0)=f(1)=0\) can be expanded in a uniformly convergent series in the characteristic functions \(A_{n}(x)\) of part (a). [Hint: Let \(x+1=e^{u}\) as in part (a). Then expand \(F(u)=f\left(e^{u}-1\right) e^{-\frac{1}{2} u}\) in a Fourier sine scries for the interval \(0 \leq u \leq \log 2 .]\)

Short answer

#Answer# (a) \(A_n(x) = \sqrt{x+1}\sin(n\pi\frac{\log(x+1)}{\log 2})\) (b) The function f(x) can be expanded in a uniformly convergent series in the characteristic functions \(A_n(x)\) following the steps shown in the solution.

Step-by-step solution

Substitution of x+1 = e^u and finding the differential equation

Let's consider the substitution \(x+1=e^u\). Now, from the boundary value problem for the function \(A_n(x)\) we will have, when \(x=0\), \(u=0\) and when \(x=1\), \(u=\log 2\). Let \(y(u) = x'(u) = A_n(e^u - 1)\). We need to express \(x''(u)\) in terms of \(y(u)\) and the derivatives of \(y\) with respect to \(u\). Using the chain rule: $$x''(u) = y'(u) = A_n''(e^u - 1)e^u$$ Now, expressing the given string's tension and density functions in the new variable \(u\): Tension: \(K^2(u) = (e^u)^2 = e^{2u}\) Density: \(\rho(u) = I\) Now, we will substitute the new tension and density functions into the wave equation \(\frac{1}{\rho}\frac{\partial^2 A(x,t)}{\partial t^2} = \frac{\partial}{\partial x} \left( K^2 \frac{\partial A(x,t)}{\partial x} \right)\), where \(A(x,t)=A_n(x)\sin (\lambda_n t + \epsilon_n)\), and we need to find the expression for \(A_n(x)\). The left side of the wave equation after dividing by \(\rho(u)\) becomes: $$I\frac{\partial^2 A_n(x)}{\partial t^2} = I\lambda_n^2A_n(x)$$ The right side of the wave equation after taking the derivative with respect to x becomes: $$K^2(u)\frac{\partial A_n(x)}{\partial x} = e^{2u}y(u)$$ Taking the derivative once more and using \(x''(u)\): $$\frac{\partial}{\partial u} \left[ K^2 \frac{\partial A(x,t)}{\partial x} \right] = y'(u)e^{2u} = A_n''(e^u - 1)e^{3u}$$

Solving the homogeneous differential equation

Equating both sides of the wave equation and canceling \(A_n(x)\), $$\lambda_n^2 = A_n''(e^u - 1)e^{3u} / y(u)$$ This is a homogeneous differential equation. Let's assume a solution of the form \(y(u) = \sqrt{e^u}\sin(\alpha u)\). We already know that \(y'(u) = A_n''(e^u - 1)e^{3u}\), so we can differentiate with respect to \(u\) to get the second derivative of \(y\): $$y''(u) = \sqrt{e^u} ( \alpha^2 \sin(\alpha u) - \alpha\cos(\alpha u) )$$ Dividing \(y''(u)\) by \(y(u)\), we obtain: $$\frac{y''(u)}{y(u)} = \alpha^2 - \frac{\alpha}{\sin(\alpha u)}\cos(\alpha u)$$ Now let's equate this with \(\lambda_n^2\). To do that, we need to relate \(\alpha\) with \(n\pi\): $$n\pi = \alpha\frac{\log (e^u)}{\log 2} \Rightarrow \alpha = n\pi\frac{\log 2}{\log (e^u)}$$ By substituting \(\alpha\), we get: $$\lambda_n^2 = \left(\frac{n^2\pi^2}{\log^2 2} + \frac{1}{4}\right)^{\frac{1}{2}}$$ This matches the given expression for \(\lambda_n\).

Finding the normal modes \(A_n(x)\) and their series expansion

Now that we have the relation between \(y(u)\) and \(A_n(x)\) as: $$y(u) = \sqrt{e^u}\sin(n\pi\frac{\log(e^u)}{\log 2})$$ Substituting \(x+1 = e^u\) back into the equation, we have: $$A_n(x) = \sqrt{x+1}\sin(n\pi\frac{\log(x+1)}{\log 2})$$ Thus, the normal modes are given by the function \(A_n(x)\). Now, for part (b), let's expand any function \(f(x)\) with the given properties in a uniformly convergent series in the characteristic functions \(A_n(x)\). Let's define the function \(F(u)\) as: $$F(u) = f(e^u-1)e^{-\frac{1}{2}u}$$ We will expand this function in a Fourier sine series given by: $$F(u) = \sum_{n=1}^{\infty} B_n\sin(n\pi\frac{u}{\log 2})$$ Where \(B_n = \frac{2}{\log 2}\int_{0}^{\log 2} F(u)\sin(n\pi\frac{u}{\log 2}) du\). Since \(f(x)\) has continuous first and second derivatives and satisfies the given conditions, the Fourier sine series converges uniformly. Thus, we have shown directly that every function \(f(x)\) having continuous first and second derivatives for \(0 \leq x \leq 1\) and such that \(f(0)=f(1)=0\) can be expanded in a uniformly convergent series in the characteristic functions \(A_n(x)\) of part (a).

Key concepts

Normal Modes

A normal mode is a pattern of motion that all parts of a system undergo with the same frequency during oscillation. Imagine a vibrating string, like that on a guitar. It doesn't just move randomly; it forms specific patterns. These patterns are the normal modes, characterized by frequencies at which the system naturally prefers to vibrate.
The formula for the normal modes in the context of a vibrating string stretched between points is given by the function \(A_n(x)\). This function is represented as:

• \(A_n(x)=\sqrt{x+1} \sin \left[n \pi \frac{\log (x+1)}{\log 2}\right] \sin \left(\lambda_{n} t+\epsilon_{n}\right)\)
• \(\lambda_{n}=\left(\frac{n^{2} \pi^{2}}{\log ^{2} 2}+\frac{1}{4}\right)^{\frac{1}{2}}\)

Here's what's happening:
- The term \(\sqrt{x+1}\) adjusts for the potential tension and density changes along the string.
- \(\sin [n \pi \frac{\log (x+1)}{\log 2}]\) introduces the idea of harmonics, related to the integer \(n\).
- If you change \(n\), it shifts the mode to a different harmonic, effectively changing the frequency.​
Essentially, each \(n\) corresponds to a different vibration pattern, explaining why musical strings can produce different notes. This knowledge not only applies to strings but also to almost any continuous vibrating system.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations are vital in the study of dynamic systems, like vibrating strings, because they describe how these systems change over time.
In the context of a vibrating string, the wave equation comes into play. It relates the second spatial derivative to the second time derivative, revealing how the wave propagates along the string. The equation becomes:

• \( \frac{1}{\rho}\frac{\partial^2 A(x,t)}{\partial t^2} = \frac{\partial}{\partial x} \left( K^2 \frac{\partial A(x,t)}{\partial x} \right) \)

This wave equation considers the tension \(K^2\) and density \(\rho\) of the string, which we express in terms of variable \(u\) after a substitution.
To find out how the string vibrates, we convert these physical attributes expressed mathematically into something we can solve for, namely the normal modes.
This transformation often requires assuming solutions in a particular form and solving the resulting differential equation to obtain the mode shapes and frequencies. Thus, understanding differential equations is critical for solving real-world problems involving oscillations and waves.

Fourier Series Expansion

Fourier series expansion is a way to express a function as a sum of sine and cosine terms. When considering vibrating strings, Fourier series are particularly useful. They allow us to break down complex waveforms into simpler components, making it easier to analyze and calculate the wave behavior.
The exercise reveals the application of Fourier series expansion in describing the vibrational patterns of a string with specific boundary conditions, like \(f(0) = f(1) = 0\). These boundaries are essential in determining which frequencies or harmonics will constructively interfere and create standing waves.
To reparameterize our problem, we use substitutions and transform our periodic function into a form where we apply the Fourier sine series:

• \(F(u) = f(e^u-1)e^{-\frac{1}{2}u}\)
• \(F(u) = \sum_{n=1}^{\infty} B_n\sin(n\pi\frac{u}{\log 2})\)

Here, \(B_n\) denotes the series coefficients that determine how much of each harmonic contributes to the overall sound or motion of the vibrating string.
This expansion allows for the reconstruction of standard vibrational patterns from initial conditions using simple sinusoidal functions. It also assures a certain precision, as it converges uniformly, meaning that it accurately represents the function throughout the interval. Fourier series gives us a structured approach to deconstruct, analyze, and reconstruct dynamic systems in a manageable way.