ODE: Vibrating String and Normal Modes

The vibration of a string, from a guitar to a suspension bridge cable, is governed by a fundamental physical law expressed mathematically as the wave equation. Understanding its solutions—normal modes—is not just an academic exercise; it explains why musical instruments produce specific notes, how energy propagates in structures, and provides a cornerstone for solving more complex partial differential equations across physics and engineering.

Deriving the Wave Equation for a String

We begin by modeling a perfectly flexible string with uniform linear density $\rho$ under constant tension $T$. Consider a tiny element of the string. Applying Newton's second law in the vertical direction, the net force is the difference in the vertical components of tension at each end of the element. For small displacements $u(x,t)$, the slope $\frac{\partial u}{\partial x}$ is small, allowing the approximations $\sin \theta \approx \tan \theta \approx \frac{\partial u}{\partial x}$. The vertical force is then $T\left(\frac{\partial u}{\partial x}(x+\Delta x,t)-\frac{\partial u}{\partial x}(x,t)\right)$.

This net force equals the mass of the element ($\rho \Delta x$) times its vertical acceleration ($\frac{{\partial }^{2}u}{\partial t^2}$). Dividing by $\Delta x$ and taking the limit as $\Delta x\to 0$ yields the one-dimensional wave equation: $$\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}$$ where $c=\sqrt{T/\rho }$ is the wave speed. For a string of length $L$ fixed at both ends, we have the boundary conditions: $u(0,t)=0$ and $u(L,t)=0$ for all time $t$.

Solution by Separation of Variables

The method of separation of variables assumes the solution can be written as a product of a function of space and a function of time: $u(x,t)=X(x)T(t)$. Substituting into the wave equation and dividing by $c^{2}X(x)T(t)$ gives: $$\frac{1}{c^2}\frac{T^{\prime \prime }(t)}{T(t)}=\frac{X^{\prime \prime }(x)}{X(x)}$$ Since the left side depends only on $t$ and the right side only on $x$, both must equal a constant, which we call $-\lambda$. This splits the PDE into two ordinary differential equations (ODEs):

1. Spatial ODE (Eigenvalue Problem): $X^{\prime \prime }(x)+\lambda X(x)=0$, with boundary conditions $X(0)=0$ and $X(L)=0$.
2. Temporal ODE: $T^{\prime \prime }(t)+\lambda c^{2}T(t)=0$.

The spatial problem is crucial. For non-trivial solutions $X(x)\not\equiv 0$, the constant $\lambda$ must take specific values called eigenvalues. Solving $X^{\prime \prime }+\lambda X=0$:

• If $\lambda <0$, the solution is exponential and cannot satisfy both zero boundary conditions.
• If $\lambda =0$, the solution is linear, $X(x)=Ax+B$, and $X(0)=X(L)=0$ forces $A=B=0$ (trivial).
• If $\lambda >0$, the solution is sinusoidal: $X(x)=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. Applying $X(0)=0$ forces $A=0$. Applying $X(L)=0$ gives $B\sin (\sqrt{\lambda }L)=0$. For a non-trivial solution ($B\ne 0$), we require $\sin (\sqrt{\lambda }L)=0$. Hence, $\sqrt{\lambda }L=n\pi$ for $n=1,2,3,...$.

Thus, the eigenvalues are ${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2$, and the corresponding eigenfunctions or normal mode shapes are: $$X_n(x)=\sin \left(\frac{n\pi x}{L}\right)$$

Normal Mode Frequencies and the General Solution

With ${\lambda }_n$ known, the temporal ODE becomes $T_n^{\prime \prime }(t)+{\left(\frac{n\pi c}{L}\right)}^2{T}_n(t)=0$. This is the harmonic oscillator equation. Its general solution is: $$T_n(t)=C_n\cos \left({\omega }_{n}t\right)+D_n\sin \left({\omega }_{n}t\right)$$ where ${\omega }_n=\frac{n\pi c}{L}=n{\omega }_1$ is the natural frequency of the $n$th mode. The fundamental frequency is ${\omega }_1=\frac{\pi c}{L}$.

Each product $u_n(x,t)=X_n(x)T_n(t)$ is a normal mode—a standing wave pattern where all points vibrate at the same frequency ${\omega }_n$. The general solution to the wave equation with fixed ends is a superposition of modes: $$u(x,t)=\sum _{n=1}^{\infty }\sin \left(\frac{n\pi x}{L}\right)\left[C_n\cos \left({\omega }_{n}t\right)+D_n\sin \left({\omega }_{n}t\right)\right]$$ This infinite series represents any possible free vibration of the string.

Applying Initial Conditions: Plucked and Struck Strings

The constants $C_n$ and $D_n$ are determined by the initial conditions: the initial shape $u(x,0)$ and the initial velocity $\frac{\partial u}{\partial t}(x,0)$.

1. Plucked String (Guitar): The string is pulled into a shape $f(x)$ and released from rest. Initial conditions: $u(x,0)=f(x)$ and $\frac{\partial u}{\partial t}(x,0)=0$.

Since $u_t(x,0)=0$, all $D_n=0$. The shape condition gives: $$u(x,0)=\sum _{n=1}^{\infty }{C}_n\sin \left(\frac{n\pi x}{L}\right)=f(x)$$ This is a Fourier sine series. The coefficients $C_n$ are found using orthogonality of sine functions: $$C_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \left(\frac{n\pi x}{L}\right)dx$$

1. Struck String (Piano): The string is initially flat but given an initial velocity $g(x)$ by a hammer. Initial conditions: $u(x,0)=0$ and $\frac{\partial u}{\partial t}(x,0)=g(x)$.

Here, $u(x,0)=0$ forces all $C_n=0$. The velocity condition gives: $$\frac{\partial u}{\partial t}(x,0)=\sum _{n=1}^{\infty }{\omega }_n{D}_n\sin \left(\frac{n\pi x}{L}\right)=g(x)$$ Solving for $D_n$: $$D_n=\frac{2}{{\omega }_{n}L}{\int }_0^{L}g(x)\sin \left(\frac{n\pi x}{L}\right)dx$$ The resulting sound is richer in higher harmonics if the hammer strike is sharp.

Energy in a Vibrating String

The total mechanical energy $E$ of the string is the sum of kinetic and potential energy. Kinetic energy ($KE$) comes from motion: $d(KE)=\frac{1}{2}\rho {\left(\frac{\partial u}{\partial t}\right)}^{2}dx$. Potential energy ($PE$) is the work done by tension in stretching the string: $d(PE)\approx \frac{1}{2}T{\left(\frac{\partial u}{\partial x}\right)}^{2}dx$. The total energy is: $$E=\frac{1}{2}{\int }_0^L\left[\rho {\left(\frac{\partial u}{\partial t}\right)}^2+T{\left(\frac{\partial u}{\partial x}\right)}^2\right]dx$$ A key result for normal modes is that they are orthogonal in energy. If you substitute the general superposition solution into this integral, cross-terms between different modes integrate to zero. This means the total energy is simply the sum of the energies contained in each individual mode: $E={\sum }_{n=1}^{\infty }{E}_n$. Each mode's energy is constant in time for free vibration, illustrating the conservation of total energy.

Common Pitfalls

1. Misapplying Boundary Conditions to the Temporal ODE: A common error is to apply the spatial boundary conditions (like $u(0,t)=0$) to the time function $T(t)$. Remember, separation of variables applies these conditions to the spatial eigenproblem $X(x)$ only. The function $T(t)$ is governed solely by its ODE and initial conditions.

1. Incorrect Eigenvalue Determination: Assuming the separation constant is positive without justification can lead to missing the trivial cases. You must systematically test $\lambda <0$, $\lambda =0$, and $\lambda >0$ to show that only positive $\lambda$ yields non-trivial solutions, leading to the condition $\sin (\sqrt{\lambda }L)=0$.

1. Confusing Wave Speed with Particle Velocity: The constant $c=\sqrt{T/\rho }$ is the phase speed of a wave traveling along the string. It is not the transverse velocity of a point on the string, which is $\frac{\partial u}{\partial t}$ and is a function of both $x$ and $t$. Conflating these will derail the physical interpretation of solutions.

1. Neglecting the Role of Initial Conditions in Energy: When calculating total energy from the general solution, students often forget that the orthogonality of modes simplifies the expression drastically. Trying to square the infinite series without exploiting orthogonality leads to an intractable mess of cross-terms that should—and do—integrate to zero.

Summary

• The transverse vibrations of a taut, flexible string are governed by the wave equation $\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}$, where $c=\sqrt{T/\rho }$ is the wave speed.
• Using separation of variables, the PDE reduces to a spatial eigenvalue problem whose solutions are the normal mode shapes $X_n(x)=\sin (n\pi x/L)$ and corresponding natural frequencies ${\omega }_n=n\pi c/L$.
• The general motion is a superposition of modes, a Fourier series in space with time-varying coefficients. The specific initial conditions (plucked shape or initial velocity) determine these coefficients via Fourier sine series integrals.
• The total vibrational energy is conserved and is the sum of the energies independently stored in each excited normal mode, a consequence of the orthogonality of the eigenfunctions.