Massive string

We have considered systems of infinite masses connected by massless string. We will now consider what happens to a massive string. We start with the familiar system of masses:

We write the equation of motion in terms of $x$ instead of $j$ .

\ddot y(x) = \frac{T}{ma} (y(x-a) -2y(x) + y(x+a)).

As each mass decreases and the distance between them shrinks this system begins to approximate a massive string. We substitute in $\rho_L = \frac{\text{mass}}{\text{length}} = \frac ma$ .

\ddot y(x) = \frac{T}{\rho_L a^2} (y(x-a) - 2y(x) +y(x+a)).

Now let $m,a \to 0$ such that $\rho_L$ remains constant. We Taylor expand $y(x-a)$ and $y(x+a)$ . Since $a \to 0$ the first few terms of the Taylor series are a very good approximation.

\begin{align*} y(x-a) &\approx y(x) - \frac{ay'(x)}{1!} + \frac{a^2y''(x)}{2} \\ y(x+a) &\approx y(x) + \frac{ay'(x)}{1!} + \frac{a^2y''(x)}{2}. \end{align*}

We substitute our approximations in and see that things cancel nicely.

\begin{align*} \ddot y(x) &= \frac{T}{\rho_La^2} (y-ay' + \frac12a^2y''-2y+y+ay'+\frac12a^2y'') \\ &= \frac{T}{\rho_L} y''(x) \\ \frac{\partial^2 y}{\partial t^2} &= \frac{T}{\rho_L} \frac{\partial^2y}{\partial x^2}. \tag{1} \end{align*}

Let’s consider the dispersion relation $\omega(\mathcal K)$ as $a \to 0$ .

\begin{align*} \omega^2 &= 2\omega_0^2(1-\cos\mathcal Ka) \\ \cos\mathcal Ka &\approx 1-\frac{(\mathcal Ka)^2}{2} \\ \omega^2 &\approx \omega_0^2 \mathcal K^2 a^2 = \frac{T\mathcal K^2}{\rho_L} \\ \omega &\approx \sqrt{\frac{T}{\rho_L}} \mathcal K. \end{align*}

We now generalize (1) by substituting $v_p^2 = \frac{T}{\rho_L}$ and replacing $y$ with $\psi$ . This gives us the general wave equation:

\frac{\partial^2\psi}{\partial t^2} = v_p^2 \frac{\partial^2\psi}{\partial x^2}.

We guess a solution $\psi(x,t) = A(x)B(t)$ .

\begin{align*} \frac{\partial^2 (A(x) B(t))}{\partial t^2} &= v_p^2 \frac{\partial^2 (A(x)B(t))}{\partial x^2} \\ A(x) \frac{\partial^2 B(t)}{\partial t^2} &= v_p^2 B(t) \frac{\partial^2 A(x)}{\partial x^2} \\ \frac{1}{v_p B(t)} \frac{\partial^2 B(t)}{\partial t^2} &= \frac{1}{A(x)} \frac{\partial^2 A(x)}{\partial x^2}. \tag{2} \end{align*}

Notice that the left side of (2) is a function only of $t$ and the right side is a function only of $x$ . The only way this equality can hold for all $t$ and $x$ is if it is constant. We call this constant value $-\mathcal K_m^2$ since it will make our result useful later. We can now separate equation (2) and set each side equal to $-\mathcal K_m^2$ .

\begin{align*} \frac{1}{v_p^2 B(t)} \frac{\partial^2B}{\partial t^2} &= -\mathcal K_m^2 \\ \ddot B &= -\mathcal K_m^2 v_p^2 B \\ B &= B_m \sin(v_p \mathcal K_m t + \beta_m). \end{align*}

Similarly

\begin{align*} \frac{1}{A(x)} \frac{\partial^2 A(x)}{\partial x^2} &= -\mathcal K_m^2 \\ A'' &= -\mathcal K_m^2 A \\ A &= A_m \sin(\mathcal K_m x + \alpha_m). \end{align*}

We plug $A$ and $B$ into our guess to find the general solution:

\begin{align*} \psi_m (x,t) &= A_m \sin(\mathcal K_m x + \alpha_m) \sin(\omega_m t + \beta_m) \\ \omega_m &= v_p \mathcal K_m \end{align*}

where $m$ stands for the $m^\text{th}$ normal mode.

Boundary conditions

We have found the general solution for infinite massive string, now we will consider how this solution applies to finite systems. Consider a string fixed between two walls as pictured below.

From the boundary conditions

\begin{align*} y_m(x,t) &= A_m \sin (\mathcal K_m x + \alpha_m) \sin(\omega_m t + \beta_m) \\ y_m(0,t) &= 0 = A_m \sin(\alpha_m) \sin(\omega_m t + \beta_m) \\ \alpha_m &= 0 \\ y_m(L,t) &= 0 = A_m \sin (\mathcal K_m L) \sin(\omega_m t + \beta_m) \\ \mathcal K_m &= \frac{m \pi}{L}. \end{align*}

Which means our solution for a string of length $L$ fixed between two walls is

y_m(x,t) = A_m \sin\left(\frac{m \pi x}{L}\right) \sin(\omega_m t + \beta_m).

Where $A_m$ and $\beta_m$ are defined by initial conditions.

Initial conditions

Consider a string released from rest in the initial configuration shown below. $f(x)$ models the initial displacement of the string.

f(x) = \begin{cases} h & \text{if } aL \le x \le bL \\ 0 & \text{otherwise} \end{cases}

Since the string is released from rest, the initial velocity is zero.

\left. \frac{\partial y}{\partial t} \right|_{t=0} = 0 = \omega_m A_m \sin\left(\frac{m \pi x}{L}\right) \cos(\beta_m)

So $\beta_m = \frac\pi2 + l \pi$ . However, all $l$ values will give the same solution, so we can just say $\beta_m = \frac\pi2$ . We find our solution is the sum of the normal modes

y(x,t) = \sum_{m=1}^\infty A_m \sin \left(\frac{m \pi x}{L}\right) \cos(\omega_m t).

Now to plug in our initial condition we want $y(x,0) = f(x) = \sum_{m=1}^\infty A_m \sin\left(\frac{m\pi x}{L}\right)$ . Note that this is a Fourier series. Following the derivation on the Fourier series page, we find

A_n = \frac{2}{L} \int_0^L \sin\left(\frac{n\pi x}{L}\right) f(x) dx.

Using the definition of $f(x)$ above, we evaluate the integral. When $x \notin [aL, bL]$ , $f(x) = 0$ so only the part of the integral from $aL$ to $bL$ contributes.

\begin{align*} A_n &= \frac{2h}{L} \int_{aL}^{bL} \sin\left(\frac{n\pi x}{L}\right) dx \\ &= \frac{2h}{n\pi}\left( \cos(n \pi a) - \cos(n \pi b) \right). \end{align*}

Which gives us the solution

y = \sum_{m=1}^\infty \frac{2h}{m\pi} \left( \cos(n \pi a) - \cos(n \pi b) \right) \sin \left(\frac{m\pi x}{L}\right) \cos(\omega_m t).

We visualize this solution (with the sum up to 100 terms) below.

Traveling wave

As shown here, a traveling wave $f(x+vt)$ solves the wave equation. A traveling wave makes it much easier to incorporate initial conditions than a series of infinite standing waves as we used above. Let’s consider how we can use this.

We saw from the visualization above that our initial wave split into two waves of half the amplitude traveling in opposite directions. Let $f(x)$ be the initial displacement of the string. Then we can model the traveling wave as

y = \frac{f(x+vt) + f(x-vt)}{2}.

Energy in oscillating massive string

Since we have not considered any damping so far, energy must be conserved in our system. Let us consider how energy is stored along an oscillating string.

We first consider the kinetic energy along the string. Kinetic energy $K = \frac12 mv^2$ , where $dm = \rho \, dx$ and $v = \frac{\partial y}{\partial t}$ . This gives us $dK = \frac12 \rho \, dx \left(\frac{\partial y}{\partial t}\right)^2$ .

We integrate to find

K = \int_0^L \frac12 \rho \left(\frac{\partial y}{\partial t}\right)^2 dx.

Now consider potential energy $U = W = Fds$ .

From the figure above we see that the displacement in string length $ds = \sqrt{dx^2 + dy^2} - dx$ .

\begin{align*} dU &= T(\sqrt{dx^2 + dy^2} - dx) \\ &= T\left(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} - 1\right)dx \\ &\approx \frac12 T \left(\frac{dy}{dx}\right)^2 dx \\ U &= \frac{T}{2}\int_0^L \left(\frac{dy}{dx}\right)^2 dx. \end{align*}

Power flowing across an oscillating massive string

We want to know how much power is flowing across a point on the wave. Power has units of energy over time, or force times velocity. Consider the force from the left acting on an infinitesimal mass along the string. We only care about the force from the left and not the net force because we want to find the power flowing from left to right.

Because $\theta$ is very small (since we are considering an infinitesimal slice of the string), we can approximate $\sin \theta \approx \tan \theta \approx \frac{dy}{dx}$ . This gives us the force in the vertical direction $F \approx -T \frac{dy}{dx}$ . Multiplying $F \cdot v$ we get

P = -T \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}.

For a more detailed explanation of energy and power in a string see this page by Mathew Schwartz at Harvard.