16.4 Advanced Topic: The Wave Equation on a String

Why do waves propagate along a string? We’ve described wave motion—the kinematics of waves—but not explained why it occurs. The motion of a string, like that of a baseball, is governed by Newton’s second law. But a baseball can be modeled as a moving particle. To explain waves, we need to see how Newton’s laws apply to a continuous object that is spread out in space.

This section will be significantly more mathematical than any analysis that we’ve done so far, so it will be good to get an overview of where we’re going. We have two primary goals:

To use Newton’s second law to find an equation of motion for displacements on a string. This is called the wave equation. We’ll see that Equation 16.14, the displacement of a sinusoidal wave, is a solution to the wave equation.
To predict the wave speed on a string.

Although we’ll derive the wave equation for a string, the equation itself occurs in many other contexts in science and engineering. Wherever this equation arises, the solutions are traveling waves.

FIGURE 16.17 shows a small piece of string that is displaced from its equilibrium position. This piece is at position x and has a small horizontal width $Δ x$ . We’re going to apply Newton’s second law, the familiar $F_{net} = m a$ , to this little piece of string. Notice that it’s curved, so the tension forces at the ends are not opposite each other. This is essential in order for there to be a net force.

The figure illustrates Newton’s second law to a small piece of string.

FIGURE 16.17 Full Alternative Text

We’ll begin by making the realistic assumption that the wave amplitude A is much smaller than the wavelength $λ$ . On average, the string “rises” distance A over a “run” of $λ / 4$ . If $A ≪ λ$ , then the slope of the string is always very small. That is, the string itself and the tension vectors are always very close to being horizontal. (Our drawings greatly exaggerate the amplitude for clarity.)

This assumption has two immediate implications. First, a small-amplitude wave doesn’t noticeably increase the length of the string. With no additional stretching, the string tension $T_{s}$ is not altered by the wave. Second, because the string is always very close to being horizontal, there’s virtually no difference between the actual length of the piece of string in Figure 16.17 and its horizontal width $Δ x$ . Thus the mass of this little piece of string is $m = μ Δ x$ , where, you’ll recall, $μ$ is the string’s linear density (mass per unit length).

Because we know the mass, let’s start with the $m a$ side of Newton’s second law. For a transverse wave, this little piece of string oscillates perpendicular to the direction of wave propagation. For wave motion along the x-axis, the string accelerates in the y-direction. If we had only this little piece of string, not an entire string, we would model it as a particle and write its acceleration as $a_{y} = d v_{y} / d t = d^{2} y / d t^{2}$ . The acceleration of a particle, as you’ve learned, is the second derivative of its position with respect to time.

But the string isn’t a particle. At any instant of time, different pieces of the string have different accelerations. To find the acceleration at a specific point on the string, we want to know how the displacement varies with t at that specific value of x. Or, because the displacement $D (x, t)$ is a function of two variables, we want to know the rate at which $D (x, t)$ changes with respect to t at a specific value of x. In multivariable calculus, the rate of change of a function with respect to one variable while all other variables are held fixed is called a partial derivative. Partial, because by holding other variables constant we’re only partially examining the many ways in which the function could change.

Partial derivatives have a special notation, using a “curly d.” The velocity of our little piece of string is written

v_{y} = \frac{\partial D}{\partial t}

(16.18)

and its acceleration is

a_{y} = \frac{\partial^{2} D}{\partial t^{2}}

(16.19)

Don’t panic if you’ve not reached partial derivatives in calculus. They are evaluated exactly like regular derivatives, but the partial-derivative notation means “treat all the other variables as if they were constants.” Using the partial derivative, we find the first half of Newton’s second law for our little piece of string is

m a_{y} = μ Δ x \frac{\partial^{2} D}{\partial t^{2}}

(16.20)

Now we can turn our attention to finding the net force on this little piece of string. The type of analysis we’re going to do may be new to you, but it is widely used in more advanced science and engineering courses. FIGURE 16.18 shows our little piece of the string, this time with the tension forces—one at each end of the string—resolved into x- and y-components.

The figure illustrates the method of finding the net force on a string.

FIGURE 16.18 Full Alternative Text

Strictly speaking, the tension force $T_{s}$ is tangent to the string. However, our small-amplitude assumption, requiring this piece of string to be almost horizontal, means there is virtually no distinction between $T_{s}$ and its horizontal component. (This is the small-angle approximation $cos θ \approx 1$ if $θ ≪ 1 rad$ .) Thus we’ve identified the two horizontal components as $T_{s}$ . Because they are equal but opposite, the net horizontal force is zero. This has to be true because each piece of this transverse wave accelerates only in the y-direction.

The net force on this little piece of string in the transverse direction (the y-direction) is

F_{net y} = T_{y} (x + Δ x) + T_{y} (x)

(16.21)

The notation $T_{y} (x)$ means “the y-component of the string tension at position x on the string.” And we’re adding, not subtracting, because this is a formal statement that the net force is the sum of all forces.

You can see, from the force triangle in Figure 16.18, that the ratio $T_{y} (x + Δ x) / T_{s}$ —rise over run—is the slope of the string at position $x + Δ x$ . This is a key part of the analysis, so make sure you understand it. The slope is the derivative of the string’s displacement with respect to x at this specific instant of time. We’re holding t constant while looking at the spatial variation of the string, so this is another partial derivative:

string slope = \frac{\partial D}{\partial x}

(16.22)

At the right end of this little piece of string, at position $x + Δ x$ , the y-component of the tension is

T_{y} (x + Δ x) = (slope at x + Δ x) \times T_{s} = T_{s} {\frac{\partial D}{\partial x} |}_{x + Δ x}

(16.23)

where the subscript on the partial derivative means to evaluate the slope at $x + Δ x$ . The same analysis holds at the left end, position x, with one change: Because $T_{s}$ points toward the left, a “negative run,” the ratio $T_{y} (x) / T_{s}$ is the negative of the string slope. Thus

T_{y} (x) = - (slope at x) \times T_{s} = - T_{s} \frac{\partial D}{\partial x} |_{x}

(16.24)

Combining Equations 16.23 and 16.24, we find that the net force on this little piece of string is

F_{net y} = T_{y} (x + Δ x) + T_{y} (x) = T_{s} [{\frac{\partial D}{\partial x} |}_{x + Δ x} - {\frac{\partial D}{\partial x} |}_{x}]

(16.25)

If this little piece of string were straight, the two slopes would be the same and there would be no net force. As we noted above, the string has to have a curvature to have a net force.

We’re almost done. We know the net force on this little piece of string (Equation 16.25) and we know its mass and acceleration (Equation 16.20). Because $F_{net y} = m a_{y}$ , we can equate these two results:

T_{s} [{\frac{\partial D}{\partial x} |}_{x + Δ x} - {\frac{\partial D}{\partial x} |}_{x}] = μ Δ x \frac{\partial^{2} D}{\partial t^{2}}

(16.26)

Dividing by $μ Δ x$ , we have

\frac{\partial^{2} D}{\partial t^{2}} = \frac{T_{s}}{μ} \times \frac{{\frac{\partial D}{\partial x} |}_{x + Δ x} - {\frac{\partial D}{\partial x} |}_{x}}{Δ x}

(16.27)

Recall, from calculus, that the derivative of the function $f (x)$ is defined as

\frac{d f}{d x} = lim_{Δ x \to 0} [\frac{f (x + Δ x) - f (x)}{Δ x}]

This is exactly what we have on the right side of Equation 16.27 if we let the width of our little piece of string approach zero: $Δ x \to 0$ . The function for which we’re evaluating the difference between $x + Δ x$ and x is the partial derivative $\partial D / \partial x$ , and the derivative of a derivative is a second derivative.

Thus in the limit $Δ x \to 0$ , Equation 16.27 becomes

\frac{\partial^{2} D}{\partial t^{2}} = \frac{T_{s}}{μ} \frac{\partial^{2} D}{\partial x^{2}} (wave equation for a string)

(16.28)

Equation 16.28 is the wave equation for a string. It’s really Newton’s second law in disguise, but written for a continuous object where the displacement is a function of both position and time. Just like Newton’s second law for a particle, it governs the dynamics of motion on a string.

16.4 Advanced Topic: The Wave Equation on a String

FIGURE 16.17 Apply Newton’s second law to this small piece of string.

FIGURE 16.18 Finding the net force on the string.