16.4.1 Traveling Wave Solutions

The equation of motion for a simple harmonic oscillator turned out to be a second-order differential equation. Although there are systematic ways to solve differential equations, we noted that—because solutions are unique—we can sometimes use what we know about a situation to guess the solution. The same is true for Equation 16.28, which is a partial differential equation. We have reason to think that sinusoidal waves can travel on stretched strings, so let’s guess that a solution to Equation 16.28 is

D (x, t) = A sin (k x - ω t + ϕ_{0})

(16.29)

where the minus sign gives a wave traveling in the +x-direction and—from Equation 16.13—the wave speed is $v = ω / k$ .

To evaluate this possible solution we need its second partial derivatives. With respect to position we have

\begin{array}{l} \frac{\partial D}{\partial x} = k A \cos (k x - ω t + ϕ_{0}) \\ \frac{\partial^{2} D}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial D}{\partial x}) = - k^{2} A \sin (k x - ω t + ϕ_{o}) \end{array}

(16.30)

and with respect to time we have

\begin{array}{l} \frac{\partial D}{\partial t} = - ω A \cos (k x - ω t + ϕ_{0}) \\ \frac{\partial^{2} D}{\partial t^{2}} = \frac{\partial}{\partial t} (\frac{\partial D}{\partial t}) = - ω^{2} A \sin (k x - ω t + ϕ_{o}) \end{array}

(16.31)

Substituting the second partial derivatives into the wave equation, Equation 16.28, gives

- ω^{2} A sin (k x - ω t + ϕ_{0}) = \frac{T_{s}}{μ} (- k^{2} A sin (k x - ω t + ϕ_{0}))

(16.32)

This will be true only if

ω^{2} = \frac{T_{s}}{μ} k^{2}

(16.33)

But $ω / k$ is the wave speed v, so what we’ve found is that the sinusoidal wave of Equation 16.1 is a solution to the wave equation, but only if the wave travels with speed

v = \frac{ω}{k} = \sqrt{\frac{T_{s}}{μ}}

(16.34)

You should be able to convince yourself that we would have arrived at the same result if we had started with $D (x, t) = A sin (k x + ω t + ϕ_{0})$ for a wave traveling in the –x-direction.

Let’s summarize. We used Newton’s second law for a small piece of the string to come up with an equation for the dynamics of motion on a string. We then showed that a solution to this equation is a sinusoidal traveling wave, and we made a specific prediction for the wave speed in terms of two properties or characteristics of the string: its tension and its mass density. Thus the answer to the question with which we started this section—why do waves propagate along a string—is that wave motion is simply a consequence of Newton’s second law, the relationship between force and acceleration, when applied to a continuous object.

With Equation 16.34 in hand, we can write Equation 16.28 as

\frac{\partial^{2} D}{\partial t^{2}} = v^{2} \frac{\partial^{2} D}{\partial x^{2}} (the general wave equation)

(16.35)

We derived Equation 16.28 specifically for a string, but any physical system that obeys Equation 16.35 for some type of displacement D will have sinusoidal waves traveling with speed v. Equation 16.35 is called the wave equation, and it occurs over and over in science and engineering. We will see it again in this chapter in our analysis of sound waves. And much later, in Chapter 31, we’ll discover that electromagnetic fields also obey this equation. Thus electromagnetic waves exist, and we’ll be able to predict that all electromagnetic waves, regardless of wavelength, travel through vacuum with the same speed—the speed of light.

EXAMPLE 16.4 Generating a sinusoidal wave

A very long string with $μ = 2.0 g / m$ is stretched along the x-axis with a tension of 5.0 N. At $x = 0 m$ it is tied to a 100 Hz simple harmonic oscillator that vibrates perpendicular to the string with an amplitude of 2.0 mm. The oscillator is at its maximum positive displacement at $t = 0 s$ .

Write the displacement equation for the traveling wave on the string.
At $t = 5.0 ms,$ what is the string’s displacement at a point 2.7 m from the oscillator?

MODEL The oscillator generates a sinusoidal traveling wave on a string. The displacement of the wave has to match the displacement of the oscillator at $x = 0 m$ .

SOLVE

The equation for the displacement is

$D (x, t) = A sin (k x - ω t + ϕ_{0})$

with A, k, $ω$ , and $ϕ_{0}$ to be determined. The wave amplitude is the same as the amplitude of the oscillator that generates the wave, so $A = 2.0 mm$ . The oscillator has its maximum displacement $y_{osc} = A = 2.0 m m$ at $t = 0 s,$ thus

$D (0 m, 0 s) = A sin (ϕ_{0}) = A$

This requires the phase constant to be $ϕ_{0} = π / 2 rad$ . The wave’s frequency is $f = 100 Hz,$ the frequency of the source; therefore the angular frequency is $ω = 2 π f = 200 π rad / s$ . We still need $k = 2 π / λ$ , but we do not know the wavelength. However, we have enough information to determine the wave speed, and we can then use either $λ = v / f$ or $k = ω / v$ . The speed is

$v = \sqrt{\frac{T_{s}}{μ}} = \sqrt{\frac{5.0 N}{0.0020 kg/m}} = 50 m / s$

Using v, we find $λ = 0.50 m$ and $k = 2 π / λ = 4 π rad / m$ . Thus the wave’s displacement equation is

$D (x, t) = (2.0 mm) \times sin [2 π ((2.0 m^{- 1}) x - (100 s^{- 1}) t) + π / 2 rad]$

Notice that we have separated out the $2 π$ . This step is not essential, but for some problems it makes subsequent steps easier.
The wave’s displacement at $t = 5.0 ms = 0.0050 s$ is

$\begin{array}{l} D (x, t = 5.0 ms) & = & (2.0 mm) sin (4 π x - π rad + π / 2 rad) \\ = & (2.0 mm) sin (4 π x - π / 2 rad) \end{array}$

At $x = 2.7 m$ (calculator set to radians!), the displacement is

$D (2.7 m, 5.0 m s) = 1.6 m m$