$$\require{cancel}$$

16.3: Mathematics of Waves

Exercise 16.3

The wave function above is derived using a sine function. Can a cosine function be used instead?

Velocity and Acceleration of the Medium

As seen in Example 16.4, the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium. The particles of the medium oscillate around an equilibrium position as the wave propagates through the medium. In the case of the transverse wave propagating in the x-direction, the particles oscillate up and down in the y-direction, perpendicular to the motion of the wave. The velocity of the particles of the medium is not constant, which means there is an acceleration. The velocity of the medium, which is perpendicular to the wave velocity in a transverse wave, can be found by taking the partial derivative of the position equation with respect to time. The partial derivative is found by taking the derivative of the function, treating all variables as constants, except for the variable in question. In the case of the partial derivative with respect to time t, the position x is treated as a constant. Although this may sound strange if you haven’t seen it before, the object of this exercise is to find the transverse velocity at a point, so in this sense, the x-position is not changing. We have

$$\begin{split} y(x,t) & = A \sin (kx - \omega t + \phi) \\ v_{y} (x,t) & = \frac{\partial y(x,t)}{\partial t} = \frac{\partial}{\partial t} [A \sin (kx - \omega t + \phi)] \\ & = -A \omega \cos (kx - \omega t + \phi) \\ & = -v_{y\; max} \cos (kx - \omega t + \phi) \ldotp \end{split}$$

The magnitude of the maximum velocity of the medium is |vy max| = A$$\omega$$. This may look familiar from the Oscillations and a mass on a spring.

We can find the acceleration of the medium by taking the partial derivative of the velocity equation with respect to time,

$$\begin{split} a_{y} (x,t) & = \frac{\partial v_{y}(x,t)}{\partial t} = \frac{\partial}{\partial t} [-A \omega \cos (kx - \omega t + \phi)] \\ & = -A \omega^{2} \sin (kx - \omega t + \phi) \\ & = -a_{y\; max} \sin (kx - \omega t + \phi) \ldotp \end{split}$$

The magnitude of the maximum acceleration is |ay max| = A$$\omega^{2}$$. The particles of the medium, or the mass elements, oscillate in simple harmonic motion for a mechanical wave.

The Linear Wave Equation

We have just determined the velocity of the medium at a position x by taking the partial derivative, with respect to time, of the position y. For a transverse wave, this velocity is perpendicular to the direction of propagation of the wave. We found the acceleration by taking the partial derivative, with respect to time, of the velocity, which is the second time derivative of the position:

$$a_{y} (x,t) = \frac{\partial^{2} y(x,t)}{\partial t^{2}} = \frac{\partial^{2}}{\partial t^{2}} [A \sin(kx - \omega t + \phi)] = -A \omega^{2} \sin (kx - \omega t + \phi) \ldotp$$

Now consider the partial derivatives with respect to the other variable, the position x, holding the time constant. The first derivative is the slope of the wave at a point x at a time t,

$$slope = \frac{\partial y(x,t)}{\partial x} = \frac{\partial}{\partial x} [A \sin (kx - \omega t + \phi)] = Ak \cos (kx - \omega t + \phi) \ldotp$$

The second partial derivative expresses how the slope of the wave changes with respect to position—in other words, the curvature of the wave, where

$$curvature = \frac{\partial^{2} y(x,t)}{\partial x^{2}} = \frac{\partial^{2}}{\partial x^{2}} [A \sin (kx - \omega t + \phi)] = -Ak^{2} \sin (kx - \omega t + \phi) \ldotp$$

The ratio of the acceleration and the curvature leads to a very important relationship in physics known as the linear wave equation. Taking the ratio and using the equation v = $$\frac{\omega}{k}$$ yields the linear wave equation (also known simply as the wave equation or the equation of a vibrating string),

$$\begin{split} \frac{\frac{\partial^{2} y(x,t)}{\partial t^{2}}}{\frac{\partial^{2} y(x,t)}{\partial x^{2}}} & = \frac{-A \omega^{2} \sin (kx - \omega t + \phi)}{-Ak^{2} \sin (kx - \omega t + \phi)} \\ & = \frac{\omega^{2}}{k^{2}} = v^{2}, \end{split}$$

$$\frac{\partial^{2} y(x,t)}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}} \ldotp \tag{16.6}$$

Equation 16.6 is the linear wave equation, which is one of the most important equations in physics and engineering. We derived it here for a transverse wave, but it is equally important when investigating longitudinal waves. This relationship was also derived using a sinusoidal wave, but it successfully describes any wave or pulse that has the form y(x, t) = f(x ∓ vt). These waves result due to a linear restoring force of the medium—thus, the name linear wave equation. Any wave function that satisfies this equation is a linear wave function.

An interesting aspect of the linear wave equation is that if two wave functions are individually solutions to the linear wave equation, then the sum of the two linear wave functions is also a solution to the wave equation. Consider two transverse waves that propagate along the x-axis, occupying the same medium. Assume that the individual waves can be modeled with the wave functions y1(x, t) = f(x ∓ vt) and y2(x, t) = g(x ∓ vt), which are solutions to the linear wave equations and are therefore linear wave functions. The sum of the wave functions is the wave function

$$y_{1} (x,t) + y_{2} (x,t) = f(x \mp vt) + g(x \mp vt) \ldotp$$

Consider the linear wave equation:

$$\begin{split} \frac{\partial^{2} (f + g)}{\partial x^{2}} & = \frac{1}{v^{2}} \frac{\partial^{2} (f + g)}{\partial t^{2}} \\ \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} g}{\partial x^{2}} & = \frac{1}{v^{2}} \left(\dfrac{\partial^{2} f}{\partial t^{2}} + \frac{\partial^{2} g}{\partial t^{2}}\right) \ldotp \end{split}$$

This has shown that if two linear wave functions are added algebraically, the resulting wave function is also linear. This wave function models the displacement of the medium of the resulting wave at each position along the x-axis. If two linear waves occupy the same medium, they are said to interfere. If these waves can be modeled with a linear wave function, these wave functions add to form the wave equation of the wave resulting from the interference of the individual waves. The displacement of the medium at every point of the resulting wave is the algebraic sum of the displacements due to the individual waves.

Taking this analysis a step further, if wave functions y1 (x, t) = f(x ∓ vt) and y2 (x, t) = g(x ∓ vt) are solutions to the linear wave equation, then Ay1(x, t) + By2(x, y), where A and B are constants, is also a solution to the linear wave equation. This property is known as the principle of superposition. Interference and superposition are covered in more detail in Interference of Waves.

Example 16.4: Interference of Waves on a String

Consider a very long string held taut by two students, one on each end. Student A oscillates the end of the string producing a wave modeled with the wave function y1(x, t) = A sin(kx − $$\omega$$t) and student B oscillates the string producing at twice the frequency, moving in the opposite direction. Both waves move at the same speed v = $$\frac{\omega}{k}$$. The two waves interfere to form a resulting wave whose wave function is yR(x, t) = y1(x, t) + y2(x, t). Find the velocity of the resulting wave using the linear wave equation $$\frac{\partial^{2} y(x,t)}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}}$$.

Strategy

First, write the wave function for the wave created by the second student. Note that the angular frequency of the second wave is twice the frequency of the first wave (2$$\omega$$), and since the velocity of the two waves are the same, the wave number of the second wave is twice that of the first wave (2k). Next, write the wave equation for the resulting wave function, which is the sum of the two individual wave functions. Then find the second partial derivative with respect to position and the second partial derivative with respect to time. Use the linear wave equation to find the velocity of the resulting wave.

Solution

1. Write the wave function of the second wave: y2(x, t) = A sin(2kx + 2$$\omega$$t).
2. Write the resulting wave function: $$y_{R} (x,t) = y_{1} (x,t) + y(x,t) = A \sin (kx - \omega t) + A \sin (2kx + 2 \omega t) \ldotp$$
3. Find the partial derivatives: $$\begin{split} \frac{\partial y_{R} (x,t)}{\partial x} & = -Ak \cos (kx - \omega t) + 2Ak \cos (2kx + 2 \omega t), \\ \frac{\partial^{2} y_{R} (x,t)}{\partial x^{2}} & = -Ak^{2} \sin (kx - \omega t) - 4Ak^{2} \sin(2kx + 2 \omega t), \\ \frac{\partial y_{R} (x,t)}{\partial t} & = -A \omega \cos (kx - \omega t) + 2A \omega \cos (2kx + 2 \omega t), \\ \frac{\partial^{2} y_{R} (x,t)}{\partial t^{2}} & = -A \omega^{2} \sin (kx - \omega t) - 4A \omega^{2} \sin(2kx + 2 \omega t) \ldotp \end{split}$$
4. Use the wave equation to find the velocity of the resulting wave: $$\begin{split} \frac{\partial^{2} y(x,t)}{\partial x^{2}} & = \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}}, \\ -Ak^{2} \sin (kx - \omega t) + 4Ak^{2} \sin(2kx + 2 \omega t) & = \frac{1}{v^{2}} \left(-A \omega^{2} \sin (kx - \omega t) - 4A \omega^{2} \sin(2kx + 2 \omega t)\right), \\ k^{2} \left(-A \sin (kx - \omega t) + 4A \sin(2kx + 2 \omega t)\right) & = \frac{\omega^{2}}{v^{2}} \left(-A \sin (kx - \omega t) - 4A \sin(2kx + 2 \omega t)\right), \\ k^{2} & = \frac{\omega^{2}}{v^{2}}, \\ |v| & = \frac{\omega}{k} \ldotp \end{split}$$

Significance

The speed of the resulting wave is equal to the speed of the original waves $$\left(v = \frac{\omega}{k}\right)$$. We will show in the next section that the speed of a simple harmonic wave on a string depends on the tension in the string and the mass per length of the string. For this reason, it is not surprising that the component waves as well as the resultant wave all travel at the same speed.

Exercise 16.4

The wave equation $$\frac{\partial^{2} y(x,t)}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}}$$ works for any wave of the form y(x, t) = f(x ∓ vt). In the previous section, we stated that a cosine function could also be used to model a simple harmonic mechanical wave. Check if the wave

$$y(x,t) = (0.50\; m) \cos (0.20 \pi\; m^{-1} x - 4.00 \pi s^{-1} t + \frac{\pi}{10})$$

is a solution to the wave equation.

Any disturbance that complies with the wave equation can propagate as a wave moving along the x-axis with a wave speed v. It works equally well for waves on a string, sound waves, and electromagnetic waves. This equation is extremely useful. For example, it can be used to show that electromagnetic waves move at the speed of light.

Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).