17.2.2 The Mathematics of Standing Waves

A sinusoidal wave traveling to the right along the x-axis with angular frequency $ω = 2 π f$ , wave number $k = 2 π / λ$ , and amplitude a is

D_{R} = a sin (k x - ω t)

(17.2)

An equivalent wave traveling to the left is

D_{L} = a sin (k x + ω t)

(17.3)

We previously used the symbol A for the wave amplitude, but here we will use a lowercase a to represent the amplitude of each individual wave and reserve A for the amplitude of the net wave. For now, we’ll assume that the phase constants are zero.

A black-and-white photograph of the Tacoma Narrows suspension bridge that experienced a wind-induced standing-wave oscillation. — This photograph shows the Tacoma Narrows suspension bridge on the day in 1940 when it experienced a wind-induced standing-wave oscillation that led to its collapse. The red line shows the original line of the deck of the bridge. You can clearly see the large amplitude of the oscillation and the node at the center of the span.

According to the principle of superposition, the net displacement of the medium when both waves are present is the sum of $D_{R}$ and $D_{L} :$

D (x, t) = D_{R} + D_{L} = a sin (k x - ω t) + a sin (k x + ω t)

(17.4)

We can simplify Equation 17.4 by using the trigonometric identity

sin (α \pm β) = sin α cos β \pm cos α sin β

Doing so gives

\begin{array}{l} D (x, t) = a (sin k x cos ω t - cos k x sin ω t) + a (sin k x cos ω t + cos k x sin ω t) \\ = (2 a sin k x) cos ω t \end{array}

(17.5)

It is useful to write Equation 17.5 as

D (x, t) = A (x) cos ω t

(17.6)

where the amplitude function $A (x)$ is defined as

A (x) = 2 a sin k x

(17.7)

The amplitude reaches a maximum value $A_{max} = 2 a$ at points where $sin k x = 1$ .

The displacement $D (x, t)$ given by Equation 17.6 is neither a function of $x - v t$ nor a function of $x + v t;$ hence it is not a traveling wave. Instead, the $cos ω t$ term in Equation 17.6 describes a medium in which each point oscillates in simple harmonic motion with frequency $f = ω / 2 π$ . The function $A (x) = 2 a sin k x$ gives the amplitude of the oscillation for a particle at position x.

FIGURE 17.6 graphs Equation 17.6 at several different instants of time. Notice that the graphs are identical to those of Figure 17.5a, showing us that Equation 17.6 is the mathematical description of a standing wave.

The figure illustrates the net displacement resulting from two counter-propagating

FIGURE 17.6 Full Alternative Text

The nodes of the standing wave are the points at which the amplitude is zero. They are located at positions x for which

A (x) = 2 a sin k x = 0

(17.8)

The sine function is zero if the angle is an integer multiple of $π$ rad, so Equation 17.8 is satisfied if

\begin{matrix} k x_{m} = \frac{2 π x_{m}}{λ} = m π & m = 0, 1, 2, 3, \dots \end{matrix}

(17.9)

Thus the position $x_{m}$ of the mth node is

x_{m} = m \frac{λ}{2} m = 0, 1, 2, 3, \dots

(17.10)

You can see that the spacing between two adjacent nodes is $λ / 2$ , in agreement with Figure 17.5b. The nodes are not spaced by $λ$ , as you might have expected.

EXAMPLE 17.1 Node spacing on a string

A very long string has a linear density of $5.0 g / m$ and is stretched with a tension of 8.0 N. 100 Hz waves with amplitudes of 2.0 mm are generated at the ends of the string.

What is the node spacing along the resulting standing wave?
What is the maximum displacement of the string?

MODEL Two counter-propagating waves of equal frequency create a standing wave.

VISUALIZE The standing wave looks like Figure 17.5a.

SOLVE

The speed of the waves on the string is

$v = \sqrt{\frac{T_{s}}{μ}} = \sqrt{\frac{8.0 N}{0.0050 kg / m}} = 40 m / s$

and the wavelength is

$λ = \frac{v}{f} = \frac{40 m / s}{100 Hz} = 0.40 m = 40 cm$

Thus the spacing between adjacent nodes is $λ / 2 = 20 cm$ .
The maximum displacement is $A_{max} = 2 a = 4.0 mm$ .