FIGURE 17.8 shows a string of length L tied at $x=0$ and $x=L$. If you wiggle the string in the middle, sinusoidal waves travel outward in both directions and soon reach the boundaries. Because the speed of a reflected wave does not change, the wavelength and frequency of a reflected sinusoidal wave are unchanged. Consequently, reflections at the ends of the string cause two waves of equal amplitude and wavelength to travel in opposite directions along the string. As we’ve just seen, these are the conditions that cause a standing wave!

FIGURE 17.8 Reflections at the two boundaries cause a standing wave on the string.

To connect the mathematical analysis of standing waves in Section 17.2 with the physical reality of a string tied down at the ends, we need to impose boundary conditions. A boundary condition is a mathematical statement of any constraint that must be obeyed at the boundary or edge of a medium. Because the string is tied down at the ends, the displacements at $x=0$ and $x=L$ must be zero at all times. Thus the standing-wave boundary conditions are $D(x=0\text{,}t)=0$ and $D(x=L\text{,}t)=0$. Stated another way, we require nodes at both ends of the string.

We found that the displacement of a standing wave is $D\left(x\text{,}t\right)=(2a\text{sin}kx)\text{cos}\omega t$. This equation already satisfies the boundary condition $D(x=0\text{,}t)=0$. That is, the origin has already been located at a node. The second boundary condition, at $x=L$, requires $D(x=L\text{,}t)=0$. This condition will be met at all times if

Equation 17.11 will be true if $\text{sin}kL=0$, which in turn requires

kL must be an integer multiple of $\pi$, but $m=0$ is excluded because L can’t be zero.

For a string of fixed length L, the only quantity in Equation 17.12 that can vary is $\text{λ}$. That is, the boundary condition is satisfied only if the wavelength has one of the values

A standing wave can exist on the string only if its wavelength is one of the values given by Equation 17.13. The mth possible wavelength ${\text{λ}}_m=2L\text{/}m$ is just the right size so that its mth node is located at the end of the string (at $x=L)$.

If standing waves are possible only for certain wavelengths, then only a few specific oscillation frequencies are allowed. Because $\text{λ}f=v$ for a sinusoidal wave, the oscillation frequency corresponding to wavelength ${\text{λ}}_m$ is

The lowest allowed frequency

which corresponds to wavelength ${\text{λ}}_1=2L$, is called the fundamental frequency of the string. The allowed frequencies can be written in terms of the fundamental frequency as

The allowed standing-wave frequencies are all integer multiples of the fundamental frequency. The higher-frequency standing waves are called harmonics, with the $m=2$ wave at frequency $f_2$ called the second harmonic, the $m=3$ wave called the third harmonic, and so on.

FIGURE 17.9 graphs the first four possible standing waves on a string of fixed length L. These possible standing waves are called the modes of the string, or sometimes the normal modes. Each mode, numbered by the integer m, has a unique wavelength and frequency. Keep in mind that these drawings simply show the envelope, or outer edge, of the oscillations. The string is continuously oscillating at all positions between these edges, as we showed in more detail in Figure 17.5a.

FIGURE 17.9 The first four modes for standing waves on a string of length $L$.

There are three things to note about the modes of a string.

1. m is the number of antinodes on the standing wave, not the number of nodes. You can tell a string’s mode of oscillation by counting the number of antinodes.

2. The fundamental mode, with $m=1$, has ${\text{λ}}_1=2L$, not ${\text{λ}}_1=L$. Only half of a wavelength is contained between the boundaries, a direct consequence of the fact that the spacing between nodes is $\text{λ}\text{/}2$.

3. The frequencies of the normal modes form a series: $f_1$, $2f_1$, $3f_1$, $4f_1\text{,}\dots$. The fundamental frequency $f_1$ can be found as the difference between the frequencies of any two adjacent modes. That is, $f_1=\text{Δ}f=f_{m+1}-f_m$.

EXAMPLE 17.2 Measuring g

Standing-wave frequencies can be measured very accurately. Consequently, standing waves are often used in experiments to make accurate measurements of other quantities. One such experiment, shown in FIGURE 17.10, uses standing waves to measure the free-fall acceleration g. A heavy mass is suspended from a 1.65-m-long, 5.85 g steel wire; then an oscillating magnetic field (because steel is magnetic) is used to excite the $m=3$ standing wave on the wire. Measuring the frequency for different masses yields the data given in the table. Analyze these data to determine the local value of g.

FIGURE 17.10 An experiment to measure $g$.

FIGURE 17.10 Full Alternative Text

Mass (kg)	$f{}_3$ (Hz)
2.00	68
4.00	97
6.00	117
8.00	135
10.00	152

MODEL The hanging mass creates tension in the wire. This establishes the wave speed along the wire and thus the frequencies of standing waves. Masses of a few kg might stretch the wire a mm or so, but that doesn’t change the length L until the third decimal place. The mass of the wire itself is insignificant in comparison to that of the hanging mass. We’ll be justified in determining g to three significant figures.

SOLVE The frequency of the third harmonic is

$$f_3={\displaystyle \frac{3}{2}}{\displaystyle \frac{v}{L}}$$

The wave speed on the wire is

$$v=\sqrt{{\displaystyle \frac{T_{\text{s}}}{\mu }}}=\sqrt{{\displaystyle \frac{Mg}{m\text{/}L}}}=\sqrt{{\displaystyle \frac{MgL}{m}}}$$

where $Mg$ is the weight of the hanging mass, and thus the tension in the wire, while m is the mass of the wire. Combining these two equations, we have

$$f_3={\displaystyle \frac{3}{2}}\sqrt{{\displaystyle \frac{Mg}{mL}}}={\displaystyle \frac{3}{2}}\sqrt{{\displaystyle \frac{g}{mL}}}\sqrt{M}$$

Squaring both sides gives

$$f_3{}^2={\displaystyle \frac{9g}{4mL}}M$$

A graph of the square of the standing-wave frequency versus mass M should be a straight line passing through the origin with slope $9g\text{/}4mL$. We can use the experimental slope to determine g.

FIGURE 17.11 is a graph of $f_3{}^2$ versus M. The slope of the best-fit line is $2289{\text{kg}}^{-1}{\text{s}}^{-2}$, from which we find

$$\begin{array}{ll}g\hfill & =\text{slope}\times {\displaystyle \frac{4mL}{9}}\hfill \\ \hfill & =2289{\text{kg}}^{-1}{\text{s}}^{-2}\times {\displaystyle \frac{4\left(0.00585\text{kg}\right)\left(1.65\text{m}\right)}{9}}=9.82{\text{m/s}}^{\text{2}}\hfill \end{array}$$

FIGURE 17.11 Graph of the data.

FIGURE 17.11 Full Alternative Text

REVIEW The fact that the graph is linear and passes through the origin confirms our model. This is an important reason for having multiple data points rather than using only one mass.