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Next: Reflection of Waves Up: Solutions to the Wave Previous: Harmonic Waveforms Propagating to   Contents

Stationary Waves

The third special case of solutions to the wave equation is that of standing waves. They are especially apropos to waves on a string fixed at one or both ends. There are two ways to find these solutions from the solutions above. A harmonic wave travelling to the right and hitting the end of the string (which is fixed), it has no choice but to reflect. This is because the energy in the string cannot just disappear, and if the end point is fixed no work can be done by it on the peg to which it is attached. The reflected wave has to have the same amplitude and frequency as the incoming wave. What does the sum of the incoming and reflected wave look like in this special case?

Suppose one adds two harmonic waves with equal amplitudes, the same wavelengths and frequencies, but that are travelling in opposite directions:

$\displaystyle y(x,t)$ $\textstyle =$ $\displaystyle y_0\left( \sin(kx - \omega t) + \sin(kx + \omega t)\right)$ (121)
  $\textstyle =$ $\displaystyle 2 y_0 \sin(kx) \cos(\omega t)$ (122)
  $\textstyle =$ $\displaystyle A \sin(kx) \cos(\omega t)$ (123)

(where we give the standing wave the arbitrary amplitude $A$). Since all the solutions above are independent of the phase, a second useful way to write stationary waves is:
y(x,t) = A \cos(kx) \cos(\omega t)
\end{displaymath} (124)

Which of these one uses depends on the details of the boundary conditions on the string.

In this solution a sinusoidal form oscillates harmonically up and down, but the solution has some very important new properties. For one, it is always zero when $x =
0$ for all possible $\lambda$:

y(0,t) = 0
\end{displaymath} (125)

For a given $\lambda$ there are certain other $x$ positions where the wave vanishes at all times. These positions are called nodes of the wave. We see that there are nodes for any $L$ such that:
y(L,t) = A \sin(k L) \cos(\omega t) = 0
\end{displaymath} (126)

which implies that:
k L = \frac{2 \pi L}{\lambda} = \pi,2\pi,3\pi,\ldots
\end{displaymath} (127)

\lambda = \frac{2 L}{n}
\end{displaymath} (128)

for $n = 1,2,3,...$

Only waves with these wavelengths and their associated frequencies can persist on a string of length $L$ fixed at both ends so that

y(0,t) = y(L,t) = 0
\end{displaymath} (129)

(such as a guitar string or harp string). Superpositions of these waves are what give guitar strings their particular tone.

It is also possible to stretch a string so that it is fixed at one end but so that the other end is free to move - to slide up and down without friction on a greased rod, for example. In this case, instead of having a node at the free end (where the wave itself vanishes), it is pretty easy to see that the slope of the wave at the end has to vanish. This is because if the slope were not zero, the terminating rod would be pulling up or down on the string, contradicting our requirement that the rod be frictionless and not able to pull the string up or down, only directly to the left or right due to tension.

The slope of a sine wave is zero only when the sine wave itself is a maximum or minimum, so that the wave on a string free at an end must have an antinode (maximum magnitude of its amplitude) at the free end. Using the same standing wave form we derived above, we see that:

k L = \frac{2 \pi L}{\lambda} = \pi/2,3\pi/2,5\pi/2\ldots
\end{displaymath} (130)

for a string fixed at $x =
0$ and free at $x = L$, or:
\lambda = \frac{4 L}{2n - 1}
\end{displaymath} (131)

for $n = 1,2,3,...$

There is a second way to obtain the standing wave solutions that particularly exhibits the relationship between waves and harmonic oscillators. One assumes that the solution $y(x,t)$ can be written as the product of a fuction in $x$ alone and a second function in $t$ alone:

y(x,t) = X(x) T(t)
\end{displaymath} (132)

If we substitute this into the differential equation and divide by $y(x,t)$ we get:
$\displaystyle \frac{d^2 y}{dt^2} = X(x)\frac{d^2 T}{dt^2}$ $\textstyle =$ $\displaystyle v^2\frac{d^2 y}{dx^2} =
v^2 T(t)\frac{d^2 X}{dx^2}$ (133)
$\displaystyle \frac{1}{T(t)}\frac{d^2 T}{dt^2}$ $\textstyle =$ $\displaystyle v^2 \frac{1}{X(x)}\frac{d^2
X}{dx^2}$ (134)
  $\textstyle =$ $\displaystyle -\omega^2$ (135)

where the last line follows because the second line equations a function of $t$ (only) to a function of $x$ (only) so that both terms must equal a constant. This is then the two equations:
\frac{d^2 T}{dt^2} + \omega^2 T = 0
\end{displaymath} (136)

\frac{d^2 X}{dt^2} + k^2 X = 0
\end{displaymath} (137)

(where we use $k = \omega/v$).

From this we see that:

T(t) = T_0\cos(\omega t + \phi)
\end{displaymath} (138)

X(x) = X_0\cos(kx + \delta)
\end{displaymath} (139)

so that the most general stationary solution can be written:
y(x,t) = y_0 \cos(kx + \delta) \cos(\omega t + \phi)
\end{displaymath} (140)

next up previous contents
Next: Reflection of Waves Up: Solutions to the Wave Previous: Harmonic Waveforms Propagating to   Contents
Robert G. Brown 2004-04-12