7.6 Normal Modes and Fourier Series

Consider a standing wave $y(x,t)=Ae^{i(kx-\omega t)}+Be^{i(kx+\omega t)}$ on a string clamped at both ends. Suppose the string runs from $x=0$ to $x=L$ , then at the boundaries we have $y(0,t)=y(L,t)=0$ . Applying the left boundary condition to the standing wave, we get

	$\displaystyle y(0,t)=\operatorname{\mathfrak{Re}}(Ae^{-i\omega t}+Be^{i\omega t% })=0$		(7.82)
	$\displaystyle A\cos(\omega t)+B\cos(\omega t)=0$		(7.83)
	$\displaystyle\implies A=-B.$		(7.84)

This makes physical sense because as we have seen above, wave pulses invert at fixed boundaries. For the right boundary condition, we get

$\displaystyle y(L,t)$	$\displaystyle=\operatorname{\mathfrak{Re}}(A[e^{i(kL-\omega t)-e^{i(kL+\omega t% )}}])$	(7.85)
	$\displaystyle=\operatorname{\mathfrak{Re}}(Ae^{ikL}[\underbrace{e^{-i\omega t}% -e^{i\omega t}}_{-2i\sin(\omega t)}])$	(7.86)
	$\displaystyle=2A\sin(kL)\sin(\omega t)=0$	(7.87)
	$\displaystyle\implies\sin(kL)=0$	(7.88)

$\sin(kL)=0$ implies that $kL=n\pi$ , so we can write the standing wave in the form

y(x,t)=A\sin(k_{n}x)\cos(\omega_{n}t+\phi_{0}),

(7.90)

where

k_{n}=\frac{n\pi}{L},\quad\omega_{n}=\frac{n\pi v}{L}.

(7.91)

Note that wavelength is related to angular wavenumber by $\lambda=\frac{2\pi}{k}$ , so $\lambda_{n}=\frac{2\pi}{k_{n}}=\frac{2L}{n}$ . These allowed wavenumbers/frequencies are the normal modes of vibration for the string clamped at both ends.

7.6.1 Fourier Series

Recall in chapter 6 we discussed exciting all the frequencies of a system at once using the impulse method. If we pluck the string, multiple normal modes are excited. The resulting motion of the string is a superposition of the normal modes, given by

y(x,t)=\sum_{n}A_{n}\sin(k_{n}x)\cos(\omega_{n}t).

(7.92)

We have chosen $\cos$ instead of $\sin$ here for the temporal part because the displacement is maximal at $t=0$ . It makes no difference to the physics which one we choose. At $t=0$ , we have

y(x,0)=\sum_{n}A_{n}\sin\left(\frac{n\pi}{L}x\right).

(7.93)

If we can calculate all the $A_{n}$ ’s, we can determine the subsequent motion of the string. But there may be infinitely many $A_{n}$ ’s! Luckily, we can use a mathematical tool called Fourier series which makes the calculation of all of them straightforward.

If we have a periodic function (for simplicity we can assume the period is $2\pi$ , since we can stretch or shrink it otherwise), it is (almost) always possible to represent it as an infinite series of sines and cosines.

f(x)=\frac{1}{2}a_{0}+\sum_{j=1}^{\infty}a_{j}\cos(jx)+\sum_{j=1}^{\infty}b_{j% }\sin(jx).

(7.94)

Using some convenient properties of integrals of sines and cosines, there is a simple method to calculate what the coefficients $a_{j}$ and $b_{j}$ should be for any function $f(x)$ . We can see how this will help us solve our problem of representing a wave pulse as a sum of normal modes.

Specifically, we will make use of the following three integrals:

	$\displaystyle\int_{-\pi}^{\pi}\sin(mx)\sin(nx)\operatorname{\mathrm{d}}\!x=% \begin{cases}0\text{ if }m\neq n\\ \pi\text{ if }m=n\end{cases}$		(7.95)
	$\displaystyle\int_{-\pi}^{\pi}\cos(mx)\cos(nx)\operatorname{\mathrm{d}}\!x=% \begin{cases}0\text{ if }m\neq n\\ \pi\text{ if }m=n\end{cases}$		(7.96)
	$\displaystyle\int_{-\pi}^{\pi}\sin(mx)\cos(nx)\operatorname{\mathrm{d}}\!x=0% \quad\forall m,n.$		(7.97)

Now consider multiplying equation 7.94 by $\cos(nx)$ and integrating from $-\pi$ to $\pi$ . We get

	$\displaystyle\begin{split}\int_{-\pi}^{\pi}\cos(nx)f(x)\operatorname{\mathrm{d% }}\!x&=\frac{1}{2}a_{0}\int_{-\pi}^{\pi}\cos(nx)\operatorname{\mathrm{d}}\!x+% \sum_{j=1}^{\infty}a_{j}\int_{-\pi}^{\pi}\cos(nx)\cos(jx)\operatorname{\mathrm% {d}}\!x\\ &+\underbrace{\sum_{j=1}^{\infty}b_{j}\int_{-\pi}^{\pi}\sin(nx)\cos(jx)% \operatorname{\mathrm{d}}\!x}_{=0}\end{split}$			(7.98)
		$\displaystyle=\frac{1}{2}a_{0}\int_{-\pi}^{\pi}\cos(nx)\operatorname{\mathrm{d% }}\!x+\sum_{j=1}^{\infty}a_{j}\int_{-\pi}^{\pi}\cos(nx)\cos(jx)\operatorname{% \mathrm{d}}\!x.$		(7.99)

If $n=0$ we get

\int_{-\pi}^{\pi}\cos(0)f(x)\operatorname{\mathrm{d}}\!x=\frac{1}{2}a_{0}\int_% {-\pi}^{\pi}\cos(0)\operatorname{\mathrm{d}}\!x=\pi a_{0}.

(7.100)

Whereas if $n>0$ we find

\int_{-pi}^{\pi}\cos(nx)f(x)\operatorname{\mathrm{d}}\!x=\sum_{j=1}^{\infty}a_% {j}\int_{-\pi}^{\pi}\cos(nx)\cos(jx)\operatorname{\mathrm{d}}\!x.

(7.101)

All the terms on the right-hand side are zero except the one where $j=n$ , where the result will be $\pi a_{n}$ . So we have found formula for $a_{0}$ and $a_{n}$ , they are

	$\displaystyle a_{0}$	$\displaystyle=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\operatorname{\mathrm{d}}\!x$		(7.102)
	$\displaystyle a_{n}$	$\displaystyle=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(nx)f(x)\operatorname{\mathrm{% d}}\!x.$		(7.103)

Similarly, if we multiply by $\sin(nx)$ and integrate from $-\pi$ to $\pi$ we find

b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(nx)f(x)\operatorname{\mathrm{d}}\!x.

(7.104)

More generally, if a function has a period $P$ , then we can adjust the periods of sine and cosine and the formulae for the coefficients as follows:

$\displaystyle f(x)$	$\displaystyle=\frac{1}{2}a_{0}+\sum_{j=1}^{\infty}a_{j}\cos\left(\frac{2\pi n}% {P}x\right)+\sum_{j=1}^{\infty}b_{j}\sin\left(\frac{2\pi n}{P}x\right)$	(7.105)
$\displaystyle a_{0}$	$\displaystyle=\frac{2}{P}\int_{-\frac{P}{2}}^{\frac{P}{2}}f(x)\operatorname{% \mathrm{d}}\!x$	(7.106)
$\displaystyle a_{n}$	$\displaystyle=\frac{2}{P}\int_{-\frac{P}{2}}^{\frac{P}{2}}\cos\left(\frac{2\pi n% }{P}x\right)f(x)\operatorname{\mathrm{d}}\!x$	(7.107)
$\displaystyle b_{n}$	$\displaystyle=\frac{2}{P}\int_{-\frac{P}{2}}^{\frac{P}{2}}\sin\left(\frac{2\pi n% }{P}x\right)f(x)\operatorname{\mathrm{d}}\!x.$	(7.108)

In fact, the integrals don’t necessarily have to be from $-\frac{P}{2}$ to $\frac{P}{2}$ , as long as they go over one full period. We can choose the most convenient range to integrate over.

Note that if a function is odd ( $f(-x)=-f(x)$ ), the Fourier series will only contain sine terms. Likewise, if a function is even ( $f(-x)=f(x)$ ) its Fourier series will only contain cosine terms. This is because sine is an odd function and cosine is even.

Example 7.6.

Find the Fourier series of a square wave of amplitude $d$ and period $P$ , which has the form

f(x)=\begin{cases}-d,\quad-\frac{P}{2}<x<0\\ d,\quad 0<x<\frac{P}{2}\end{cases}

(7.109)

on the domain $\left[-\frac{P}{2},\frac{P}{2}\right]$ . Outside of this domain it repeats periodically.

We have defined the square wave above as an odd function, so there should be no cosine terms. We can show this explicitly by calculating the $a_{n}$ coefficients:

$\displaystyle a_{n}$	$\displaystyle=\frac{2}{P}\int_{-\frac{P}{2}}^{\frac{P}{2}}\cos\left(\frac{2\pi n% }{P}x\right)f(x)\operatorname{\mathrm{d}}\!x$	(7.110)
	$\displaystyle=\frac{2}{P}\left[\int_{-\frac{P}{2}}^{0}-d\cos\left(\frac{2\pi n% }{P}x\right)\operatorname{\mathrm{d}}\!x+\int_{0}^{\frac{P}{2}}d\cos\left(% \frac{2\pi n}{P}x\right)\operatorname{\mathrm{d}}\!x\right]$	(7.111)
	$\displaystyle=\frac{2d}{P}\underbrace{\left[\int_{0}^{\frac{P}{2}}\cos\left(% \frac{2\pi n}{P}x\right)\operatorname{\mathrm{d}}\!x-\int_{-\frac{P}{2}}^{0}% \cos\left(\frac{2\pi n}{P}x\right)\operatorname{\mathrm{d}}\!x\right]}_{=0% \text{ since }\cos(-x)=\cos(x)}.$	(7.112)

Now calculating the $b_{n}$ coefficients, we find

$\displaystyle b_{n}$	$\displaystyle=\frac{2}{P}\int_{-\frac{P}{2}}^{\frac{P}{2}}\sin\left(\frac{2\pi n% }{P}x\right)f(x)\operatorname{\mathrm{d}}\!x$	(7.113)
	$\displaystyle=\frac{2}{P}\left[\int_{-\frac{P}{2}}^{0}-d\sin\left(\frac{2\pi n% }{P}x\right)\operatorname{\mathrm{d}}\!x+\int_{0}^{\frac{P}{2}}d\sin\left(% \frac{2\pi n}{P}x\right)\operatorname{\mathrm{d}}\!x\right]$	(7.114)
	$\displaystyle=\frac{2d}{P}\left[\int_{0}^{\frac{P}{2}}\sin\left(\frac{2\pi n}{% P}x\right)\operatorname{\mathrm{d}}\!x-\int_{-\frac{P}{2}}^{0}\sin\left(\frac{% 2\pi n}{P}x\right)\operatorname{\mathrm{d}}\!x\right]$	(7.115)
	$\displaystyle=\frac{2d}{P}\left(\left[\frac{P}{2\pi n}\cos\left(\frac{2\pi n}{% P}x\right)\right]_{-\frac{P}{2}}^{0}-\left[\frac{P}{2\pi n}\cos\left(\frac{2% \pi n}{P}x\right)\right]_{0}^{\frac{P}{2}}\right)$	(7.116)
	$\displaystyle=\frac{d}{\pi n}([1-\cos(-\pi n)]-[\cos(\pi n)-1])$	(7.117)
	$\displaystyle=\frac{2d}{\pi n}[1-\cos(\pi n)].$	(7.118)

If $n$ is even then $\cos(\pi n)=1$ and if $n$ is odd then $\cos(\pi n)=-1$ , so we get

b_{n}=\begin{cases}0\text{ if $n$ is even}\\ \frac{4d}{\pi n}\text{ if $n$ is odd.}\end{cases}

(7.119)

Therefore the Fourier series for the square wave consists only of sine terms with odd $n$ . The first few terms are

f(x)=\frac{4d}{\pi}\sin\left(\frac{2\pi}{P}x\right)+\frac{4d}{3\pi}\sin\left(% \frac{6\pi}{P}x\right)+\frac{4d}{5\pi}\sin\left(\frac{10\pi}{P}x\right)+\dots.

(7.120)

7.6.2 Motion of a Plucked String

Let’s now apply this amazing mathematical technique to the motion of a plucked string clamped at both ends. We know that the initial displacement of the string is given by equation 7.93, and we want to find the $A_{n}$ ’s. However, the displacement of the string is not periodic in $x$ because the string is finite in length! How we can find a Fourier series? We can just find a Fourier series for a periodic function that matches our string in the range $0$ to $L$ . Really, what we are asking is “what normal modes are excited when we pluck the string”, so we actually want our Fourier series to contain only the normal modes. Since $\lambda_{1}=2L$ , our periodic representation of the string must have period $2L$ otherwise the Fourier series won’t contain the fundamental mode. Note that it must also be an odd function, because the initial conditions for the string only contains sines. With these constraints, we end up with the following function:

y(x,0)=\begin{cases}4d\frac{x}{L},\quad 0<x<\frac{L}{4}\\ \frac{4d}{3}\left(1-\frac{x}{L}\right),\quad\frac{L}{4}<x<L,\end{cases}

(7.121)

where $d$ is the maximum initial displacement of the string. Using the formula for the Fourier coefficients with $y^{\prime}(x)$ as the periodic extension of $y(x,0)$ , we get

$\displaystyle A_{n}$	$\displaystyle=\frac{1}{L}\int_{-L}^{L}\underbrace{\sin\left(\frac{2\pi}{% \lambda_{n}}x\right)}_{\text{odd}}\underbrace{y^{\prime}(x)}_{\text{odd}}% \operatorname{\mathrm{d}}\!x$	(7.122)
	$\displaystyle=\frac{2}{L}\int_{0}^{L}\sin\left(\frac{2\pi}{\lambda_{n}}x\right% )y^{\prime}(x)\operatorname{\mathrm{d}}\!x$	(7.123)
	$\displaystyle=\frac{2}{L}\int_{0}^{L}\sin\left(\frac{2\pi}{\lambda_{n}}x\right% )y(x,0)\operatorname{\mathrm{d}}\!x.$	(7.124)

The second line follows because two odd functions multiplied together make an even function, so we can cut the range of the integral in half and double the result. The last line follows because $y^{\prime}(x)=y(x,0)$ on the domain $[0,L]$ , so actually only the initial displacement between $0$ and $L$ matters! Now we can find the $A_{n}$ ’s:

A_{n}=\frac{8d}{L^{2}}\int_{0}^{\frac{L}{4}}x\sin\left(\frac{n\pi}{L}x\right)% \operatorname{\mathrm{d}}\!x+\frac{8d}{3L}\int_{\frac{L}{4}}^{L}\sin\left(% \frac{n\pi}{L}x\right)\operatorname{\mathrm{d}}\!x-\frac{8d}{3L^{2}}\int_{% \frac{L}{4}}^{L}x\sin\left(\frac{n\pi}{L}x\right)\operatorname{\mathrm{d}}\!x.

(7.125)

We can solve these integrals using the following results

	$\displaystyle\int\sin\left(\frac{n\pi}{L}x\right)\operatorname{\mathrm{d}}\!x$	$\displaystyle=-\frac{L}{n\pi}\cos\left(\frac{n\pi}{L}x\right)+c$		(7.126)
	$\displaystyle\int x\sin\left(\frac{n\pi}{L}x\right)\operatorname{\mathrm{d}}\!x$	$\displaystyle=\frac{L^{2}}{n^{2}\pi^{2}}\sin\left(\frac{n\pi}{L}x\right)-\frac% {Lx}{n\pi}\cos\left(\frac{n\pi}{L}x\right).$		(7.127)

So we find

\begin{split}A_{n}&=8d\left[\frac{1}{n^{2}\pi^{2}}\sin\left(\frac{n\pi}{4}% \right)-\frac{1}{4\pi n}\cos\left(\frac{n\pi}{4}\right)\right]\\ &+\frac{8d}{3}\left[\frac{1}{n\pi}\cos\left(\frac{n\pi}{4}\right)-\frac{1}{n% \pi}\cos(n\pi)\right]\\ &+\frac{8d}{3}\left[\frac{1}{n\pi}\cos(n\pi)+\frac{1}{n^{2}\pi^{2}}\sin\left(% \frac{n\pi}{4}\right)-\frac{1}{4\pi n}\cos\left(\frac{n\pi}{4}\right)\right].% \end{split}

(7.128)

The cosine terms all cancel out, and we are left with

A_{n}=\frac{32d}{3n^{2}\pi^{2}}\sin\left(\frac{n\pi}{4}\right).

(7.129)

$\sin\left(\frac{n\pi}{4}\right)$ follows the repeating sequence $\frac{1}{\sqrt{2}}$ , $1$ , $\frac{1}{\sqrt{2}}$ , $0$ , $-\frac{1}{\sqrt{2}}$ , $-1$ , $-\frac{1}{\sqrt{2}}$ , $0$ …, so the first few terms of the Fourier series are

\begin{split}y(x,0)=\frac{32d}{3\pi^{2}}\biggl{[}&\frac{1}{\sqrt{2}}\sin\left(% \frac{\pi}{L}x\right)+\frac{1}{4}\sin\left(\frac{2\pi}{L}x\right)+\frac{1}{9% \sqrt{2}}+\sin\left(\frac{3\pi}{L}x\right)\\ &-\frac{1}{25\sqrt{2}}\sin\left(\frac{5\pi}{L}x\right)-\frac{1}{36}\sin\left(% \frac{6\pi}{L}x\right)-\frac{1}{49\sqrt{2}}\sin\left(\frac{7\pi}{L}x\right)+% \dots\biggr{]}\end{split}

(7.130)

We can plot the frequency spectrum of $y(x,0)$ , which is a plot showing $A_{n}$ against $n$ . We see that the lowest $n$ are the largest components, and they can smaller as $n$ increases. The gaps in the frequency spectrum are where the modes have a node at the point where the string was plucked. This means that because of where we chose to pluck the string, the resulting vibration will not contain any of those frequencies at all.

With damping effects, we find that all of the normal modes decay away, with the higher frequency modes decaying faster.