(III) The manner in which a string is plucked determines
the mixture of harmonic amplitudes in the resulting wave.
Consider a string exactly $\frac{1}{2}$ -m longthat is fixed at both its ends
located at $x=0.0$ and $x=\frac{1}{2} \mathrm{m} .$ The first five harmonics
of this string have wavelengths $\lambda_{1}=1.0 \mathrm{m}, \quad \lambda_{2}=\frac{1}{2} \mathrm{m},$ $\lambda_{3}=\frac{1}{3} \mathrm{m}, \lambda_{4}=\left\{\mathrm{m},$ and $\lambda_{5}=\frac{1}{5} \mathrm{m}$ . According to Fourier's \right. theorem, any shape of this string can be formed by a sum of its harmonics, with each harmonic having its own unique
amplitude $A .$ We limit the sum to the first five harmonics
in the expression
$$D(x)=A_{1} \sin \left(\frac{2 \pi}{\lambda_{1}} x\right)+A_{2} \sin \left(\frac{2 \pi}{\lambda_{2}} x\right)$$
$$+A_{3} \sin \left(\frac{2 \pi}{\lambda_{3}} x\right)+A_{4} \sin \left(\frac{2 \pi}{\lambda_{4}} x\right)+A_{5} \sin \left(\frac{2 \pi}{\lambda_{5}} x\right)$$
and $D$ is the displacement of the string at a time $t=0$ .
Imagine plucking this string at its midpoint (Fig. 45$a )$ or at a
point two-thirds from the left end (Fig. 45b). Using a graphing calculator or computer program, show that the above expression can fairly accurately represent the shape
in: $(a)$ Fig. 45$a$ , if
$$\begin{aligned} A_{1} &=1.00 \\ A_{2} &=0.00 \\ A_{3} &=-0.11 \\ A_{1} &=0.00, \text { and } \end{aligned}$$
$$A_{5}=0.040 ;$ and in
(b) Fig. 45 $\mathrm{b}$ , if
$A_{1}=0.87$$
$$\begin{aligned} A_{2} &=-0.22 \\ A_{3} &=0.00 \\ A_{4} &=0.054, \text { and } \\ A_{5} &=-0.035 \end{aligned}$$
