Standing Waves and Resonance — Question 1

Question 1

Two fixed ends quantize the allowed wave numbers

A uniform string of length $L$ is fixed at both ends. Seek a nonzero normal mode $y=B\sin(kx)\cos\omega t$, with $\omega=vk$.

1. Derive the allowed wave numbers and frequencies.

2. For $L=1.20\,\mathrm{m}$ and $v=24\,\mathrm{m/s}$, find the third-mode frequency and all its interior nodes. Explain why $n=0$ is not an oscillating string mode here.

Question 1 - Solution

Two fixed ends quantize the allowed wave numbers

Method and mathematical working

Satisfy the fixed endpoint at $x=0$ with a sine spatial factor. Enforcing the second fixed endpoint then quantizes the wave number and frequency.$$\begin{aligned}Y(x)&=B\sin kx,\qquad Y(L)=0\Rightarrow kL=n\pi,\\f_n&=\frac{vk_n}{2\pi}=\frac{nv}{2L},\qquad n=1,2,\ldots.\\[4pt]f_3&=\frac{3(24)}{2(1.20)}=30\,\mathrm{Hz},\qquad x_{\mathrm{interior}}=L/3,2L/3=0.40,0.80\,\mathrm m.\end{aligned}$$

(a) Calculation and conclusion

The left boundary is already satisfied. The right requires $\sin(kL)=0$, so $k_n=n\pi/L$, $n=1,2,\ldots$, and $f_n=nv/(2L)$. The allowed shape contains $n$ half-wavelengths.

(b) Interpretation and limits

$f_3=3(24)/(2.40)=30\,\mathrm{Hz}$. Its interior nodes are $x=L/3=.40$ and $2L/3=.80\,\mathrm{m}$. With $n=0$ the assumed sine shape is identically zero; it provides no nonzero oscillation satisfying both fixed boundaries.

Physical check

The third mode has three half-wavelengths within the length and two interior nodes. Taking $n=0$ makes the entire sine shape vanish, so it is not a nonzero oscillation mode.