Standing Waves and Resonance — Question 8

Question 8

A tension sweep reverses the order in which mode numbers appear

A string of length $L=1.0\,\mathrm{m}$ and density $.010\,\mathrm{kg/m}$ is driven at fixed frequency $f_d$. As tension increases, consecutive fixed-end resonances occur at $16$ and $25\,\mathrm{N}$. Assume no mode is skipped.

1. Infer the two mode numbers and drive frequency.

2. Find the next higher tension resonance and explain why its mode number is smaller.

Question 8 - Solution

A tension sweep reverses the order in which mode numbers appear

Method and mathematical working

Solve the resonance equation for tension rather than frequency. Consecutive increasing tensions correspond to consecutive decreasing positive mode numbers.$$\begin{aligned}F_n&=\mu\left(\frac{2Lf_d}{n}\right)^2,\\\sqrt{25/16}&=\frac n{n-1}=\frac54,\qquad n=5.\\[4pt]f_d&=\frac5{2(1)}\sqrt{16/0.010}=100\,\mathrm{Hz},\\F_3&=0.010\left(\frac{200}{3}\right)^2=44.44\,\mathrm N.\end{aligned}$$

(a) Calculation and conclusion

Resonance requires $f_d=n\sqrt{F/\mu}/(2L)$, so $F_n=\mu(2Lf_d/n)^2$. Thus $25/16=[n/(n-1)]^2$, giving $n=5$ at $16\,\mathrm{N}$ and $n=4$ at $25\,\mathrm{N}$. Then $f_d=5\sqrt{16/.010}/2=100\,\mathrm{Hz}$.

(b) Interpretation and limits

The next is $n=3$, at $F_3=.010(200/3)^2=44.44\,\mathrm{N}$. Greater tension increases speed and wavelength at fixed drive frequency, so fewer half-wavelengths fit the same length. Mode number decreases as tension rises.

Physical check

The next higher tension belongs to $n=3$, not $n=6$. At fixed frequency an increased wave speed lengthens the wavelength, reducing the number of half-wavelengths that fit.