Standing Waves and Resonance — Question 2

Question 2

Adjacent resonances determine a missing fundamental

A fixed-end uniform string of length $.80\,\mathrm{m}$ and density $.0050\,\mathrm{kg/m}$ has adjacent normal-mode frequencies $120$ and $150\,\mathrm{Hz}$. Assume every intervening mode would have been detectable.

1. Infer the fundamental frequency, mode numbers, wave speed and tension.

2. Why is the adjacency qualification essential if a detector can miss modes?

Question 2 - Solution

Adjacent resonances determine a missing fundamental

Method and mathematical working

For a fixed-end ideal string, consecutive modes differ by exactly the fundamental frequency. Use that spacing before identifying the mode numbers and mechanical parameters.$$\begin{aligned}f_1&=150-120=30\,\mathrm{Hz},\\n_a&=120/30=4,\quad n_b=150/30=5,\\v&=2Lf_1,\qquad F=\mu v^2.\\[4pt]v&=2(0.80)(30)=48\,\mathrm{m/s},\quad F=(0.0050)(48)^2=11.52\,\mathrm N.\end{aligned}$$

(a) Calculation and conclusion

Adjacent fixed-end frequencies differ by $f_1$, so $f_1=30\,\mathrm{Hz}$. The observed modes are $n=4,5$. Then $v=2Lf_1=48\,\mathrm{m/s}$ and $F=\mu v^2=11.52\,\mathrm{N}$.

(b) Interpretation and limits

If unobserved modes lie between the peaks, the gap can be an integer multiple of the fundamental. For example, $f_1=15\,\mathrm{Hz}$ also places both observed frequencies on the spectrum, at modes $8$ and $10$. Absence of a detected peak does not automatically establish spectral adjacency.

Physical check

The inferred tension has units $(\mathrm{kg/m})(\mathrm{m^2/s^2})=\mathrm N$. If modes could be missing, the same observed gap might span several fundamental intervals.