Standing Waves and Resonance — Question 4

Question 4

Length tuning and tension tuning reach the same frequency by different changes

A fixed-fixed string has fundamental $f_1=(2L)^{-1}\sqrt{F/\mu}$. The target is a frequency $4/3$ times its initial value. Treat density as unchanged in each comparison.

1. Find the required active-length ratio if tension stays fixed, and the required tension ratio if active length stays fixed.

2. Explain why the fixed-density assumption is reasonable for shortening the vibrating portion but may fail when substantially stretching the entire string.

Question 4 - Solution

Length tuning and tension tuning reach the same frequency by different changes

Method and mathematical working

Take a ratio of the fundamental-frequency formula. Treat shortening at fixed density and retensioning at fixed length as separate controlled changes.$$\begin{aligned}\frac{f_1^{\prime}}{f_1}&=\frac{L}{L^{\prime}}\sqrt{\frac{F^{\prime}}F}\quad(\mu\ \text{fixed}),\\L^{\prime}&=\frac34L\quad\text{or}\quad F^{\prime}=\frac{16}{9}F.\\[4pt]\frac{F^{\prime}}F&=\left(\frac{f_1^{\prime}}{f_1}\right)^2=\left(\frac43\right)^2=\frac{16}{9}.\end{aligned}$$

(a) Calculation and conclusion

At fixed tension, $f\propto1/L$, so $L_{\mathrm{new}}/L=3/4$. At fixed length and density, $f\propto\sqrt F$, so $F_{\mathrm{new}}/F=(4/3)^2=16/9$.

(b) Interpretation and limits

A movable contact can shorten the active vibrating portion without materially stretching the string, leaving mass per length nearly unchanged. Stretching a fixed total mass changes its current length and hence density; that correction must be included when appreciable. The two tuning procedures have different controlled variables.

Physical check

The target frequency is higher, so the length must decrease or the tension must increase. Stretching the same material also changes its density and would require a modified comparison.