Standing Waves and Resonance — Question 3

Question 3

A fixed-free string supports an odd-harmonic sequence

A uniform string of length $L$ is fixed at $x=0$ and has a massless force-free transverse endpoint at $x=L$. Use $y(0,t)=0$ and $y_x(L,t)=0$.

1. Derive its allowed normal-mode frequencies.

2. For $L=.50\,\mathrm{m}$ and $v=40\,\mathrm{m/s}$, find the first three frequencies. Compare its fundamental with a fixed-fixed string of the same length and speed.

Question 3 - Solution

A fixed-free string supports an odd-harmonic sequence

Method and mathematical working

Use zero displacement at the fixed end and zero transverse force, hence zero slope, at the free end. Apply the latter to the derivative of the sine mode.$$\begin{aligned}Y(x)&=B\sin kx,\qquad Y^{\prime}(L)=Bk\cos kL=0,\\kL&=(n-\tfrac12)\pi,\qquad f_n=\frac{(2n-1)v}{4L}.\\[4pt]f_1&=\frac{40}{4(0.50)}=20\,\mathrm{Hz},\qquad f_2=3f_1=60\,\mathrm{Hz},\quad f_3=5f_1=100\,\mathrm{Hz}.\end{aligned}$$

(a) Calculation and conclusion

The fixed end selects a spatial sine. The force-free condition requires $\cos(kL)=0$, so $k_n=(2n-1)\pi/(2L)$ and $f_n=(2n-1)v/(4L)$, $n=1,2,\ldots$.

(b) Interpretation and limits

The frequencies are $20,60,100\,\mathrm{Hz}$. The fundamental is half the fixed-fixed value $v/(2L)=40\,\mathrm{Hz}$. The free endpoint is a displacement antinode, so the lowest pattern fits a quarter wavelength, not a half wavelength.

Physical check

The fundamental accommodates one quarter-wavelength. Successive allowed shapes add half-wavelengths, producing odd multiples of the fundamental rather than every integer multiple.