Standing Waves and Resonance — Question 10

Question 10

Bending stiffness makes higher modes depart from exact harmonics

A stiff-string model is supplied: $\mu y_{tt}=F y_{xx}-D y_{xxxx}$, with $D>0$ a bending coefficient. At both ends take $y=0$ and $y_{xx}=0$, so sine modes remain admissible. Define $\eta=D\pi^2/(FL^2)$.

1. Derive $\omega_n$ and the ratio $f_n/(nf_1)$.

2. For $\eta=.010$, compute $f_2/(2f_1)$ and interpret the deviation. Why is specifying the extra boundary condition necessary here?

Question 10 - Solution

Bending stiffness makes higher modes depart from exact harmonics

Method and mathematical working

Differentiate a sine mode twice and four times, keeping the change in sign. Substitute into the supplied equation to obtain the modified dispersion relation.$$\begin{aligned}Y^{\prime\prime}&=-k^2Y,\qquad Y^{(4)}=k^4Y,\\\omega_n^2&=\frac F\mu k_n^2+\frac D\mu k_n^4,\qquad k_n=n\pi/L,\\\frac{f_n}{nf_1}&=\sqrt{\frac{1+\eta n^2}{1+\eta}}.\\[4pt]\frac{f_2}{2f_1}&=\sqrt{\frac{1+4(0.010)}{1+0.010}}=\sqrt{\frac{1.04}{1.01}}=1.01474.\end{aligned}$$

(a) Calculation and conclusion

For $\sin(n\pi x/L)$, $y_{xx}=-k_n^2y$ and $y_{xxxx}=k_n^4y$. Hence $\omega_n^2=(F/\mu)k_n^2+(D/\mu)k_n^4$, where $k_n=n\pi/L$. Therefore $f_n/(nf_1)=\sqrt{(1+\eta n^2)/(1+\eta)}$.

(b) Interpretation and limits

The ratio is $\sqrt{1.04/1.01}=1.01474$, so the second mode lies about $1.47\%$ above twice the fundamental. Higher spatial derivatives penalize shorter-scale curvature more strongly. A fourth-order spatial equation needs additional boundary conditions; clamped ends would not generally have the same sine-mode spectrum.

Physical check

Positive bending stiffness raises the higher modes proportionally more than the fundamental. The ratio tends to one as $D\to0$, recovering the ideal harmonic spectrum.