Standing Waves and Resonance — Question 6

Question 6

A symmetric pluck selects odd modes and weights their energies

A fixed-end string is plucked at its midpoint to height $h$ and released from rest. Its initial shape is $H(x)=2hx/L$ for $0\le x\le L/2$ and $2h(1-x/L)$ for $L/2\le x\le L$. Use $B_n=(2/L)\int_0^L H(x)\sin(n\pi x/L)dx$.

1. Find the mode coefficients and explain the missing even modes.

2. Find the third-to-first amplitude-magnitude ratio and energy ratio, using $E_n=\mu L\omega_n^2B_n^2/4$ and $\omega_n=n\omega_1$.

Question 6 - Solution

A symmetric pluck selects odd modes and weights their energies

Method and mathematical working

Project the symmetric triangular initial displacement onto the sine modes. Midpoint symmetry suppresses even modes; integration of each linear half supplies the inverse-square decay.$$\begin{aligned}k_n&=n\pi/L,\qquad B_n=\frac2L\int_0^L H(x)\sin(k_nx)\,dx,\\B_n&=\frac{4h}{L^2}\left[\int_0^{L/2}x\sin(k_nx)\,dx+\int_{L/2}^L(L-x)\sin(k_nx)\,dx\right]\\&=\frac{8h}{L^2k_n^2}\sin(k_nL/2)=\frac{8h}{n^2\pi^2}\sin(n\pi/2),\\\frac{E_3}{E_1}&=\frac{\omega_3^2 B_3^2}{\omega_1^2B_1^2}.\\[4pt]B_{2j}&=0,\qquad B_3/B_1=-1/9,\qquad E_3/E_1=3^2(1/9)^2=1/9.\end{aligned}$$

(a) Calculation and conclusion

Integrating the two linear pieces gives $B_n=8h\sin(n\pi/2)/(n^2\pi^2)$. Even coefficients vanish because the symmetric shape has zero overlap with modes antisymmetric about the midpoint. The signs of successive odd coefficients alternate.

(b) Interpretation and limits

$B_1=8h/\pi^2$ and $B_3=-8h/(9\pi^2)$, so $|B_3/B_1|=1/9$. Energy includes an extra frequency-squared factor: $E_3/E_1=3^2(1/9)^2=1/9$, not $1/81$. Small displacement amplitudes at high mode number can still carry appreciable energy.

Physical check

Although the third displacement coefficient is only $-1/9$ of the first, its frequency is three times larger. The energy ratio is therefore $9/81=1/9$, and the negative coefficient describes phase rather than negative energy.