Standing Waves — Question 10

Question 10

Adding two modes usually removes their separate interior nodes

A fixed-end string has a motion $y=B\sin(2\pi x/L)\cos\omega_2t+C\sin(3\pi x/L)\cos\omega_3t$, with $B,C\ne0$ and distinct frequencies $\omega_2,\omega_3$.

1. Find all permanent nodes in $0\le x\le L$.

2. Why is $x=L/2$, a node of the second spatial harmonic, not a node of the full motion? Distinguish a permanent node from a cancellation at one time.

Question 10 - Solution

Adding two modes usually removes their separate interior nodes

Method and mathematical working

Independent temporal frequencies cannot cancel permanently at one location unless each nonzero mode has a zero there. Intersect the two sets of spatial zeros.$$\begin{aligned}\sin(2\pi x/L)&=0,\qquad \sin(3\pi x/L)=0,\\\frac{x}{L}&=\frac m2=\frac n3.\\[4pt]3m=2n&\Rightarrow m=2j,\ n=3j,\ x=jL,\\0\le x\le L&\Rightarrow x=0\ \text{or}\ L.\end{aligned}$$

(a) Calculation and conclusion

For the sum to vanish for every time, the coefficients of the two distinct cosine frequencies must each vanish. The zeros of $\sin(2\pi x/L)$ and $\sin(3\pi x/L)$ coincide only at $x=0,L$.

(b) Interpretation and limits

At $x=L/2$, the first component is zero but the second is $-C\cos\omega_3t$, generally nonzero. Two components can cancel at isolated events without producing a permanent node. A multimode motion need not retain the interior nodes of either mode separately.

Physical check

Within $0\le x\le L$, the only common zeros are the endpoints. At the midpoint the second mode vanishes but the third does not, so an instantaneous cancellation there is not a permanent node.