Standing Waves and Resonance — Question 5

Question 5

Drive and detector positions can hide different modes

For fixed-end mode $n$, take the shape $\varphi_n(x)=\sin(n\pi x/L)$. A point driver at $a$ couples in proportion to $\varphi_n(a)$, and a displacement detector at $b$ reads in proportion to $\varphi_n(b)$. Let $a=L/3$ and $b=L/2$.

1. Identify the modes the driver cannot excite and the modes the detector cannot see.

2. Among modes $n=1$ through $6$, which can be both excited and detected? Explain why a sparse measured spectrum need not mean that other normal modes are absent.

Question 5 - Solution

Drive and detector positions can hide different modes

Method and mathematical working

The excitation strength and measured signal each contain a mode-shape factor at their respective positions. A mode is visible only if both factors are nonzero.$$\begin{aligned}C_n&\propto\sin(n\pi/3)\sin(n\pi/2),\\\sin(n\pi/3)=0&\quad(n\ \text{multiple of }3),\\\sin(n\pi/2)=0&\quad(n\ \text{even}).\\[4pt]\{1,2,3,4,5,6\}\setminus(\{2,4,6\}\cup\{3,6\})&=\{1,5\}.\end{aligned}$$

(a) Calculation and conclusion

The driver factor is $\sin(n\pi/3)$, zero for multiples of three. The detector factor is $\sin(n\pi/2)$, zero for even $n$. These are distinct selection conditions.

(b) Interpretation and limits

Only $n=1$ and $5$ survive both conditions among the first six. Modes $2,4$ are missed by the detector, mode $3$ is not driven, and mode $6$ fails both. The system still possesses all normal modes; spatial coupling determines which appear in this measurement.

Physical check

Of the first six modes, only $n=1$ and $n=5$ survive both filters. A missing measured peak therefore need not mean that the corresponding normal mode is absent from the string.