# Appendix A: Standing Wave Resonance from First Principle (Beyond Syllabus)

What does the H2 syllabus require to know? Not much really. Just that repeated reflections at both ends of the string causes incident and reflected waves to superpose. At resonant frequencies, a standing wave of large amplitude is formed, with nodes at both ends of the string.

I have noticed that even university textbooks do not provide much detail about what’s really happening in the string. So I have had to form my own thoughts, which I share in this section.

Let’s call the wave that has just departed from the vibrator the 1st generation incident wave. When it returns after the reflection, it becomes the 1st generation reflected wave. The 1st generation incident wave superposes with the 1st generation reflected wave to form, well, the 1st generation standing wave. Assuming no energy loss, the amplitudes of the incident, reflected and standing wave would be A, A and 2A respectively.

Of course, the 1G reflected wave will undergo reflection and sets off as the 2G incident wave, returns as the 2G reflected wave, superposing to form the 2G standing wave. As long as the vibrator keeps on oscillating the string, the waves in the string will keep piling up. So we have the 1G, 2G, 3G, 4G … NG standing waves all in the string at the same time. [1]

And they all superpose of course. At most frequencies, all these generations of standing waves would have a progressive phase difference between generations. The resulting destructive interference produces a standing wave of zero (or close to zero) amplitude.

However, at resonant frequencies, all these generations of standing waves would be in-phase with one another. The resulting constructive interference produces a standing wave of amplitude much larger than A. Assuming that there are N generations in the string, and no attenuation occurs during reflections, no energy lost to dissipative force, the amplitude of the standing wave would be 2NA.

So what’s the condition for resonance to occur? Since every incident wave must go forth and back along the length L of the string before setting off as the next generation incident wave, the path difference between generations is simply 2L.

For the generations of waves to be in phase,

$2L=n\lambda ,\text{ }n=1,2,3,...\text{ }$

This means that the resonant wavelengths are

$\displaystyle \lambda =\frac{{2L}}{n},\text{ }n=1,2,3,...\text{ }$

Since $v=f\lambda$, the resonant frequencies are

$\displaystyle f=n\frac{v}{{2L}},\text{ }n=1,2,3,...\text{ }$

Closed Pipe

Actually, the reflection at a fixed end adds a phase change of π rad. But on a guitar string, an incident wave must undergo two reflections before setting off as the next generation incident wave, so a total phase change of 2π rad phase per round trip. This allows use to ignore the phase change caused by reflections. If we are talking about a closed pipe, where the total phase change caused by reflections is only π rad (because the reflection at the open end incurs no phase change), the condition for resonance would have been

$\displaystyle 2L=(n-\frac{1}{2})\lambda ,\text{ }n=1,2,3,...\text{ }$

This means that the resonant wavelengths are

$\displaystyle \lambda =\frac{{4L}}{{2n-1}},\text{ }n=1,2,3,...\text{ }$

Since $v=f\lambda$, the resonant frequencies are

$\displaystyle f=(2n-1)\frac{v}{{4L}},\text{ }n=1,2,3,...\text{ }$

[1] I created this “wave generations” nomenclature myself. You won’t read about iterations of waves described as 1G, 2G, 3G and so on in any textbook.