## Guitar Analogy

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

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Isaac Wen
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Joined: Wed Sep 30, 2020 9:35 pm
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### Guitar Analogy

Hi Everyone!

In the textbook, there was this analogy comparing Schrodinger's equation to a guitar string...I don't really understand it haha. Does anyone mind explaining?

"Think of a guitar string: because it is tied down at each end, it can support only shapes like the ones shown in Fig. 1C.3, which have zero displacement at each end. The shapes of the wavefunctions for the particle in the one-dimensional box are the same as the displacements of a vibrating string."

Marcus Lagman 2A
Posts: 91
Joined: Wed Sep 30, 2020 9:40 pm
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### Re: Guitar Analogy

Hello!

So this is my take on it. Basically, the comparison is saying that when you have a guitar, its strings are confined to its endpoints: the headstock and the bridge. When you play a string, the vibrating string is similar to a wave's shape in which that its restricted between two specific points. From this comparison, it is saying that just like a guitar's string being confined to two points, a wavefunction of a particle will be confined to a one-dimensional box, giving you a distance/displacement.

Sorry if this explanation does not help. If anyone has a better explanation, please do tell!

allyssa bradley 1H
Posts: 76
Joined: Wed Sep 30, 2020 9:47 pm

### Re: Guitar Analogy

I agree with you, Marcus! In lecture last week, Lavelle showed a diagram where he compared waves that were non-sinusoidal or out out of sync with a correct wave. The two ends of the wave correct have to end at the same point in the wavelength, otherwise they would be mismatched when you connect them to the rest of the repeating wave! Basically, the guitar analogy is taking this wavy line and applying it to a real-life, observable wavy line (guitar string) that has actual anchors on each side so you can definitely see how they end at the same place in the wave.

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