Understanding SHM: Calculating the Wave Function of a Simple Harmonic Wave

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The wave function of a simple harmonic wave is represented by the equation y(x,t)=Asin(ωt+kx), derived from solving Newton's second law of motion. This involves setting up a differential equation, where ω is defined as √(k/m). However, there is confusion regarding the context, as the original question pertains to classical physics rather than quantum mechanics. Some participants argue that the Schrödinger equation is not relevant for classical simple harmonic motion. The discussion highlights the ambiguity in the question and the distinction between classical and quantum harmonic oscillators.
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What is the wave function of a simple harmonic wave?

y(x,t)=Asin(ωt+kx)
 
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y(x,t)=Asin(ωt+kx) is the equation of motion for a simple harmonic oscillator.

You get this by solving Newton's force law..

F=ma=-kx \\<br /> ma+kx=0 \\<br /> a+\frac{k}{m}x=0
Or you can write

\ddot{x}+\frac{k}{m}x=0

This is a differential equation, solved by Asin(ωt+kx), where \omega = \sqrt{\frac{k}{m}}.

I'm not sure if this answers your question?
 
Last edited:
sleepycoffee said:
y(x,t)=Asin(ωt+kx) is the equation of motion for a simple harmonic oscillator.

You get this by solving Newton's force law..

F=ma=-kx \\<br /> ma+kx=0 \\<br /> a+\frac{k}{m}x=0
Or you can write

\ddot{x}+\frac{k}{m}x=0

This is a differential equation, solved by Asin(ωt+kx), where \omega = \sqrt{\frac{k}{m}}.

I'm not sure if this answers your question?
He was asking for the wave function. You need to solve it with the Schrodinger equation, not Newtons laws.
 
This is posted in classical physics, however.. and in any case if it is undergoing simple harmonic motion then it isn't a quantum harmonic oscillator, so I don't see any reason to be messing around with Schrodingers.
 
sleepycoffee said:
This is posted in classical physics, however.. and in any case if it is undergoing simple harmonic motion then it isn't a quantum harmonic oscillator, so I don't see any reason to be messing around with Schrodingers.
Fair enough, it is a bit ambiguous eh?
 
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