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asdf1

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why is phase velocity v=w/k ?

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- Thread starter asdf1
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- #1

asdf1

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why is phase velocity v=w/k ?

- #2

arildno

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Let us look at a typical harmonic wave, [tex]f(x,t)=a\sin(kx-\omega{t})[/tex]

Here, a,k,[tex]\omega[/tex] are constants, where a is the amplitude, k the wave number and [tex]\omega[/tex] the frequency.

Rewrite this in the following form:

[tex]f(x,t)=a\sin(k(x-\frac{\omega}{k}t))[/tex]

Do you see that if [tex]x-\frac{\omega}{k}t=s[/tex] where s is a CONSTANT, makes the value of f constant as well (equal to [tex]a\sin(ks)[/tex])?

But that means, that the signal value [tex]a\sin(ks)[/tex] can be regarded as MOVING ALONG THE POSITIVE X-DIRECTION WITH VELOCITY [tex]\frac{\omega}{k}[/tex]!

For, (remembering that s is constant) we have the trajectory for our signal:

[tex]x=s+\frac{\omega}{k}t[/tex]

and this simply shows what the propagation velocity of our signal is..

Here, a,k,[tex]\omega[/tex] are constants, where a is the amplitude, k the wave number and [tex]\omega[/tex] the frequency.

Rewrite this in the following form:

[tex]f(x,t)=a\sin(k(x-\frac{\omega}{k}t))[/tex]

Do you see that if [tex]x-\frac{\omega}{k}t=s[/tex] where s is a CONSTANT, makes the value of f constant as well (equal to [tex]a\sin(ks)[/tex])?

But that means, that the signal value [tex]a\sin(ks)[/tex] can be regarded as MOVING ALONG THE POSITIVE X-DIRECTION WITH VELOCITY [tex]\frac{\omega}{k}[/tex]!

For, (remembering that s is constant) we have the trajectory for our signal:

[tex]x=s+\frac{\omega}{k}t[/tex]

and this simply shows what the propagation velocity of our signal is..

Last edited:

- #3

asdf1

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"But that means, that the signal value can be regarded as MOVING ALONG THE POSITIVE X-DIRECTION WITH VELOCITYw/k !"

can you explain that a little clearer?

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