Speed of wave from wave equation.

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back2square1
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Homework Statement


A traveling pulse is given by [itex]f(x,t)=A{ e }^{ \frac { 2abxt-{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ t }^{ 2 } }{ { c }^{ 2 } } }[/itex] where [itex]A, a, b, c[/itex] are positive constants of appropriate dimentions. The speed of pulse is:
a) [itex]\frac { b }{ a }[/itex]
b) [itex]\frac { 2b }{ a }[/itex]
c) [itex]\frac { cb }{ a }[/itex]
d) [itex]\frac { b }{ 2a }[/itex]

Homework Equations


[itex]\tilde { f } (x,t)=\tilde { A } { e }^{ i(kx-\omega t) }[/itex] and
[itex]\omega=kv[/itex]
[itex]\tilde { f }[/itex] is complex wave function.
[itex]\tilde { A }[/itex] is complex amplitude.
[itex]k[/itex] is the wave number.
[itex]\omega[/itex] is angular frequency.
[itex]v[/itex] is the velocity of pulse.

The Attempt at a Solution


The strategy was reduce the equation into the standard wave equation and see which part corresponds with [itex]\omega[/itex] and [itex]k[/itex]. Once we find the expression for [itex]\omega[/itex] and [itex]k[/itex], its not very difficult to find the velocity using the 2nd equation.

The equation doesn't contain [itex]i[/itex]. But reducing the equation, we get something like this,
[itex]f(x,t)={ e }^{ { (i\frac { ax-bt }{ c } ) }^{ 2 } }[/itex]. The only problem is how to get rid of the square? Using logarithms wasn't helpful.
 
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back2square1 said:

Homework Statement


A traveling pulse is given by [itex]f(x,t)=A{ e }^{ \frac { 2abxt-{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ t }^{ 2 } }{ { c }^{ 2 } } }[/itex] where [itex]A, a, b, c[/itex] are positive constants of appropriate dimentions. The speed of pulse is:
a) [itex]\frac { b }{ a }[/itex]
b) [itex]\frac { 2b }{ a }[/itex]
c) [itex]\frac { cb }{ a }[/itex]
d) [itex]\frac { b }{ 2a }[/itex]
Your wave function is a function of position and time. What is speed, with respect to these two variables? :-p
 
back2square1 said:
[itex]\tilde { f } (x,t)=\tilde { A } { e }^{ i(kx-\omega t) }[/itex]
This equation is for a "sinusoidal" or "harmonic" wave that extends from -∞ to ∞.

More generally, any well-behaved function f(x,t) of the form f(x-vt) would represent a wave traveling in the positive x direction with speed v. (You do not need to have ##i## in the expression).

Can you express your function in the form f(x-vt)?
 
TSny said:
You do not need to have ##i## in the expression.

I don't understand. ##{ e }^{ i\theta }=cos(\theta )+i sin(\theta )## is the reason why we can write sinusoidal wave in Euler's from. Without ##i##, how can we say its a sinusoidal wave?

TSny said:
Can you express your function in the form f(x-vt)?
Expressing it as a function of ##x-vt##, it looks like this: ##f(x-vt)=A{ e }^{ -\frac { a }{ c } { (x-\frac { b }{ a } t) }^{ 2 } }##, which indicates that velocity of pulse is ##v=\frac { b }{ a }##. So, are you trying to say that is the answer?
 
back2square1 said:
I don't understand. ##{ e }^{ i\theta }=cos(\theta )+i sin(\theta )## is the reason why we can write sinusoidal wave in Euler's from. Without ##i##, how can we say its a sinusoidal wave?

Note that you are talking about "sinusoidal" waves. Non-sinusoidal waves do not have to be in that form, naturally. Yet they must still have a propagation velocity, so that can't depend on some property of trigonometric functions.

The only reason why we study sinusoidal waves extensively is because they are the simplest solutions of the wave equation and because practically any wave can be represented as an (infinite) sum of sinusoidal waves.
 
back2square1 said:
Expressing it as a function of ##x-vt##, it looks like this: ##f(x-vt)=A{ e }^{ -\frac { a }{ c } { (x-\frac { b }{ a } t) }^{ 2 } }##, which indicates that velocity of pulse is ##v=\frac { b }{ a }##. So, are you trying to say that is the answer?

Yes, that's essentially right. I think the factor of a/c should also be squared, but that doesn't change the answer.
 
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