Speed of wave from wave equation.

[back2square1]

Homework Statement

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

Homework Equations

\tilde { f } (x,t)=\tilde { A } { e }^{ i(kx-\omega t) } and
\omega=kv
\tilde { f } is complex wave function.
\tilde { A } is complex amplitude.
k is the wave number.
\omega is angular frequency.
v 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 \omega and k. Once we find the expression for \omega and k, its not very difficult to find the velocity using the 2nd equation.

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

 

[Mandelbroth]

Homework Statement

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

Your wave function is a function of position and time. What is speed, with respect to these two variables?

 

[TSny]

\tilde { f } (x,t)=\tilde { A } { e }^{ i(kx-\omega t) }

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)?

 

[back2square1]

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?

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?

 

[voko]

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.

 

[TSny]

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.