When Is the Superposition of These Plane Waves Zero?

[GreenPrint]

Homework Statement

Two plane waves are given by

$E_{1} = \frac{5E_{0}}{((3 \frac{1}{m})x - (4 \frac{1}{s})t)^{2} + 2}$ and
$E_{2} = -\frac{5E_{0}}{((3 \frac{1}{m})x + (4 \frac{1}{s})t)^{2} - 6}$

a) Describe the motion of the two waves.
b) At what instant is their superposition everywhere zero?
c) At what point is their superposition always zero?

Homework Equations

The Attempt at a Solution

For part b)
$E_{R} = E_{1} + E_{2} = 5E_{0}(\frac{1}{((3 \frac{1}{m})x - (4 \frac{1}{s})t)^{2} + 2} - \frac{1}{((3 \frac{1}{m})x + (4 \frac{1}{s})t)^{2} - 6}) = 0$

Since I'm looking for the instant in which the superposition is zero I set x = 0 and solve for t

$\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 36 + 2} = 0$
$\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 38} = 0$
$\frac{1}{16t^{2} + 2} = \frac{1}{16t^{2} - 48t + 38}$
$\frac{16t^{2} - 48t + 38}{16t^{2} + 2} = 1$
$16t^{2} - 48t + 38 = 0$
This provides me with an imaginary time. I'm not exactly sure what it is I'm doing wrong here. Thanks for any help you can provide me.

 

[BOYLANATOR]

x is not the superposition of the waves here. It is just the spatial coordinate or position.

 

[GreenPrint]

So how would I go about solving the problem then?

 

[BOYLANATOR]

Keep the x in your expression for the resultant amplitude and see if you can find some sort of simplified expression for t.
Why did the s term disappear in your working?

 

[GreenPrint]

I thought that the $\frac{1}{s}$ and $\frac{1}{m}$ where units? So I removed them to make my expressions less messy. When I leave x and t in the expression and solve I get

$\frac{16t^{2} - 24tx + 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1$

I'm not sure how that really helps or how I'm supposed to simplify this further. Thanks for the help.

 

[BOYLANATOR]

I'm not sure how to solve this. Someone else may have to help. In your original solution you can multiply both sides by the denominator of the LHS to get a real time but I still don't think setting x=0 is the right thing to do.

Since I'm looking for the instant in which the superposition is zero I set x = 0 and solve for t

$\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 36 + 2} = 0$
$\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 38} = 0$
$\frac{1}{16t^{2} + 2} = \frac{1}{16t^{2} - 48t + 38}$
$\frac{16t^{2} - 48t + 38}{16t^{2} + 2} = 1$

This provides me with an imaginary time. I'm not exactly sure what it is I'm doing wrong here. Thanks for any help you can provide me.

 

[BOYLANATOR]

I thought that the $\frac{1}{s}$ and $\frac{1}{m}$ where units? So I removed them to make my expressions less messy. When I leave x and t in the expression and solve I get

$\frac{16t^{2} - 24tx + 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1$

I'm not sure how that really helps or how I'm supposed to simplify this further. Thanks for the help.

When I leave x in I do get a neat answer (although I'm not sure if it answers the question). Try the algebra again and if you reach a point similar to this again - multiply by the denominator.

 

[GreenPrint]

I start off with this
$E = E_{1} + E_{2} = \frac{5E_{0}}{(3x - 4t)^{2} + 2} - \frac{5E_{0}}{(3x + 4t - 6)^{2} + 2} = 0$
I factor
$E = E_{1} + E_{2} = 5E_{0}(\frac{1}{(3x - 4t)^{2} + 2} - \frac{1}{(3x + 4t - 6)^{2} + 2} = 0$
I expand
$5E_{0}(\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}) = 0$
simplify
$\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38} = 0$
move second term to right hand side
$\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} = \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}$
multiply both sides by the denominator of the RHS term
$\frac{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1$
I set the numerator equal to zero
$16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38 = 0$

And from here it doesn't look like I can solve for a real answer. I don't see what it is I'm doing wrong here.

The back of my book says that the answer is
$t = \frac{3}{4} s$

I find it interesting that

$|\frac{∂(16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38)}{∂t}| = \frac{3}{4}$

I'm not sure if this is just a coincidence or not though.

 

[BOYLANATOR]

Looking at your first line you have a -6 inside the squared brackets. This is not what you showed originally. Did you make a mistake in your first post?

 

[BOYLANATOR]

You must have written it wrong originally as it now works.

I start off with this
$E = E_{1} + E_{2} = \frac{5E_{0}}{(3x - 4t)^{2} + 2} - \frac{5E_{0}}{(3x + 4t - 6)^{2} + 2} = 0$
I factor
$E = E_{1} + E_{2} = 5E_{0}(\frac{1}{(3x - 4t)^{2} + 2} - \frac{1}{(3x + 4t - 6)^{2} + 2} = 0$
I expand
$5E_{0}(\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}) = 0$
simplify
$\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38} = 0$
move second term to right hand side
$\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} = \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}$
multiply both sides by the denominator of the RHS term
$\frac{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1$
I set the numerator equal to zero
$16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38 = 0$

And from here it doesn't look like I can solve for a real answer. I don't see what it is I'm doing wrong here.

Setting the numerator to zero is wrong. You have already set the displacement to zero at the start so now you just need to solve it as it appears. Remember what I said to do earlier when you reached a point like this?

 

[nasu]

You don't need to do all this common denominator thing.

As the two fractions have the same numerator (Eo) it is sufficient to equate their denominators in order to have zero amplitude.