(Waves) superposition of 2 pulses along a rope.

[Lavabug]

Homework Statement

Two pulses of the form

e^-(x + 4)^2

e^-(x - 2)^2

travel in opposite directions along a tensed non dispersive rope. If the speed of propagation of both pulses is 2 cm/s, find the instant of time in which the wave's amplitude goes to zero, if it exists.

Homework Equations

I am supposed to replace x with x - vt & x + vt for each of the pulses, respectively (I would have never guessed), then set their sum equal to zero.

The Attempt at a Solution

I've got e^-(x -2t + 4)^2 + e^-(x +2t -2)^2 = 0
and need to find the value of t that satisfies the equality (if it exists, which it does btw)

I feel like I don't know enough about exponentials and logarithms to solve this. They're gaussian curves not complex exponentials so I can't expand them into cos and sin arguments and play around with those to see what gives me a 0.

I tried expanding the polynomial in the exponents but it didn't seem to get me anywhere. How do I solve this type of problem?

 

[diazona]

One thing you could do is think about it graphically. You know what a Gaussian pulse looks like on a graph, right? Imagine these two pulses moving toward each other and think about what will happen when they overlap.

Incidentally, how do you know a solution exists?

 

[Lavabug]

Because it was given to me by my professor. t = 1.5 satisfies that inequality.

I guess there's no other choice than to churn out the polynomials in the exponents?

 

[aim1732]

Exponential terms are always positive so two of them adding to zero should mean both of them are separately zero.That happens only when they are raised to power of negative infinity.

 

[Lavabug]

I thought of that actually, but it's not the answer, it asks for the instant of time where their sum cancels (1.5s).

 

[aim1732]

Two waves that are always above the x-axis for all t and x can never cancel.

 

[Lavabug]

Well I mentioned all of this to my prof since it seems to defy what I know of math and general physics but there is an answer to the question and that's not it. :(

 

[aim1732]

Well I think you forgot a negative sign for either of the curves because on plugging t=1.5 in the equation the two give the same expression.Now if you notice that would have given zero displacement for all possible x if on of them had negative amplitude.

 

[lepton5]

Homework Statement

Two pulses of the form

e^-(x + 4)^2

e^-(x - 2)^2

travel in opposite directions along a tensed non dispersive rope. If the speed of propagation of both pulses is 2 cm/s, find the instant of time in which the wave's amplitude goes to zero, if it exists.

Homework Equations

I am supposed to replace x with x - vt & x + vt for each of the pulses, respectively (I would have never guessed), then set their sum equal to zero.

The Attempt at a Solution

I've got e^-(x -2t + 4)^2 + e^-(x +2t -2)^2 = 0
and need to find the value of t that satisfies the equality (if it exists, which it does btw)

I feel like I don't know enough about exponentials and logarithms to solve this. They're gaussian curves not complex exponentials so I can't expand them into cos and sin arguments and play around with those to see what gives me a 0.

I tried expanding the polynomial in the exponents but it didn't seem to get me anywhere. How do I solve this type of problem?

maybe this way, just equate y1 and y2 to find condition where A = 0 like below,

$$e^{-(x_0 - 2t + 4)^2} = e^{-(x_0 + 2t - 2)^2}$$, if initially x₀ same for both pulse, then you get t = 1.5 s

 

[aim1732]

lepton5,isn't it the same thing I said.Maybe it is a typo.

 

[Lavabug]

maybe this way, just equate y1 and y2 to find condition where A = 0 like below,

$$e^{-(x_0 - 2t + 4)^2} = e^{-(x_0 + 2t - 2)^2}$$, if initially x₀ same for both pulse, then you get t = 1.5 s

Wow I can't believe it would boil down to just that, thanks!

I was thinking of adding the two pulses (of amplitude 1) and setting it to 0, but I couldn't take the logs of my equality to work with the expressions in the exponents, as I had a 0 on one side.

What I don't understand is: how does setting both gaussian curves equal to each other represent where/when they cancel each other out? Wouldn't the resulting amplitude be twice as high when they overlap?

 

[diazona]

What I don't understand is: how does setting both gaussian curves equal to each other represent where/when they cancel each other out? Wouldn't the resulting amplitude be twice as high when they overlap?

Yes, it would. I was hoping you'd see this and come to the conclusion yourself that the problem as you have stated it does not have a solution. Either your professor is wrong about there being an answer, or the problem statement as you've given it is not correctly constructed.

 

[lepton5]

well, actually it comes from $$\Delta y = y_1 - y_2 = 0$$.

it was based on your information that other wave move oppositely from other one, so the substraction is there.

 

[Lavabug]

I trust the prof doesn't do it intentionally but from experience it sounds a lot like the former. :(

 

[diazona]

well, actually it comes from $$\Delta y = y_1 - y_2 = 0$$.

it was based on your information that other wave move oppositely from other one, so the substraction is there.

That doesn't mean you subtract the value of one pulse from the other, though. That subtraction shows up in the exponent.