Wave problem - when they meet given their equations of motion

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Part of question

So what I did was made the two equations equal each other. A lot of stuff cancels out and I end up with x=-vt. My issue is that t isn't given and I am not entirely sure how to get it. I don't think taking the partitial derivative of time will be any help nor the partial derivative of displacement.

I got t=1/2v by ignoring the first half of the exponent and just making -vt=vt, but this doesn't really look right.

 

[haruspex]

Homework Statement:: See attached
Relevant Equations:: Part of question

what I did was made the two equations equal each other

Doesn't sound like the right thing to do. Why would that be where their sum is a maximum?

 

[JoeyBob]

Doesn't sound like the right thing to do. Why would that be where their sum is a maximum?

Well the question says where the two wave pulses meet. I am assuming waves that meet can either have a damping or amplification effect on each other. As the question says, "where the two wave pulses meet"

 

[haruspex]

Well the question says where the two wave pulses meet. I am assuming waves that meet can either have a damping or amplification effect on each other. As the question says, "where the two wave pulses meet"

But these are isolated pulses, not persistent sine waves, so meeting means their peaks coinciding in position at some instant.
Indeed, it also provides an alternate wording, but it's perhaps still not very clear: find where the maximum pressure occurs "over all times".

 

[JoeyBob]

But these are isolated pulses, not persistent sine waves, so meeting means their peaks coinciding in position at some instant.
Indeed, it also provides an alternate wording, but it's perhaps still not very clear: find where the maximum pressure occurs "over all times".

Idk man, I hate this chapter so much. I figured out how to do the question though. First Ill show a way I feel like I am suppose to do it. Then Ill show a way that seems completely wrong yet gave me the right answer.

Way that looks right:

So when I make the two equations equal each other a lot of stuff cancels and I end up with this overall

(x-vt)^2=(x-d+vt)^2

Now when I expand these things and cancel stuff I am left with

0=d^2-2xd-2dvt.

The issue here is I have two unknown (time and displacement). So I am stuck.

Way that seems completely wrong yet gave me the right answer:

This is dumb but I decided to try dropping the exponent so I am left with

x-vt=x-d+vt

Stuff cancels and I am left with

t=d/v

Now I plug in t into (x-vt)^2=(x-d+vt)^2 (added the exponents back)

It for whatever reason gives me the right answer. This seems really stupid. Furthermore, the next question asks when they meet, and guess what, my t value is now incorrect. What fixes the t value for the next question? Dividing it by 2 for whatever reason.

 

[haruspex]

Idk man, I hate this chapter so much. I figured out how to do the question though. First Ill show a way I feel like I am suppose to do it. Then Ill show a way that seems completely wrong yet gave me the right answer.

Way that looks right:

So when I make the two equations equal each other a lot of stuff cancels and I end up with this overall

(x-vt)^2=(x-d+vt)^2

Now when I expand these things and cancel stuff I am left with

0=d^2-2xd-2dvt.

The issue here is I have two unknown (time and displacement). So I am stuck.

Way that seems completely wrong yet gave me the right answer:

This is dumb but I decided to try dropping the exponent so I am left with

x-vt=x-d+vt

Stuff cancels and I am left with

t=d/v

Now I plug in t into (x-vt)^2=(x-d+vt)^2 (added the exponents back)

It for whatever reason gives me the right answer. This seems really stupid. Furthermore, the next question asks when they meet, and guess what, my t value is now incorrect. What fixes the t value for the next question? Dividing it by 2 for whatever reason.

As I hinted, merely setting them equal has no rational basis. E.g., suppose you were to double the amplitude of one. That won't affect when and where they meet, but will surely affect when they are equal.

The sum of the two waves is a function of (x,t). You are looking for the peak of this in the (x,t) plane. How do you solve that?

 

[JoeyBob]

As I hinted, merely setting them equal has no rational basis. E.g., suppose you were to double the amplitude of one. That won't affect when and where they meet, but will surely affect when they are equal.

The sum of the two waves is a function of (x,t). You are looking for the peak of this in the (x,t) plane. How do you solve that?

but then why does it give the right answer when I make them equal, drop the exponents to find t, then add the exponents back and use t to find x? Doesnt that process (which gives the right answer) sound super dumb? I feel like I am getting dumber just thinking about how that solved the question.

Anyways, I guess youre saying that when you add the two waves you get the total effect on the location. So if the sum is a function (x, t), where I add the two equations, how does that help me get x? t is still a variable there after all. Does it involve the use of the linear wave equations with the partial derivatives and all that?

I mean I am trying to read the part of the textbook that seems to talk about this sort of stuff, but unfortunately it uses cos/sin instead of exponents and all that. It also doesn't seem to go over something like this specific question.

 

[haruspex]

but then why does it give the right answer when I make them equal

It is a bit like the question "find the max value of x(L-x) wrt x".
Setting x=L-x gives the right answer, and in that example it's not hard to see why. The same happens in this question, except that it is much harder to see why.

Note that it does not work with x(L-2x). In short, you got lucky.

I feel it is important that you understand what the trustworthy procedure is,

x-vt=x-d+vt
Stuff cancels and I am left with
t=d/v

It gives me t=d/(2v). I assume the above is a typo.

drop the exponents to find t, then add the exponents back and use t to find x?

I didn't understand that part. Given the (invalid) path taken so far, you could legitimately have dropped the exponents writing: |x-vt|=|x-d+vt|; or, equivalently,
"Either x-vt=x-d+vt or x-vt=-(x-d+vt)"
giving two solutions.
Having chosen the x-vt=x-d+vt solution, you say you plugged that back into the equation with exponents and got the right answer, but when I do that I get (x-d/2)² = (x-d/2)². So maybe you really meant t=d/v.
Super lucky.

 

[JoeyBob]

[QUOTE
Having chosen the x-vt=x-d+vt solution, you say you plugged that back into the equation with exponents and got the right answer, but when I do that I get (x-d/2)² = (x-d/2)². So maybe you really meant t=d/v.
Super lucky.
[/QUOTE]

I did mean t=d/v. I guess I screwed that up too. Yet somehow all those mistakes gave the right answer.

I do want to solve it the right way though, but my problem with taking the sum is that I still have 2 unknown variables, position and time. Without either finding one the variables or finding another equation to represent them, I have no way of finding x.

Adding the two functions seem familiar to the method where I would find the natural oscillations of a two mass coupled system. But unknown variables didnt matter then because I was just finding w, which involved taking the sqare root of what was in front of (y1+y2) or (y1-y2) if that makes sense to you.

Or perhaps its nothing similar to the above and involves using the linear wave equation where that involves taking partial derivatives. But my notes only involve that when there's one system involved. I suppose if I am adding the equations, the max pressure would be where the amplitude is largest. Since the equations are added, this would be 2P. That would mean I would want each exponent to equal 0.

So this involves saying

-(x-vt)^2 / b^2 = 0 and -(x-d+vt)^2/b^2 = 0

Is this the right idea?

 

[haruspex]

the max pressure would be where the amplitude is largest. Since the equations are added, this would be 2P

In general there need not be any combination of x and t where both waves peak, but certainly if there is such a place then it must be the overall peak, so that's worth a try.

taking partial derivatives

That would be the standard approach. The overall (local) max of f(x,y) would be where both partial derivatives are zero. That gives the two equations you need for your two unknowns.

 

[JoeyBob]

In general there need not be any combination of x and t where both waves peak, but certainly if there is such a place then it must be the overall peak, so that's worth a try.

That would be the standard approach. The overall (local) max of f(x,y) would be where both partial derivatives are zero. That gives the two equations you need for your two unknowns.

Thanks for the help, I think I got it now.