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Find wave speed by using relation x-y at given time.

  • Thread starter NihalSh
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Homework Statement


The displacement due to a wave moving in the positive x-direction is given by ##y=\frac{1}{1+x^2}## at time t=0 s and by ##y=\frac{1}{1+(x-1)^2}## at t=2 s, where x and y are in meters. Find the velocity of the wave in m/s.


Homework Equations


## \frac{∂^2y}{(∂x)^2}=\frac{1}{v^2}.\frac{∂^2y}{(∂t)^2}##


The Attempt at a Solution


Since we have to deal with partial derivative wrt x on L.H.S., we can simply take the derivative of given x-y relations.
For case, t=0 we have:

##y=\frac{1}{1+x^2}##

or ##y.(1+x^2)=1##

differentiating on both sides wrt x, we get

##(1+x^2).\frac{∂y}{∂x}=-2.x.y##

differentiating again and substituting value of y, we get

## \frac{∂^2y}{(∂x)^2}=\frac{2.(3.x^2-1)}{(1+x^2)^3}##

Similarly for case t=2, we get:

## \frac{∂^2y}{(∂x)^2}=\frac{2.(3.(x-1)^2-1)}{(1+(x-1)^2)^3}##

The only common factor they seem to have is 2 so that must be equal to ##\frac{1}{v^2}##.

##\frac{1}{v^2}=2##
that means ##v= \sqrt{\frac{1}{2}} m/s≈0.71 m/s##
But my book says the answer is ##v=0.5 m/s##, I haven't got a clue what went wrong or how to approach it with some other method.

Any help would be greatly appreciated, thanks.
 
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Answers and Replies

  • #2
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Hi NihalSh!

The only common factor they seem to have is 2 so that must be equal to ##\frac{1}{v^2}##.

##\frac{1}{v^2}=2##
I can't follow this.

Anyways, you have a notation problem in your equation. ##\frac{∂^2y}{(∂x)^2}## is not same as ##\frac{∂^2y}{∂x^2}## and I think this is what you mean.

To solve the problem, you need to find ##y(x,t)##. Find the partial derivatives and substitute in the linear wave equation to find v.
 
  • #3
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I can't follow this.
from my point of view (as explained in the textbook), the velocity in a given material doesn't change with time unless something physical is being changed like tension. so I compared the partial derivative with the equation mentioned :

## \frac{∂^2y}{∂x^2}=\frac{1}{v^2}.\frac{∂^2y}{∂t^2}##
this part ,##\frac{∂^2y}{∂t^2}## will change in both cases but the factor ##\frac{1}{v^2}## remains the same and it should be constant. so I did equated the constant factor with ##\frac{1}{v^2}##

Anyways, you have a notation problem in your equation. ##\frac{∂^2y}{(∂x)^2}## is not same as ##\frac{∂^2y}{∂x^2}## and I think this is what you mean.
##\frac{∂}{∂x}.\frac{∂y}{∂x}= \frac{∂^2y}{(∂x)^2} =\frac{∂^2y}{∂x^2}##
I think it makes no difference, I used a rather unconventional notation but both are correct I believe. I'll be careful next time

To solve the problem, you need to find ##y(x,t)##. Find the partial derivatives and substitute in the linear wave equation to find v.
I have no clue how to proceed, the wave could have any shape since its not mentioned that it's a sinusoidal function. Please show how to proceed further.
 
  • #4
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so I compared the partial derivative with the equation mentioned :

this part ,##\frac{∂^2y}{∂t^2}## will change in both cases but the factor ##\frac{1}{v^2}## remains the same and it should be constant. so I did equated the constant factor with ##\frac{1}{v^2}##
I am still not able to understand.

I have no clue how to proceed, the wave could have any shape since its not mentioned that it's a sinusoidal function. Please show how to proceed further.
You need to find ##y(x,t)## which satisfies the given conditions. Do you see it has to be of the following form:
$$y(x,t)=\frac{1}{1+(x-kt)^2}$$
Look for an appropriate value of k (where k is some constant) and find the partial derivatives.
 
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  • #5
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You need to find ##y(x,t)## which satisfies the given conditions. Do you see it has to be of the following form:
$$y(x,t)=\frac{1}{1+(x-kt)^2}$$
Look for an appropriate value of k (where k is some constant) and find the partial derivatives.
I'll give your method a try and then post the results.

Edit : I gave it a try, but it is turning out to be to complex. The question was a MCQ. This is as far as I could get.

if I assume the wave has equation of the form,

##y(x,t)=\frac{1}{1+(x-kt)^2}## here k should be 1/2 by trial and error.

$$\frac{∂^2y}{∂x^2}=\frac{2.(3.(x-kt)^2-1)}{(1+(x-kt)^2)^3}$$

$$\frac{∂^2y}{∂t^2}=\frac{2.((x-kt)^2-1)}{(1+(x-kt)^2)^2}$$

if I proceed further by taking ##k=\frac{1}{2}##, then

##4v^2=\frac{x^4-1}{3x^2-1}## for t=0

##4v^2=\frac{(x-1)^4-1}{3(x-1)^2-1}## for t=2

But ##v## should be constant, right?....but here it is a function of x. Did I do something wrong?.....or did I miss something crucial?.....or maybe this isn't the way?!?!?!?:confused:
 
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  • #6
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$$\frac{∂^2y}{∂t^2}=\frac{2.((x-kt)^2-1)}{(1+(x-kt)^2)^2}$$
Doesn't look right to me.
##4v^2=\frac{x^4-1}{3x^2-1}## for t=0

##4v^2=\frac{(x-1)^4-1}{3(x-1)^2-1}## for t=2
No, that's not right. Don't substitute the values, you have to substitute the expressions for derivatives directly into the linear wave equation. t=0 and t=2 were given in the problem statement so that you can deduce ##y(x,t)##.
 
  • #7
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Hi NihalSh :smile:

You are putting way too much effort than is actually required .

I will try and help you out.

Do you understand that wave equation is a function of both x and t and not x alone i.e y=f(x,t) .

First try and express y as a function of x and t i.e y=f(x,t) .A little hit and trial will be okay , no calculus required .

Let me know what you get.
 
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  • #8
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First try and express y as a function of x and t i.e y=f(x,t) .A little hit and trial will be okay , no calculus required :wink:.
Sounds interesting, I am curious as to how you would do that without calculus.
 
  • #9
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Doesn't look right to me.

No, that's not right. Don't substitute the values, you have to substitute the expressions for derivatives directly into the linear wave equation. t=0 and t=2 were given in the problem statement so that you can deduce ##y(x,t)##.
I appreciate your help but I haven't got a clue where I am being led!:tongue:
It doesn't seem reasonable.

Hi NihalSh :smile:

You are putting way too much effort than is actually required .

I will try and help you out.

Do you understand that wave equation is a function of both x and t and not x alone i.e y=f(x,t) .

First try and express y as a function of x and t i.e y=f(x,t) .A little hit and trial will be okay , no calculus required .

Let me know what you get.
I know, right!!! This question is an MCQ for god's sake. But I've managed to figure it out, thanks to Walter Lewin. I'll post the solution in a minute for anyone's future reference.
 
  • #10
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Sounds interesting, I am curious as to how you would do that without calculus.
:smile: .I think Nihal has figured it out.
 
  • #11
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You don't need any derivatives. Anyway you don't have y as a function of t so you cannot calculate derivatives in respect to t.

If you take an arbitrary point at distance x1, the displacement at t=0 is y1=1/(1+x1^2)
At t=2, a point with the same displacement, y1, will have to be at a distance x1+1.
So in 2 seconds the point with the same displacement moves by 1 unit.
The wave speed is then....
 
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  • #12
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Sounds interesting, I am curious as to how you would do that without calculus.
I can now guarantee you since I have figured it out, it can be done without trial and error and without calculus. There is an appropriate way.
 
  • #13
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If you want to change a function (ex. ##y=x^2 +1##) to traveling/moving function. Then all we have to do is replace ##x→x-vt##, where ##v## is the velocity and ##t## is the time.

Since I have been given two value of ##y##, first at t=0. This value is simply the function which I want to convert to traveling function. So replacing ##x→x-vt##, in the given t=0 case.

I get ##y=\frac{1}{1+(x-vt)^2}##, then comparing with the second case I simply get:

##x-vt=x-1## here t=2, so

##v=+\frac{1}{2}##

+sign indicates the function travels in the +ve ##x## direction.

P.S. : Source first 2-3 minutes of the video explains the method.
thanks to everyone for helping out!!!
I know the solution looks messy, but it is very very easy.
 
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  • #14
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:thumbs:

Just check if you can apply the same technique if y=1/[1+(1-x)2] at t=2 instead of what is given in the original question.
 
  • #15
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:thumbs:

Just check if you can apply the same technique if ##y=\frac{1}{1+(1-x)^2}## at t=2 instead of what is given in the original question.
for the technique to apply we should know what ##y## is at t=0. Otherwise, it becomes trial and error.

since ##(1-x)^2## is essentially same as ##(x-1)^2## so it should be same for ##y=\frac{1}{1+(1-x)^2}## and ##y=\frac{1}{1+(x-1)^2}##

assuming ##y=\frac{1}{1+x^2}## at t=0, then it simply follows replace ##x→x-vt##.

the answer would still come out to be ##v=+\frac{1}{2}##

Edit: Alternate method:

##y=\frac{1}{1+x^2}=\frac{1}{1+(-x)^2}## at t=0 and ##y=\frac{1}{1+(1-x)^2}## at t=2

##x→x-vt##

##-x→vt-x##

Solving gives same answer, that is ##v=+\frac{1}{2}##
 
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  • #16
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Similar questions I found in other textbooks can also be solved by this method:

The displacement function of a wave travelling along positive x-direction is ##y=\frac{1}{2+3x^2}## at t=0 s and by ##y=\frac{1}{2+3(x-2)^2}## at t=2 s, where y and x are in meters. Find the velocity of the wave.
Answer: +1 m/s

If at t=0 s, a travelling wave pulse on a string is described by the function ##y=\frac{6}{x^2+3}##. What will be the wave function representing the pulse at time t, if the pulse is propagating along positive x-axis with speed 4 m/s?
Answer: ##y=\frac{6}{(x-4t)^2+3}##

At t=0 s, a transverse wave pulse in wire is described by the function ##y=\frac{6}{x^2+3}## where x and y are in meters. Write the function y(x,t) that describes this wave if it is travelling in the positive x-direction with a speed of 4.50 m/s.
Answer: ##y(x,t)=\frac{6}{(x-4.50t)^2+3}##

P.S. : It works like a charm every time because as I found out it is among the very basic equations normally using to derive wave equations from oscillations/wave pulse.
 
  • #17
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Excellent :smile:

What about y(x,t=0) = 1/(1+x3) and y(x,t=2) = 1/(1+(1-x)3) ?
 
  • #18
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Excellent :smile:

What about ##y(x,t=0) = \frac{1}{1+x^3}## and ##y(x,t=2) = \frac{1}{1+(1-x)^3}## ?
They couldn't be of the same wave function. I believe they both are part of different wave functions. Hence, my method doesn't apply.:wink:

P.S. : ##y(x,t)## function of ##y(x,t=0) = \frac{1}{1+x^3}## is

##y(x,t) = \frac{1}{1+(x-vt)^3}##

which doesn't match the second given function.

I believe you are just testing the limit of method's applicability, but I can assure it would work for every real case (unlike this one).
 
  • #19
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Sorry...that was a typo

I meant y(x,t=0) = 1/(1-x3) ,not y(x,t=0) = 1/(1+x3) .

I am not testing the method.Just trying to understand this technique by applying different functions.
 
  • #20
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##x→x-vt##

##y(x,t)## function of ##y(x,t=0) = \frac{1}{1-x^3}## is

##y(x,t) = \frac{1}{1-(x-vt)^3}##

since ##y(x,t=2) = \frac{1}{1+(1-x)^3}=\frac{1}{1-(x-1)^3}##

solving them would give ##v=+\frac{1}{2}##
 

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