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

## 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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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##

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.

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.

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.

1 person
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?!?!?!?

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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)##.

Hi NihalSh

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

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.

1 person
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 .
Sounds interesting, I am curious as to how you would do that without calculus.

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

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

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.

Sounds interesting, I am curious as to how you would do that without calculus.
.I think Nihal has figured it out.

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....

2 people
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.

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}##

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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1 person
: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.

: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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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.

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?

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.

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.

Excellent

What about y(x,t=0) = 1/(1+x3) and y(x,t=2) = 1/(1+(1-x)3) ?

Excellent

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.

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).

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.

##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}##