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

1. Oct 12, 2013

### NihalSh

1. The problem statement, all variables and given/known data
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.

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

3. 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.

Last edited: Oct 12, 2013
2. Oct 12, 2013

### Saitama

Hi NihalSh!

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. Oct 12, 2013

### NihalSh

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 :

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 $\frac{1}{v^2}$

$\frac{∂}{∂x}.\frac{∂y}{∂x}= \frac{∂^2y}{(∂x)^2} =\frac{∂^2y}{∂x^2}$
I think it makes no difference, I used a rather unconventional notation but both are correct I believe. I'll be careful next time

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. Oct 12, 2013

### Saitama

I am still not able to understand.

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.

5. Oct 12, 2013

### NihalSh

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?!?!?!?

Last edited: Oct 12, 2013
6. Oct 12, 2013

### Saitama

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 $y(x,t)$.

7. Oct 12, 2013

### Tanya Sharma

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.

8. Oct 12, 2013

### Saitama

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

9. Oct 12, 2013

### NihalSh

I appreciate your help but I haven't got a clue where I am being led!:tongue:
It doesn't seem reasonable.

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. Oct 12, 2013

### Tanya Sharma

.I think Nihal has figured it out.

11. Oct 12, 2013

### nasu

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....

12. Oct 12, 2013

### NihalSh

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. Oct 12, 2013

### NihalSh

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.

Last edited by a moderator: Sep 25, 2014
14. Oct 12, 2013

### Tanya Sharma

: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. Oct 12, 2013

### NihalSh

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

Last edited: Oct 12, 2013
16. Oct 12, 2013

### NihalSh

Similar questions I found in other textbooks can also be solved by this method:

Answer: $y=\frac{6}{(x-4t)^2+3}$

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. Oct 12, 2013

### Tanya Sharma

Excellent

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

18. Oct 12, 2013

### NihalSh

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).

19. Oct 12, 2013

### Tanya Sharma

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. Oct 12, 2013

### NihalSh

$x→x-vt$

$y(x,t)$ 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 $v=+\frac{1}{2}$