# Determining the ratio of wave amplitudes along a string

1. Sep 14, 2014

### rmjmu507

1. The problem statement, all variables and given/known data
A point mass M is concentrated at a point on a string of characteristic impedance ρc. A transverse wave of frequency ω moves in the positive x-direction and is partially reflected and transmitted at the mass. The boundary conditions are that the string displacements just to the left and right of the mass are equal (yi+yr=yt) and that the difference between the transverse forces just to the left and right of the masses equal the mass times its acceleration. If Ai, Ar, and At are respectively the incident, reflected, and transmitted wave amplitudes show that

$\frac{Ar}{Ai}$=$\frac{-iq}{1+iq}$
$\frac{At}{Ai}$=$\frac{1}{1+iq}$

Where q=$\frac{Mω}{2ρc}$ and i2=-1.

2. Relevant equations
See above

3. The attempt at a solution
So the first boundary condition imposes the following: Ai+Ar=At

Let T be the tension, and let yi,r,t(x,t)=Ai,r,tei(ωt-kx)

Taking the partial derivatives w.r.t. x for all three of these, and evaluating at x=0, we end up with

Tkeiωt(-Ai+Ar+At)=Ma

I'm stuck on where to go from here...I can make the Z=T/v substitution, and then Z=ρc, but I'm still stuck with the exponential which I don't know how to deal with.

Any assistance greatly appreciated

Last edited: Sep 14, 2014
2. Sep 14, 2014

### TSny

This isn't quite correct for the reflected wave because the reflected wave travels toward -x.

I'm not sure that you got all the signs correct here. Also, did you drop a factor of i?

Can you express the acceleration of M in terms of Ai, Ar, and At? Hint: The displacement of the mass must always equal the total displacement of the string at the point where the mass is attached.

If you can do this, you will be able to handle the exponential factors.

3. Sep 14, 2014

### rmjmu507

Thanks for catching those errors - I've fixed them.

Could you explain your last point in a little more detail. Conceptually, it sort of makes sense to me...

the vertical displacement of the mass = the vertical displacement of the string at x=0 (assuming here that the mass is positioned at this coordinate).

however, I can't seem to formulate this in more mathematical terms...is it that a=the second derivative of the position...thus the acceleration of the mass=...

Sorry, I still can't follow

4. Sep 14, 2014

### TSny

Yes, the acceleration of the mass is the same as the acceleration of the string at that point. You should be able to write the displacement, y, of the string in terms of the 3 amplitudes A and their exponential factors. Then take second time derivative.

5. Sep 14, 2014

### rmjmu507

let y=(Ai-Ar+At)eiωt

Then the second time derivative is

a=i2ω2(Ai-Ar+At)eiωt

So plugging in, I get

Tk(-Ai+Ar-At)=Miω2(Ai-Ar+At)

Am I on the right track here? Do I just need to make substitutions now?

6. Sep 14, 2014

### TSny

Why the minus sign in front of Ar? Shouldn't the displacement of the string just be the sum of the contributions from each wave?

Yes, you are on the right track. I don't agree with all of your signs, including the signs on the left side of the equation.

7. Sep 14, 2014

### rmjmu507

I thought that because Ar is moving in the other direction, it's magnitude is opposite and therefore the - sign...is my logic incorrect?

Here are the equations I have for the incident, reflected, and transmitted waves:

yi=Aiei(ωt-kx)
yr=Arei(ωt+kx)
yt=Atei(ωt-kx)

So, taking the partial derivatives w.r.t. x in each case, I end up with:

$\frac{∂yi}{∂x}$=-ikAiei(ωt-kx)
$\frac{∂yr}{∂x}$=ikArei(ωt-kx)
$\frac{∂yt}{∂x}$=-ikAtei(ωt-kx)

All of these, evaluated at x=0, yield:

$\frac{∂yi}{∂x}$(x=0)=-ikAieiωt
$\frac{∂yr}{∂x}$(x=0)=ikAreiωt
$\frac{∂yt}{∂x}$(x=0)=-ikAteiωt

So T*k*i*eiωt(-Ai+Ar-At=Ma

Where do you see a sign mistake?

8. Sep 14, 2014

### TSny

Suppose the string attached to the right side of M has a positive slope at the location of M. What is the sign of the y-component of the tension force that acts on M due to this string?

Suppose the string attached to the left side of M also has a positive slope at the location of M. What is the sign of the y-component of the tension force that acts on M due to this string?

Did you get the same answer for these two questions? If not, did you take that into account when you were setting up the net force acting on M?

9. Sep 14, 2014

### rmjmu507

I realize that the slope across the interface (the mass) must be the same...

Normally, I would say that:

T∂yi/∂x + T∂yr/∂x = T∂yt/∂x across an arbitrary point on the string.

However, the problem states that the difference between the transverse forces just to the left and right of the masses equal the mass times its acceleration.

I interpret this as:

T∂yi/∂x + T∂yr/∂x - T∂yt/∂x=Ma

Is this incorrect?

10. Sep 14, 2014

### TSny

If the slope across the mass is the same, then the mass would have zero acceleration. Look at the attached figure. On the left, you can see that if the slope is the same on each side of M, then the y-components of the tension forces will cancel. So, the mass would have zero acceleration in this case. On the right, there is a sudden change in slope from one side of M to the other. This allows the net y-component of force to be nonzero and causes acceleration of the mass. (In the figure, I drew the slopes very large to make it easier to see. But for the wave equation to hold, the slope of the string should always be small.)

That looks good except I think that the signs on the left side need to be changed. From the figure you can see that when ∂y1/∂x is positive, then T1y is negative; and when ∂y2/∂x is positive, then T2y is positive. Here, "1" refers to the string on the left of M and "2" refers to the string on the right of M.

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11. Sep 15, 2014

### rmjmu507

I tried flipping the signs, but I'm still not able to derive the given ratios....

Here's what I have:

T*k*i*eiωt(Ai-Ar+At)=M*i22(Ai-Ar+At)*eiωt

Cancelling terms,

T*k*(Ai-Ar+At)=M*i*ω2(Ai-Ar+At)

I made the following substitutions: k=ω/v, Z=T/v, Z=ρc, and the substitutions necessary to have the desired ratios, but no luck...

Do you see where I am making an error?

Thanks for all your help so far.

12. Sep 15, 2014

### TSny

The right side of the equation is not quite right. The displacement of the mass is the same as the displacement of the left piece of string, yi + yr, or equivalently, it's the displacement of the right piece of string, yt. The displacement of the mass is not the sum of all 3 displacements yi, yr, and yt.

On the left side of your equation I think there is still a sign error.

$\sum$Fy = Ma.

Fy due to string 1 is -T∂y1/∂x, where y1 = yi + yr. The minus sign is due to the direction of the tension force of string 1 as shown in the figure that I posted.

Fy due to string 2 is +T∂y2/∂x = T∂yt/x.

13. Sep 15, 2014

### rmjmu507

But then this isn't the difference between the transverse forces just to the left and right of the mass, it's the sum, no?

14. Sep 15, 2014

### TSny

In Newton's second law, $\sum$Fy is the sum of all the y-forces acting on M. Since, generally, the sign of Fy is positive on one side of M and negative on the other, you can also say that $\sum$Fy is equivalent to the difference in magnitude of the force on one side and the force on the other side.

To me, it's less confusing to write Fnet,y as the sum of all forces with the possibility that some of the terms that you are summing could be negative. But, if you are careful, you should get the same result if you want to think of Fnet,y as the difference in the absolute values of the positive and negative terms. However, I think I'm more likely to make a sign error doing it that way (at least for this problem).