Determining the ratio of wave amplitudes along a string

[rmjmu507]

Homework Statement

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 (y_i+y_r=y_t) and that the difference between the transverse forces just to the left and right of the masses equal the mass times its acceleration. If A_i, A_r, and A_t are respectively the incident, reflected, and transmitted wave amplitudes show that

[itex]\frac{A_r}{A_i}[/itex]=$\frac{-iq}{1+iq}$
[itex]\frac{A_t}{A_i}[/itex]=$\frac{1}{1+iq}$

Where q=$\frac{Mω}{2ρc}$ and i²=-1.

Homework Equations

See above

The Attempt at a Solution

So the first boundary condition imposes the following: A_i+A_r=A_t

Let T be the tension, and let y_i,r,t(x,t)=A_i,r,te^i(ωt-kx)

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

Tke^iωt(-A_i+A_r+A_t)=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

 

[TSny]

let y_i,r,t(x,t)=A_i,r,te^i(ωt-kx)

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

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

Tke^iωt(-A_i+A_r+A_t)=Ma

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

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.

Can you express the acceleration of M in terms of A_i, A_r, and A_t? 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.

 

[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

 

[TSny]

...is it that a=the second derivative of the position...thus the acceleration of the mass=...

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.

 

[rmjmu507]

let y=(A_i-A_r+A_t)e^iωt

Then the second time derivative is

a=i²ω²(A_i-A_r+A_t)e^iωt

So plugging in, I get

Tk(-A_i+A_r-A_t)=Miω²(A_i-A_r+A_t)

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

 

[TSny]

let y=(A_i-A_r+A_t)e^iωt

Then the second time derivative is

a=i²ω²(A_i-A_r+A_t)e^iωt

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

So plugging in, I get

Tk(-A_i+A_r-A_t)=Miω²(A_i-A_r+A_t)

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

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.

 

[rmjmu507]

I thought that because A_r 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:

y_i=A_ie^i(ωt-kx)
y_r=A_re^i(ωt+kx)
y_t=A_te^i(ωt-kx)

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

[itex]\frac{∂y_i}{∂x}[/itex]=-ikA_ie^i(ωt-kx)
[itex]\frac{∂y_r}{∂x}[/itex]=ikA_re^i(ωt-kx)
[itex]\frac{∂y_t}{∂x}[/itex]=-ikA_te^i(ωt-kx)

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

[itex]\frac{∂y_i}{∂x}[/itex](x=0)=-ikA_ie^iωt
[itex]\frac{∂y_r}{∂x}[/itex](x=0)=ikA_re^iωt
[itex]\frac{∂y_t}{∂x}[/itex](x=0)=-ikA_te^iωt

So T*k*i*e^iωt(-A_i+A_r-A_t=Ma

Where do you see a sign mistake?

 

[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?

 

[rmjmu507]

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

Normally, I would say that:

T∂y_i/∂x + T∂y_r/∂x = T∂y_t/∂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∂y_i/∂x + T∂y_r/∂x - T∂y_t/∂x=Ma

Is this incorrect?

 

[TSny]

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

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.)

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∂y_i/∂x + T∂y_r/∂x - T∂y_t/∂x=Ma

Is this incorrect?

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 T_1y is negative; and when ∂y₂/∂x is positive, then T_2y is positive. Here, "1" refers to the string on the left of M and "2" refers to the string on the right of M.

 

[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*e^iωt(A_i-A_r+A_t)=M*i²*ω²(A_i-A_r+A_t)*e^iωt

Cancelling terms,

T*k*(A_i-A_r+A_t)=M*i*ω²(A_i-A_r+A_t)

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.

 

[TSny]

T*k*(A_i-A_r+A_t)=M*i*ω²(A_i-A_r+A_t)

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, y_i + y_r, or equivalently, it's the displacement of the right piece of string, y_t. The displacement of the mass is not the sum of all 3 displacements y_i, y_r, and y_t.

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

$\sum$F_y = Ma.

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

F_y due to string 2 is +T∂y₂/∂x = T∂y_t/x.

 

[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?

 

[TSny]

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?

In Newton's second law, $\sum$F_y is the sum of all the y-forces acting on M. Since, generally, the sign of F_y is positive on one side of M and negative on the other, you can also say that $\sum$F_y 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 F_net,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 F_net,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).