Help with transverse wave motion question

[ruku320]

Trying to start my homework and stuck on this first problem...

A point mass M is concentrated at a point on a string of characteristic impedance pc. A transverse wave of frequency w 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 and that the difference between the transverse forces just to the left and right of the mass equal the mass times its acceleration.

Then the problem goes on to stating variables for the wave amplitudes and how I have to show that the reflected amplitude ratio and the transmitted amplitude ratio equal these values that it set.

Okay, the problem I seem to be having with this problem is the part where it says "the difference between the transverse forces just to the left and right of the mass equal the mass times its acceleration". As I tried to work out the boundary conditions, it seems to be to be saying that T(d/dx(yi + yr)) - T(d/dx(yt)) = Ma where T is tension, d/dx is partial differentation with respect to x, yi is the incident wave, yr is the reflected wave, yt is the transmitted wave and a is acceleration of the mass. The part which I seem to be stuck on is the Ma part. How exactly am I suppose to determine what the acceleration of the mass is in terms of the wave equations? It can't just be the second derivatives of my wave equations since those are the accelerations of the waves themselves and not of the mass. Is there some relation here that I'm missing or am I totally off in writing down the equation for the boundary condition?

 

[CarlB]

For transverse waves, Ma is just the second derivative of y with respect to t. [Forgive me, the LaTex was ugly.] This applies at the mass point just like anywhere else.

Maybe what's bothering you is that you're not taking note of the fact that the above force and derivative is valid only at the mass point. Thus it is not the wave itself, just the motion of the mass point.

Carl

 

[quicknote]

Dear student,

Thank you for reaching out for help with your homework problem. It seems like you are on the right track but are struggling with understanding the boundary conditions for the transverse wave motion. Let me try to explain it in a simpler way.

First, let's define the variables:

- T: tension in the string
- x: position along the string
- y: displacement of the string at position x
- M: mass at a point on the string
- w: frequency of the transverse wave
- pc: characteristic impedance of the string
- yi: incident wave
- yr: reflected wave
- yt: transmitted wave

Now, the boundary conditions given state that the displacements just to the left and right of the mass (at position x) are equal. This means that y(x) = y(x-δx) = y(x+δx), where δx is a very small distance. This can also be written as d/dx(y) = 0, which is the first boundary condition.

The second boundary condition states that the difference between the transverse forces just to the left and right of the mass is equal to the mass times its acceleration. In other words, the net force acting on the mass is equal to its mass times its acceleration. Mathematically, this can be written as T(d/dx(yi + yr)) - T(d/dx(yt)) = Ma, where a is the acceleration of the mass.

To determine the acceleration of the mass, we need to use the wave equations. In a transverse wave, the displacement y at position x and time t can be described by the equation y(x,t) = A sin(kx - wt), where A is the amplitude, k is the wave number, and w is the angular frequency. The wave number is related to the frequency and the characteristic impedance by the equation k = w/pc.

Now, to find the acceleration of the mass, we need to take the second derivative of the wave equation with respect to time, since acceleration is the second derivative of displacement with respect to time. This gives us a = -Akw^2 sin(kx - wt). Substituting this into our boundary condition equation, we get T(d/dx(yi + yr)) - T(d/dx(yt)) = -Mkw^2 sin(kx - wt).

Solving for a, we get a = -Tk/M sin