# Transverse Oscillations with 3 beads

1. Nov 17, 2015

### spacetimedude

1. The problem statement, all variables and given/known data
Consider a light elastic string of unstretched length 4a0, stretched horizontally between two fixed points distance 4a apart (a>a0). There are particles of mass m attached so as to divide the string into four equal sections. We enumerate the segments from left to right, i=1 through 4. The tension Ti is proportional to its extension (a-a0), with the elastic constant being c>0.
Suppose that the particles can only move in a perpendicular direction.
Write down the equations of motion for the vertical displacements xi under the assumption that the displacements are small. Keep only linear terms in xi/a. Show that in this approximation the equation takes the form: x''+n^2Ax=0 and determine the constant coefficient n and the numerical matrix A.

2. Relevant equations

3. The attempt at a solution
First, I' d like to know if we can assume that the tensions are equal since the displacements are small.
If so, for each mass with amplitude xi, I can set up the equation of motion.

mx1''=-T*x1/a+T*(x2-x1)/a
mx2''=-T(x2-x1)/a+T(x3-x2)/a
mx3''=-T(x3-x2)/a-T*x3/a

Rearranging the equation and setting n^2=T/ma:
x1''+2n^2x1-n^2x2=0
x2''-n^2+2n^2x2-n^2x3=0
x3''-n^2x2+2n^2x3=0

which is in the form we wanted. Factoring out the n^2, we can easily find the matrix A.
In class, our professor told our class that tension T is approximately -k(xi-xi-1) by taylor expanding and that li=√[(xi-xi-1)+a^2] (l is the hypotenuse of the triangle created by the oscillation). Why are these equations relevant to the question?

The next question asks us to diagonalise A and find the general solution for x(t) but I do not want to proceed unless if I am sure that my work so far is correct.

Any comment will be appreciated!

EDIT: It seems like I get the same solution when we find Ti where Ti=xi-xi-1-(a-a0), which leads to the equation of motion mx1''=T1-T2, mx2''=T2-T3, and mx3''=T3-T4

Last edited: Nov 17, 2015
2. Nov 19, 2015

### Geofleur

When the masses are displaced, the string becomes longer than $4a$. The new length is $l_1 + l_2 + l_3 + l_4$. But you can write this new length in terms of displacements divided by $a$ and Taylor expand. Keeping the terms linear in these ratios gives the approximate expression for the new length.

3. Nov 20, 2015

### Staff: Mentor

In my judgement, you set the problem up correctly by recognizing that the directions of the tensions changed but, to a good approximation, not their magnitudes. I think your professor wanted you to recognize that there was a geometric approximation involved in doing what you did when you essentially replaced the sine of an angle with the tangent of the angle (for a small angle).

Chet

4. Nov 20, 2015

### Geofleur

I agree - I just wanted to make sure the OP realized that the Taylor expanded length reduces to $4a$ again.