Transverse Oscillations with 3 beads

[spacetimedude]

Homework Statement

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 x_i/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.

Homework Equations

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 x_i, 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(x_i-x_i-1) by taylor expanding and that l_i=√[(x_i-x_i-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 T_i where T_i=x_i-x_i-1-(a-a₀), which leads to the equation of motion mx₁''=T1-T2, mx₂''=T2-T3, and mx₃''=T3-T4

 

[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.

 

[Chestermiller]

Homework Statement

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 x_i/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.

Homework Equations

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 x_i, 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(x_i-x_i-1) by taylor expanding and that l_i=√[(x_i-x_i-1)+a^2] (l is the hypotenuse of the triangle created by the oscillation). Why are these equations relevant to the question?

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

 

[Geofleur]

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