Problem 1: Normal Coordinates The normal coordinates of a coupled oscillator pro
ID: 1884783 • Letter: P
Question
Problem 1: Normal Coordinates The normal coordinates of a coupled oscillator problem are the coordinates that decouple the equations of motion. If you could figure out what they are by just staring at your EOMs, these problems would be very simple! Unfortunately, in practice there are very few cases where you can do this. One such case is problems with two DOFs, described by the normal coordinates qi and q2, where the system is symmetric under the exchange of qi and q: Since the EOMs are unchanged when you swap qi and q2, here's what you do to decouple them equilibrium equilibriunm Add and subtract the EOMs to give two new EOMs Identify the single linear combination of q's that appears in each of these two EOMs these are the normal coordinates +and 5. (For 1Explanation / Answer
for two masses and three springs , the equation of moition of the4 two masses is
mx1" = -kx1 + k(x2 - x1) = -2kx1 + kx2
mx2" = -k(x2 - x1) - kx2 = kx1 - 2kx2
now
adding
m(x1" + x2") = -kx1 - kx2 = -k(x1 + x2)
subtracting
m(x1" - x2") = -3kx1 + 3kx2 = -3k(x1 - x2)
let z+ = x1 + x2
z- = x1 - x2 ( normal coorinates)
then
mz+" = -kz+
mz-" = -3kz-
hecne in matrix form
m[z+"] = -k[1 0][z+]
[z-"] [0 3][z-]
or
m*[z"] = -kz * z
b. for damping
mx1" = -kx1 - bx1' + k(x2 - x1) + b(x2' - x1') = -2kx1 + kx2 - 2bx1' + bx2'
mx2" = -k(x2 - x1) - bx2' + bx1' - bx2' - kx2 = kx1 - 2kx2 - 2bx2' + bx1'
adding
m(X1" + x2") = -kx1 - kx2 - bx1' - bx2' = -k(x1 + x2) - b(x1' + x2')
subtracting
m(x1" - x2") = -3kx1 + 3kx2 - 3bx1' + 3bx2' = -3k(x1 - x2) - 3b(x1' - x2')
henceagain
normal coordinates are (x1 + x2) and (x1 - x2)
c. mz+" = -kz+ - bz+'
mz-" = -kz- - bz-'
slolution is of the form
z-, z+ = Ae^(-gamma*t)*sin(w1t - phi1)
gamma = c/2m
d. z+(0) = A
z-(0) = A
A = A*sin(-phi)
-phi = pi/2
phi = -pi/2
hence
z+ = A*cos(w1t)
simiiliarly
z- = A*cos(w2t)
w1 andd w2 are normal modes of osscilation