Coupled Oscillators Initial Conditions and Phase

[yklin_tux]

Hello I have a question about coupled oscillators and what initial conditions affect what constants of integration.

In the book I have, A.P. French Vibrations and Waves, the guesses at solutions are chosen at random and sometimes do include a phase shift, while sometimes they dont.

For solutions to a coupled oscillator example (two masses, three equal strings, no damping) there is a set of two equations

x1 = a*cos(w1*t+A1) + b*cos(w1*t + A2)
x2 = a*cos(w1*t+A1) - b*cos(w1*t + A2)

and this is based on the guess that one solution to the initial differential equation was of the form

x = C*cos(w*t + A)

I get why and how that is, but when assuming that the initial velocities are zero, the book just guesses at a solution of the form

x = C*cos(w*t), without the phase.

Which would give something of the form

x1 = a*cos(w1*t) + b*cos(w1*t)
x2 = a*cos(w1*t) - b*cos(w1*t)

I have to deal with a problem where I have initial values for the displacement, but the velocities are assumed to be zero.

Obviously, the first case, where there are phases, requires me to solve for 4 unknowns {a, b, A1, A2}

But if I use the latter case without A1, A2, the job is easier.

My question is, does the assumption that the initial velocities = 0 allow me to assume A1 = A2 = 0? and get rid of the phases?

Or is it incorrect for me to assume a solution in the form described above, without the phases present (i.e. I have to find A1, A2, from my initial conditions)

The book doesn't do a good job of explaining why they guess one solution with a phase and one without, so I sometimes get confused on that.

Cheers!

 

[kuruman]

The book is not guessing that when the initial velocities are zero, the phases are zero. Starting from

x1 = a*cos(w1*t+A1) + b*cos(w1*t + A2)
x2 = a*cos(w1*t+A1) - b*cos(w1*t + A2)

and taking derivatives to find the velocities which you evaluate at $t=0$, you get
$v_1(0)=-a \sin A_1-b \sin A_2=0$
$v_2(0)=-a \sin A_1+b \sin A_2=0$

If you add the equations, you get
$-2a \sin A_1=0$.

You have a product of three things that is equal to zero. This means that one of the following is TRUE:

1. $2=0$, not a chance even for small values of $2$
2. $a=0$, which is the trivial solution because it means there are no oscillations.
3. $\sin A_1 = 0$ which implies that $A_1=0$, yay!

Similarly, by subtracting the equations you can show that #A_2=0##.