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## Coupled oscillators - mode and mode co-ordinates

For this question I'm not going to introduce the particular problem I am working on, rather, I am merely wanting some explanation of a concept which I can't seem to find in any of my textbooks. I suspect the authors think it is just too obvious to bother explaining .

I'm revising for a test and have the full worked solutions for this problem in front of me. I can follow the mathematics, but not the reasoning behind it.

The question:

Two masses M1 and M2 are connected by springs as in my expertly drawn diagram attached.  Show that the amplitude of the displacement of the masses is described by expressions of the form: $$\psi _{1}\left( t \right)=A_{0}\cos \omega _{+}t$$ $$\psi _{2}\left( t \right)=A_{0}\cos \omega _{+}t$$
My worked solutions now say:

Notice that when the system is in mode 1, the quantity (x2 - x1) is always zero, and (x1 + x2) varies harmonically. In mode 2 the reverse is true. Let us define a set of variables:  $$q_{1}=\sqrt{\frac{m}{2}}\left( \psi _{2}+\psi _{1} \right)\; -->\; \dot{q}_{1}=\sqrt{\frac{m}{2}}\left( \dot{\psi }_{2}+\dot{\psi }_{1} \right)$$ $$q_{2}=\sqrt{\frac{m}{2}}\left( \psi _{2}-\psi _{1} \right)\; -->\; \dot{q}_{2}=\sqrt{\frac{m}{2}}\left( \dot{\psi }_{2}-\dot{\psi }_{1} \right)$$
My question:

What exactly are q1 and q2, and why should these be equal to $$\sqrt{\frac{m}{2}}\left( \psi_{2}+\psi_{1} \right)$$ etc? Why $$\sqrt{\frac{m}{2}}$$? Is there a more specific name for this law that I could look up?

I hope my question is easily understandable! Thank you for your help.

(note: for the sets of equations relating q1 and q2 to m and x, there should be a "≡" sign rather than an "=" sign - for some reason my TEX formatting comes out with "8801;" rather than a "≡" sign. Odd.)
Attached Thumbnails

PhysOrg.com  science news on PhysOrg.com  >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Formulae now fixed. I hope. Sorry if I confused anyone while I was editing things

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