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Why do we assume that in normal modes, particles oscillate in form cos (wt) ?

How do we know that the general motion of particles can be expressed as a superposition of normal modes?

In both French and Crawford, the assumption of harmonic motion is made without any proof, please help.

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In most cases, this is related to an assumption of small displacements from equilibrium. Assume that the system is described by a potential function $V(s)$, where $s$ represents the coordinate(s) associated with the normal modes. Let $s_0$ represent value of the coordinates the equilibrium state. Taylor expanding the potential about this point yields

$$ V(s-s_0) \approx V(s_0) + V'(s_0)(s-s_0) + (1/2) V''(s_0)(s-s_0)^2 + ... $$

The key feature is that we know $V'(s_0)=0$, since that is the the definition of equilibrium. We can also ignore the first term since it is independent of the state of the system.

Thus the resulting form of the equation of motion of the form

$$ 0 = \ddot{ s } + \omega^2 s^2 $$

with $\omega^2$ a function of $V''(s_0)$ and the masses/moments of inertia of the system. This equation has $\sin( \omega t), cos(\omega t)$ as its solutions.

Thus, simple harmonic motion is a generic feature of small oscillations about any mechanical equilibrium.

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  • $\begingroup$ Are there any common systems where the second derivative of the potential is also zero so that the first approximation gives us a third-order differential equation for the equation of motion? $\endgroup$ – Mark Allen Jun 20 '13 at 12:18
  • $\begingroup$ I don't know any common situations where the second derivative is also zero; but it is an important edge case that one needs to be aware of. $\endgroup$ – Dave Jun 20 '13 at 12:23
  • $\begingroup$ @MarkAllen: All positions where the first non-vanishing derivative is of uneven order are unstable, as a slight perturbation would move the system away from this point. Further, the order of the first non-vanishing term has nothing to do with the derivatives in the e.o.m. Suppose the first non-vanishing term was $V^{\rm IV}$, then the e.o.m. would read $0 = \ddot s + \xi^4 s^4$ (with a constant $\xi$) which does not describe a harmonic oscillation. Such a situation is rather ... contrived, though. $\endgroup$ – Neuneck Jun 20 '13 at 14:41

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