We all know that the phase trajectory of an undamped linear harmonic oscillator is an ellipse. But when we calculate the area of the ellipse we find it does not depend of mass of the particle. Why is it so? What is the physical significance of this?
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1$\begingroup$ I don't understand the question. Why do you expect the area to depend on mass? $\endgroup$– DanielSankCommented Sep 29, 2015 at 16:32
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$\begingroup$ Echoing @DanielSank's comment: For a classical HO, by which protocol are the initial conditions and ellipse changed (or not) as the mass is varied in order to make a non-trivial well-posed statement? $\endgroup$– Qmechanic ♦Commented Sep 29, 2015 at 18:05
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1 Answer
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So the Hamiltonian is $H = \frac12 m v^2 + \frac12 m \omega^2 x^2$and therefore we can define $u = v/\omega$ to find that the circle swept out (of radius $x = a$) has $ux$-area $\pi a^2$ or $px$-area of $\pi~m~\omega~a^2.$ Presumably you mean that this area is $2\pi E/\omega,$ independent of mass at constant energy and frequency. Other combinations will lead to $m$-dependence!
The only physical significance I know for this fact comes from the fact that a volume of phase space measures the entropy of a system. The constancy with respect to mass results in the Dulong–Petit law that the specific heat per mole of atoms (rather than per kg) for solids is generally around $3R$ where $R$ is the gas constant. Each atom acts as a 3D harmonic oscillator, hence it acts as 3 harmonic oscillators.
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$\begingroup$ But $$\omega=\sqrt{\frac{k}{m}}$$ where k is the linear spring constant. So what you've shown is it is dependent on mass, right? $\endgroup$ Commented Sep 29, 2015 at 20:19
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$\begingroup$ @docscience Indeed, if you hold E and k fixed, then it is. If you hold $\sqrt{m}~E$ and $k$ fixed, then it's not again. In general we cannot say "independent of mass" unless we say what we're holding constant, and SiddharthaDam did not specify, so the question is ill-defined. $\endgroup$– CR DrostCommented Sep 29, 2015 at 22:26
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$\begingroup$ I wanted to say, the area of the ellipse formed by H in Chris' answer does not depend explicitly depend upon mass. And yes i was holding E as a constant $\endgroup$ Commented Sep 30, 2015 at 8:27