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Consider an harmonic oscillator. Suppose that I manage to write the mechanical energy as a function of a quantity, like the angle $\theta$ in this way

$$E=\frac{1}{2} q \dot{\theta}^2+\frac{1}{2} s \theta^2$$

With $s$ and $q$ costants. Then

$$\omega=\sqrt{\frac{s}{q}}.$$

Is this correct? If so I don't understand the reason of that.

Can someone help me to understand why does this hold?

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  • $\begingroup$ Where did you get the equation $\omega = \sqrt{s/q}$? Can you try to follow the reasoning given in that source? $\endgroup$
    – David Z
    Feb 28, 2016 at 10:57
  • $\begingroup$ @DavidZ The reasoning is what I'm looking for. I saw it as a method used in a exercise on a textbook to find the frequency of the oscillator just knowing the mechanical energy expression but it doesn't give explanation of this $\endgroup$
    – Gianolepo
    Feb 28, 2016 at 11:14

1 Answer 1

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The point of the energy in any system is that it is constant in time$^1$, which means $\dot E=0$. In this specific case, $$ E=\frac{1}{2}q\dot\theta^2+\frac{1}{2}s\theta^2 $$ which upon time-differentiation$^2$ becomes $$ \dot E=q\dot\theta\ddot\theta+s\theta\dot\theta $$

If you set this equal to zero, you get $$ \ddot\theta+\frac{s}{q}\theta=0 $$ which is just the equation for harmonic motion, where you can read off$^3$ the frequency as $\omega^2=s/q$.


$^1$ the energy is constant in almost any situation, but sometimes (e.g., where the forces are not conservative, such as friction) $E$ might change in time. I assume this is familiar to you.

$^2$ note that here I use the chain rule: $\frac{\mathrm d}{\mathrm dt}f(\theta)=f'(\theta)\dot\theta$.

$^3$ the general solution of this equation is $\theta(t)=a\cos\left(\sqrt{s/q}\;t+\phi\right)$, where $a,\phi$ are constants of integration. You can check this by pluging this expression into the equation, and verify that it works. From this, it should be clear that the frequency is given by the square root $\sqrt{s/q}$, as we wanted to prove.

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    $\begingroup$ OP: note that this can easily be generalised to any quadratic energy: as long as $E$ depends on a variable $\theta$ and its time derivative $\dot\theta$ as a sum of two quadratic terms, you get oscillatory motion, where the frequency is given by the square root of the ratio of coefficients. $\endgroup$ Feb 28, 2016 at 12:34
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    $\begingroup$ Funny, never thought that mere differentiation of energy by time would immediately give us equations of motion (for a conservative system). $\endgroup$
    – Ruslan
    Feb 28, 2016 at 12:56
  • $\begingroup$ @Ruslan yes, its a nice trick (but it only works in 1D, because in several dimensions you need more than just one equation) $\endgroup$ Feb 28, 2016 at 14:00

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