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Maybe it's naive but, I have done a lot of derivations and a lot of questions on simple harmonic motion, but this one thing always strikes me , that what force us to take $\frac gl=\omega^2$ (in case of simple pendulum undergoing SHM) or $\frac km=\omega^2$ (in case of spring mass simple harmonic oscillator), where $g,l,m$ and $k$ have their usual meanings.

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  • $\begingroup$ One note: you can always go through the problem without making these definitions - it's simply a convenient way of rewriting the variables - you can otherwise carry around things like $\sqrt{g/l}$ and $\sqrt{k/m}$ $\endgroup$ Commented Sep 9, 2021 at 14:55

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Newton's 2nd law applied to a mass on a spring yields $$m \ddot x(t) = -k x(t) \implies \ddot x(t) = -\left(\frac{k}{m}\right) x(t)$$ On the other hand, simple harmonic motion arises when you have a differential equation of the form $$\ddot x(t) = -\omega^2 x(t) $$ $$\longrightarrow x(t) = c_1 \sin(\omega t) + c_2 \cos(\omega t)$$ Comparing the two forms provides the answer to your question. Essentially the same idea holds for the simple pendulum, though in that case the derivation is slightly more subtle because one needs to make use of the small angle approximation.

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  • $\begingroup$ Thanks, This is what I exactly missing. $\endgroup$ Commented Sep 10, 2021 at 13:44
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You can derive this from simple unit dimensional analysis. The problem in principle depends on the mass $m$, length of the pendulum $l$ and gravity $g$. The units are related in such a way that:

$$[g]= \mathrm{distance}/\mathrm{time}^2$$ $$[l]= \mathrm{distance} $$ $$[m] = \mathrm{mass} $$

How can you make $\omega^2$ ($[\omega^2]=\mathrm{time}^{-2}$) out of these three variables?

Well $g$ has units of inverse of time${}^2$ which is what we need, and $l$ can cancel the unit of distance in $g$. So $[g/l]=\mathrm{length}/(\mathrm{length}\cdot\mathrm{time}^2)=\mathrm{time}^{-2}$ which is what we need. It seems that we do not even need mass for this to work.

So $$\omega^2\propto \frac{g}{l}\tag{1}$$ where $\propto$ means "proportional". If you want the actual constant of proportionality you need to solve the dynamics from Newton's equations. Note that (1) is valid for simple gravity pendulums in general even if you don't use the small angle approximation, the only thing that matters here is that there are no other important factors and that the movement is periodic.

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  • $\begingroup$ Ya ,I tried the same way earlier,thanks anyway :) $\endgroup$ Commented Sep 10, 2021 at 13:45

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