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Can anyone please give me a proof for $k/m=w^2$ in simple harmonic motion?

I have tried energy conservation and Newton's laws as follows :

In the case of a mass-spring system,

$$F=ma =-kx\\ F=ma = mr\omega^2 $$

hence

$$\frac km = \omega^2$$

Or

$$\frac12 mv^2 = \frac12kx^2$$

$$\frac km = \frac{V^2}{x^2} = \omega^2$$

therefore $k/m = \omega^2$

Are these valid and correct? I have been stuck in this for the entire day.

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closed as off-topic by John Rennie, Jim, tpg2114, Pranav Hosangadi, ACuriousMind Jan 7 '15 at 17:45

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    $\begingroup$ Wikipedia shows you exactly how they get it in two lines $\endgroup$ – Jim Jan 7 '15 at 16:35
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    $\begingroup$ Yeah, slight confusion on my wording there. Sorry about that. You are thinking of things moving in a circle, which is periodic. Here you have an object that is moving up and down, which is also periodic. The use of $\omega$ is a convenient definition which can relate to the motion in a circle. $\endgroup$ – Kyle Kanos Jan 7 '15 at 16:46
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    $\begingroup$ radian is dimensionless ,probably this is the thing you were looking for $\endgroup$ – Gowtham Jan 7 '15 at 16:53
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    $\begingroup$ $k$ has units N/m=kg/s$^2$, $m$ has units of kg; divide the two and take the square root ($\omega=\sqrt{k/m}$) & you've got 1/s$\equiv$rad/s. $\endgroup$ – Kyle Kanos Jan 7 '15 at 16:54
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    $\begingroup$ I don't understand the close votes on this question. $\endgroup$ – rob Jan 7 '15 at 17:32
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The simple harmonic oscillator is governed by Hooke's Law, $F=-kx$. Since Newton's laws tell us that $F=ma=m\ddot x$ (where $\ddot x$ is the second derivative of $x$ with respect to time), we have a second-order differential equation $$ \ddot x = -\frac kmx. $$ We want to find the position as a function of time, $x(t)$, which solves this equation. A function which is proportional to its derivative usually involves an exponential, so let's guess \begin{align} x(t) &= e^{\beta t} \\ \dot x &= \beta x \\ \ddot x &= \beta^2 x \end{align} This guess does in fact solve our differential equation if and only if $$ \beta^2 = -\frac km, $$ which is permitted if $\beta$ is imaginary. There are two solutions, then: $\beta = \pm i\omega$, where $\omega$ is real and has units of $\mathrm s^{-1}$. This gives you $\omega^2 = +\frac km$, as you asked.

We can construct purely real solutions thanks to the Euler identity: \begin{align} x_\text{even}(t) &= \frac{e^{i\omega t}+e^{-i\omega t}}{2} = \cos \omega t \\ x_\text{even}(t) &= \frac{e^{i\omega t}-e^{-i\omega t}}{2i} = \sin \omega t \\ \end{align} This construction makes the interpretation of $\omega$ as angular frequency more obvious: the position repeats whenever the dimensionless product $\omega t$ increases by $2\pi$, so we can think of $\omega t$ as a "phase angle" (though no physical angle need be involved) measured in radians. The number of cycles that have occurred up to time $t$ is $\omega t/2\pi$, so the clock frequency is $f=\omega/2\pi$.

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  • $\begingroup$ maybe I'm not understanding your reasoning, but aren't you proving the relation by assuming it? How are you saying $\alpha= \pm i \omega$? Your reasoning just leads to $\alpha = \pm i \sqrt{k/m}$. The following identification $\omega= \sqrt{k/m}$ is nothing but a definition, not a logical consequence $\endgroup$ – glS Jan 7 '15 at 17:33
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    $\begingroup$ n.b. @glance I changed $\alpha$ to $\beta$ while you were commenting to avoid confusion with angular acceleration. … The "proof" here is that the differential equation has some solution and that the simplest guess involves complex numbers. If you do some algebra to make all the numbers real then you find yourself dealing with the mathematics of oscillations and the idea of a phase angle arises naturally. This is why Hooke's Law leads to oscillatory motion. I think it's a nontrivial result. $\endgroup$ – rob Jan 7 '15 at 17:44
  • $\begingroup$ that wasn't my point. I completely agree with the derivation of the solutions of the equation, but you are kind of "cheating" when you say "This gives you $\omega^2 = k/m$, as you asked". You are not "proving" that relation for $\omega$, you are simply defining it that way. $\endgroup$ – glS Jan 7 '15 at 18:04
  • $\begingroup$ @glance Well I have to call it something. I could have called it $|\beta|$ or kept writing $\sqrt{k/m}$ or called it "Larry." The point is that $\sqrt{k/m}$ acts like an angular frequency, which justifies the traditional notation. $\endgroup$ – rob Jan 7 '15 at 18:40

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