# I was learning SHM and I came across equation $k = mw^2$ [duplicate]

I tried searching for proofs here and there but couldn't find.. Some people used $$a= w^2 x$$ to prove this.. But how do you prove that?

• The picture is not legible. Please write out specifically what your question is, and use MathJax. – kaylimekay Jan 9 at 15:00
• My question is what's the derivation of k = mw^2 or what's the derivation of a = w^2x – Yash Gabra Jan 9 at 15:10
• And I thought that $g=\ell \omega^2$. :) – Bill N Jan 9 at 16:03
• @yashgabra Those are two different questions. Plus you haven't given a specific context. We can read between the lines, but that's not a good way to pose a question. You should state what the specific physical system is. Also, the character is omega, $\omega$ not $w$. – Bill N Jan 9 at 16:06

Newton's second law states the force on a system is proportional to its acceleration. For linear restoring force $$-kx$$, we have

$$F = ma$$ $$-kx = m\ddot{x}$$

where $$\ddot{x}$$ is the second time-derivative of position, i.e. acceleration. Then you have

$$\ddot{x} = a = -\frac{k}{m}x = -\omega^2 x$$

where we define $$\omega = \sqrt{\frac{k}{m}}$$.

• Is omega another constant in this and not angular frequency? – Yash Gabra Jan 9 at 15:27
• $\omega$ is the angular frequency, but for a simple harmonic oscillator, that happens to be a constant value. – zhutchens1 Jan 9 at 15:29

In Zhutchens1's force equation, the second derivative of x is a negative constant times x. The simplest solution for this differential equation is x = sin(ωt + φ) where ω is the angular frequency and φ depends on when you start the clock. If you put this function for x into the equation, you find that $$ω^2$$ = k/m.

I think what is bothering you is how did we end here: $$a=-\omega^2 x$$ form here: $$a=-(k/m) x$$

This is matter more of Mathematics than Physics. In mathematics, the standard solution of the below differential equation: $$\frac{d^2 y}{dx^2}= - p^2 y\tag1$$ is this: $$y= C_1\sin(px + C_2)$$where $$C_1$$ and $$C_2$$ are arbitrary constants.

In Physics, so we assign the value $$\omega=k/m$$ on our own "arbitrarily". And further obtain the solution: $$y= A \sin(\omega t + \phi)$$ where A and $$\phi$$ are arbitrary constants. It just so happens that we physically interpret A as amplitude and $$\phi$$ as initial phase.

Now if you want the solution of equation (1) , that was nothing to do with Physics. Try referring to a Mathematics text for the same.

• Having two $k$'s, with $k=mk^2$, is not great for clarity – fqq Jan 9 at 15:42
• It's not an arbitrary assignment. In fact, what you call $k^2$ is, in the careful physics analysis, $k/m$ for a spring-mass oscillator system. – Bill N Jan 9 at 16:00
• @BillN I used the word "arbitrarily" in quotes for the same purpose. Are you suggesting me to edit my terminology? – Tony Stark Jan 9 at 17:32
• @fqq Point noted. – Tony Stark Jan 9 at 17:35
• Please don't mess up with variables, try to make it lucid. Solve the differential equation to show the asker how it's derived without just writing it as a standard solution. – Roger Michealson Jan 9 at 17:40