Question

So if you look at ma=-kx you can fairly easily derive x=Asin(2pi*v*t). Now my question isn't about trigonometric substitution, but why is it that the original equation of setting a second derivative of x equal to a coefficient times x imply wave motion? I am just sort of amazed and baffled by this because I know that it follows because of calculus, but it just seems interesting! Can someone explain this to me?

Answer 1

\[m \frac{ \delta^2x }{ \delta t^2 }=-kx\]\[m \frac{ dv }{ dt }=m \frac{ dv }{ dx }\frac{ dx }{ dt }=mvdv=-kxdx\]\[mv^2=-kx^2+C\]solve for C by noting that in oscillatory motion, when the position is at the amplitude (x=A) the pendulum stops (v=0) so we can say C=kA^2
\[mv^2=-kx^2+kA^2\]\[v=\frac{ dx }{ dt }=\sqrt{\frac{ k }{ m }(A^2-x^2)}\]\[\frac{ dx }{ \sqrt{A^2-x^2} }=\sqrt{\frac{ k }{ m }}dt\]\[\sin^{-1}(\frac{ x }{ A })=\sqrt{\frac{ k }{ m }}t\]\[x=Asin(\sqrt{\frac{ k }{ m }}t)\]

I just thought I'd type it out because it's weird to me and if it will help someone to explain to me why ma=-kx implies oscillatory motion without calculus, but with words, I'd be really interested!

Answer 2

The second derivative of x times m (where m is mass of object) equal to a coefficient times x

Answer 3

We could easily condense mass and spring constant into one. That's not really what I'm concerned about though.

Answer 4

It's because force is proportional to displacement... if the mass moves away from equilibrium, the force always acts to restore it to equilibrium. Thus: restoring force.
And the farther away from equilibrium, the greater the force.

Answer 5

Hmm, but how does that connect to the second derivative?

Answer 6

acceleration? F=ma
aka acceleration is proportional to displacement.