Prove that the period of a SHM is 2pi*sqrt(m/k)

[ItsTheSebbe]

Homework Statement

When I was in high school I was thaught that the period of a simple harmonic oscillation (mass on spring, ball on pendulum, etc) was equal to $T=2\pi \sqrt \frac m k$ though they have never explained to me why. That's what I wanted to find out.

So for example, let's take a mass $m$ on a spring with spring constant $k$, give it a little nudge and it will start oscillating in a wave-like manner (let's assume no energy is lost due to friction).

PS: This is the first time I'm trying to use LaTeX (first post in general), so if there's anything wrong with the formatting, please let me know :)!

Homework Equations

$T=\frac {2\pi} {\omega}$
$F_{spring}=-kx$
$F_{net}=ma$

The Attempt at a Solution

$F_{net}=F_{spring}$
$ma=-kx$
$m \frac {d^2x} {dt^2} +kx=0$
$\frac {d^2x} {dt^2} +\frac k m x=0$
$\ddot x(t) + \frac k m x(t)=0$

$\text{take} \ \ x=e^{Rt}$

$R^2e^{Rt} +\frac k m e^{Rt}=0$
$R^2+\frac k m =0$
$R^2=-\frac k m$
$R=\pm i\sqrt \frac k m$

$\text{hence, } x(t)=C_1e^{i\sqrt \frac k m t}+C_2e^{-i\sqrt \frac k m t}$

Now, here I get stuck. I know I need to eventually get $x(t)=Acos(\omega t+\phi)$
I don't really understand how to get there; probably by using Euler's formula.
From here on out I can proof the equation, namely:

$\ddot x=-A\omega^2\cos(\omega t+\phi)=-x\omega^2$
$\text{plugging into EoM gives:}$
$-x\omega^2+\frac k m x=0$
$\omega^2=\frac k m$
$\omega=\sqrt \frac k m$
$T=\frac {2\pi} {\omega}=2\pi \sqrt \frac m k$

 

[James R]

$\text{hence, } x(t)=C_1e^{i\sqrt \frac k m t}+C_2e^{-i\sqrt \frac k m t}$

Now, here I get stuck. I know I need to eventually get $x(t)=Acos(\omega t+\phi)$
I don't really understand how to get there; probably by using Euler's formula.

Your expression here can be re-written as:
$$x(t) = C_3 \cos(\sqrt{\frac{k}{m}} t) + C_4 \sin(\sqrt{\frac{k}{m}} t)$$
by suitably defining $C_3$ and $C_4$ in terms of $C_1$ and $C_2$ using Euler's formula, as you say.

 

[ItsTheSebbe]

Now how can I get from $x(t) = C_3 \cos(\sqrt{\frac{k}{m}} t) + C_4 \sin(\sqrt{\frac{k}{m}} t)$ to a form of $x(t)=Acos(\omega t+\phi)$?

 

[ehild]

Now how can I get from $x(t) = C_3 \cos(\sqrt{\frac{k}{m}} t) + C_4 \sin(\sqrt{\frac{k}{m}} t)$ to a form of $x(t)=Acos(\omega t+\phi)$?

You derived that ω = √(k/m). So you have to prove that C₃cos(ωt)+C₄sin(ωt)=Acos(ωt+Φ).

Expand the right side, using the addition formula for cosine. cos(α+β)=cos(α)cos(β)-sin(α)sin(β).
α=ωt and β=Φ.
So C₃cos(ωt)+C₄sin(ωt)=Acos(ωt)cos(Φ)-Asin(ωt)sin(Φ). The equality must hold for all values of ωt which means that the coefficient of both the cosine terms and the sine terms must be equal on both sides.
Or you can think that it must be true ωt=0 and also for ωt=pi/2:
For ωt=0, C₃=Acos(Φ).
for ωt=pi/2, C₄=-Asin(Φ).

You can find both A and Φ from these equations.