# Simple Harmonic Motion proof

1. Feb 18, 2015

### Cpt Qwark

1. The problem statement, all variables and given/known data
Prove that:
$$x=8sin2t+6cos2t$$ is undergoing S.H.M.
(Not too sure about how to prove for solution.)
2. Relevant equations
Solution for S.H.M. $$x=asin(nt+α)$$ is $$\frac{d^{2}x}{dy^{2}}=-n^2x$$

3. The attempt at a solution
$$r=\sqrt{8^{2}+6^{2}}=10\\α=tan^{-1}\frac{3}{4}\\∴x=10sin(2t+tan^{-1}\frac{3}{4})$$
Differentiating with respect to time: $$\frac{dx}{dt}=20cos(2t+tan^{-1}\frac{3}{4})\\\frac{d^{2}x}{dt^{2}}=-40sin(2t+tan^{-1}\frac{3}{4})$$

2. Feb 18, 2015

### Nathanael

Right. Adding any two sinusoidal functions of the same frequency will result in another sinusoidal function, regardless of their amplitudes.

If you solve the SHM differential equation, $\frac{d^2x}{dt^2}=-kx$ you will get $x=C_1\sin(\sqrt{k}t)+C_2\cos(\sqrt{k}t)$ and it because of the above fact that you can write the solution as $x=C_3\sin(\sqrt{k}t+C_4)$

Last edited: Feb 18, 2015
3. Feb 19, 2015

### rude man

You can just take this equation, compute x', then x'', and see that ω2 must = 4 by equating sine and cosine coefficients. Both yield the same answer ω2 = 4. Had the sine & cosine coeff. yielded differing ω then x(t) would not be shm.

4. Apr 29, 2015

### Physicist123

Take the second derivative of the given expression and express it in terms of x. The result would eliminate sin and cos and will prove your answer in form of a=-nx.