Can x=8sin2t+6cos2t be proven as S.H.M. using a second derivative?

[Cpt Qwark]

Homework Statement

Prove that:
x=8sin2t+6cos2t is undergoing S.H.M.
(Not too sure about how to prove for solution.)

Homework Equations

Solution for S.H.M. x=asin(nt+α) is \frac{d^{2}x}{dy^{2}}=-n^2x

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})

 

[Nathanael]

Solution for S.H.M. x=asin(nt+α) is \frac{d^{2}x}{dy^{2}}=-n^2x

...

∴x=10sin(2t+tan^{-1}\frac{3}{4})

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)

 

[rude man]

Homework Statement

Prove that:
x = 8sin2t+6cos2t is undergoing S.H.M.

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.

 

[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.