Simple Harmonic Motion of a particle

[danago]

A particle moves in a straight line so that its position, x cm, from a fixed point O, at time t seconds is given by $$x=3+12sin(2\pi t)+5cos(2 \pi t)$$.

Prove that the particle is undergoing simple harmonic motion, and find x when its speed is 26$$\pi$$ cm/s.

I was fine with the first part of the question...

$$\begin{array}{l}<br /> x = 3 + 12\sin (2\pi t) + 5\cos (2\pi t) \\ <br /> v = \dot x = 24\pi \cos (2\pi t) - 10\pi \sin (2\pi t) \\ <br /> a = \ddot x = - 48\pi ^2 \sin (2\pi t) - 20\pi ^2 \cos (2\pi t) = - 4\pi ^2 (x - 3) = 12\pi ^2 - 4\pi ^2 x \\ <br /> \end{array}$$

With the second part, i can easily set the velocity equal to 26$$\pi$$ and solve it on my calculator, but how can i solve it algebraically?

Thanks in advance,
Dan.

 

[Dick]

If you want to solve it exactly, you can write an expression of the form A*sin(x)+B*cos(x) as a single trig function. See the wikipedia page of trig identities under 'Linear combinations'.

 

[chaoseverlasting]

Algebraically, the equation simplifies to $$x=3+13sin(2\pi t+\theta)$$ where $$\theta=sin^{-1}\frac{5}{13}$$.

 

[danago]

Ah ofcourse, i forgot about solving it like that. Thanks

 

[lihavahko]

I have a wooden log floating in the water and it is being pushed towards the water. I am to show that the force that is pushing the log back to the equilibrium position is harmonic, and i just have no idea how to do it... There are no equations attached.

 

[Dick]

I have a wooden log floating in the water and it is being pushed towards the water. I am to show that the force that is pushing the log back to the equilibrium position is harmonic, and i just have no idea how to do it... There are no equations attached.

If you have a question, don't piggy-back it on someone else's thread. It won't get the notice it deserves. You don't need to show the 'force is harmonic', to show something is a harmonic oscillator you just need to show the restoring force is proportional to displacement. If this isn't clear, DON'T REPLY HERE. Start a new thread.