Thread: Simple Harmonic Motion with Transformations

1. Simple Harmonic Motion with Transformations

Prove the following motions are simple harmonic and find the centre of motion, the period and the amplitude.

1. $x=2+cos 3t - \sqrt 3 sin 3t$

With this I tried to put the 2 over to the other side so that it becomes $x-2=cos 3t - \sqrt 3 sin 3t$and transform to $R sin (3t+\alpha)$ but I got stuck at the transformation bit.

2. $x=6 cos^2 8t -4$

First I differentiated and got $12cos 8t (-sin 8t)$ which is the same as $-12cos 8t sin 8t$ right? And I know I have to differentiate again but do I use product rule or differentiate altogether?

2. Originally Posted by nerdzor
Prove the following motions are simple harmonic and find the centre of motion, the period and the amplitude.

[snip]

2. $x=6 cos^2 8t -4$

First I differentiated and got $12cos 8t (-sin 8t)$ which is the same as $-12cos 8t sin 8t$ right? And I know I have to differentiate again but do I use product rule or differentiate altogether?
There are a number of approaches.

Substitute (from the double angle formula) $\cos^2 (8t) = \frac{\cos (16t) + 1}{2}$.

Originally Posted by nerdzor
Prove the following motions are simple harmonic and find the centre of motion, the period and the amplitude.

1. $x=2+cos 3t - \sqrt 3 sin 3t$

With this I tried to put the 2 over to the other side so that it becomes $x-2=cos 3t - \sqrt 3 sin 3t$and transform to $R sin (3t+\alpha)$ but I got stuck at the transformation bit.

[snip]
$\cos (3t) - \sqrt{3} \sin (3t) = R \sin (3t + \alpha)$:

Expand the right hand side and compare witht he left hand side:

$R \sin \alpha = 1$ .... (1)

$R \cos \alpha = - \sqrt{3}$ .... (2)

Solve equations (1) and (2) for $R$ and $\alpha$.