# Simple Harmonic Motion with Transformations

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• Jun 7th 2009, 01:52 AM
nerdzor
Simple Harmonic Motion with Transformations
Prove the following motions are simple harmonic and find the centre of motion, the period and the amplitude.

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

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

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

First I differentiated and got $\displaystyle 12cos 8t (-sin 8t)$ which is the same as $\displaystyle -12cos 8t sin 8t$ right? And I know I have to differentiate again but do I use product rule or differentiate altogether?
• Jun 7th 2009, 03:20 AM
mr fantastic
Quote:

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

[snip]

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

First I differentiated and got $\displaystyle 12cos 8t (-sin 8t)$ which is the same as $\displaystyle -12cos 8t sin 8t$ 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) $\displaystyle \cos^2 (8t) = \frac{\cos (16t) + 1}{2}$.

Quote:

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

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

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

[snip]

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

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

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

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

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