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thanks in advance

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- Aug 4th 2008, 03:54 AMMisa-CampoFast SHM question
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thanks in advance - Aug 4th 2008, 04:08 AMflyingsquirrel
Hi !

Note that $\displaystyle \sin(4t)+\sqrt{3}\cos(4t)=2\left( \frac{1}{2}\sin(4t)+\frac{\sqrt{3}}{2}\cos(4t)\rig ht)

=2\left( \cos\frac{\pi}{3}\sin(4t)+\sin\frac{\pi}{3}\cos(4t )\right)$. As $\displaystyle \sin a \cos b+\sin b \cos a =\sin(a+b)$, we get that $\displaystyle \sin(4t)+\sqrt{3}\cos(4t)=2\sin\left( 4t+\frac{\pi}{3}\right)$ hence $\displaystyle x(t)=3+2\sin\left( 4t+\frac{\pi}{3}\right)$. Can you use this to answer the two questions? - Aug 4th 2008, 04:13 AMmr fantastic
By re-writing as $\displaystyle x = 3 + 2 \sin(4t + \phi)$ it's simple to see that the centre of motion is x = 3.

i) Solve x = 3 for t: $\displaystyle \tan (4t) = -\sqrt{3}$ etc.

ii) $\displaystyle v = dx/dt = 8 \cos (4t + \phi)$. Clearly the maximum value is v = 8. - Aug 4th 2008, 04:18 AMMisa-Campo
- Aug 4th 2008, 04:24 AMmr fantastic
Since the centre of motion is x = 3, you want the times at which x = 3:

$\displaystyle 3 = 3 + \sin (4t) + \sqrt{3} \cos (4t)$

$\displaystyle \Rightarrow 0 = \sin (4t) + \sqrt{3} \cos (4t)$

$\displaystyle \Rightarrow - \sin (4t) = \sqrt{3} \cos (4t)$

$\displaystyle \Rightarrow - \tan (4t) = \sqrt{3}$

$\displaystyle \Rightarrow \tan (4t) = - \sqrt{3}$.

And you know how to solve this trig equation ....?