# Math Help - Fast SHM question

1. ## Fast SHM question

2. Hi !
Originally Posted by Misa-Campo
Note that $\sin(4t)+\sqrt{3}\cos(4t)=2\left( \frac{1}{2}\sin(4t)+\frac{\sqrt{3}}{2}\cos(4t)\rig ht)
=2\left( \cos\frac{\pi}{3}\sin(4t)+\sin\frac{\pi}{3}\cos(4t )\right)$
. As $\sin a \cos b+\sin b \cos a =\sin(a+b)$, we get that $\sin(4t)+\sqrt{3}\cos(4t)=2\sin\left( 4t+\frac{\pi}{3}\right)$ hence $x(t)=3+2\sin\left( 4t+\frac{\pi}{3}\right)$. Can you use this to answer the two questions?

3. Originally Posted by Misa-Campo

By re-writing as $x = 3 + 2 \sin(4t + \phi)$ it's simple to see that the centre of motion is x = 3.

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

ii) $v = dx/dt = 8 \cos (4t + \phi)$. Clearly the maximum value is v = 8.

4. Originally Posted by mr fantastic
By re-writing as $x = 3 + 2 \sin(4t + \phi)$ it's simple to see that the centre of motion is x = 3.

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

ii) $v = dx/dt = 8 \cos (4t + \phi)$. Clearly the maximum value is v = 8.
for i) im still stuck

what do you mean by solve x = 3 for t: $\tan (4t) = -\sqrt{3}$ etc. I've only started this topic so im not too sure

5. Originally Posted by Misa-Campo

Originally Posted by mr fantastic
By re-writing as $x = 3 + 2 \sin(4t + \phi)$ it's simple to see that the centre of motion is x = 3.

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

[snip]
Originally Posted by Misa-Campo
for i) im still stuck

what do you mean by solve x = 3 for t: $\tan (4t) = -\sqrt{3}$ etc.
Since the centre of motion is x = 3, you want the times at which x = 3:

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

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

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

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

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

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