1. ## SHM

A particle moves in SHM with acceleration given by x^(..)=-6x ms^2. When x=2, v=+-4.
a. Find the period
b. The amplitude
c. Find the maximum velocity and acceleration

2. ## Re: SHM

From the reference sheet we have $\ddot x = -n^2(x-b)$, where the period $T$ is given by

$T = \frac{2\pi}{n}$.

Clearly $n = \sqrt 6$, so

$T = \frac{2\pi}{\sqrt6} = \frac{\pi\sqrt6}{3}$.

Now,

\begin{aligned} \ddot x &= -6x \\ \frac{d}{dx} \left(\frac12v^2\right) &= -6x \\ v^2 &= -6x^2 + C\end{aligned}

Substituting in the initial conditions gives $C = 40$.

\begin{aligned}v^2 &= 40-6x^2 \\ v^2 &= 6\left(\frac{20}{3} -x^2\right)\end{aligned}

The maximum value of $v^2$ occurs when $x = 0$ ($v^2 = 40$). Therefore the maximum velocity is $2\sqrt{10}$.

Note that since $v^2$ must be non-negative, we have a restriction on $x$, which is

$-\sqrt{\frac{20}{3}} \leq x \leq \sqrt{\frac{20}{3}}$

Therefore the amplitude is $\sqrt{20/3}$.

Finally, going back to the first equation:

$\ddot x &= -6x$

The maximum acceleration occurs for the lowest possible value of $x$, which is $-\sqrt{20/3}$. This gives $\ddot x = 6\sqrt{20/3} = 4\sqrt{15}$.

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