SHM (1 Viewer)

HoldingOn · Jul 12, 2018

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

Thanks in advance

fan96 · Jul 13, 2018

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 $\ddot x = 6\sqrt{20/3} = 4\sqrt{15}$ .

SHM (1 Viewer)

HoldingOn

Active Member

fan96

617 pages

Users Who Are Viewing This Thread (Users: 0, Guests: 1)