Question

A body executes simple harmonic motion according to following equation,
$x=5.0cos[2\pi t+\frac{\pi }{4}]m$
where t = 1.5 s, calculate (i) displacement (ii) velocity (iii) acceleration.

Answer 1

(i) $X=5.0cos(2\pi t+\frac{\pi }{4})$
$\therefore$ Displacement at $t=1.5s$
$x=5.0cos(2\times \pi \times 1.5+\frac{\pi }{4})$
=$5.0cos(3\pi +\frac{\pi }{4})$
=$5.0cos(\pi +\frac{\pi }{4})=-5.0cos\frac{\pi }{4}$
=$-5.0\times \frac{1}{\sqrt{2}}$
$\therefore x=-3.536m$
(ii) From the given equation.
Amplitude a=$0.5m$; angular speed ${\omega }_0=2\pi rad{s}^{-1}$
at $t=1.5s$ , displacement$x=-\frac{5.0}{\sqrt{2}}m$
$\therefore$ Velocity of body, $v={\omega }_0\sqrt{a^2-x^2}$
=$2\pi \sqrt{(5{)}^2-{(-\frac{5}{\sqrt{2}})}^2}=2\pi \sqrt{25-\frac{25}{2}}$
=$2\pi \sqrt{\frac{25}{2}}$
=$2\pi \times \frac{5}{\sqrt{2}}=\frac{10\times 3.14}{1.414}$
$\therefore v=22.2m{s}^{-1}$
(iii)$\because$ Acceleration $f=-{\omega }_0^{2}x$
for $t=1.5s,x=-\frac{5}{\sqrt{2}}$
$\therefore f=-(2\pi {)}^2\left(\frac{-5}{\sqrt{2}}\right)=2\pi \times 2\pi \times \frac{5}{\sqrt{2}}$
=$2{\pi }^2\times \sqrt{2}\times 5=2\times 9.86\times 1.414\times 5$
$f=139.42m{s}^{-2}$