Question

A partical having mass 10 g oscillates according to the equation $x=\left(2.0cm\right)\sin \left[\right(100s^{-1})t+\frac{6}{\pi }]$. Find
(a) the amplitude, the time period and the force constant
(b) the position, the velocity and the acceleration at $t=0$.

Answer 1

$x=\left(2.0cm\right)Sin(100s^{-1}t+\frac{\pi }{6})$

$m=10g$

a) $A$ =Amplitude = $2cm$

Angular frequency, $\omega =100rad/sec$

$\therefore$ Time period, $T=\frac{2\pi }{\omega }=\frac{2\pi }{100}=\frac{\pi }{50}sec=0.063sec$

We know, $T=2\pi \sqrt{\frac{m}{k}}$

$\frac{T}{2\pi }=\sqrt{\frac{m}{k}}$

$\Rightarrow \frac{1}{\omega }=\sqrt{\frac{m}{k}}$

$\omega =\sqrt{\frac{k}{m}}$

${\omega }^2=\frac{k}{m}\Rightarrow k=m\times {\omega }^2=\frac{10}{1000}kg\times {100}^2{s}^{-2}=100N/m$

b) At $t=0$,

$x=\left(2\right)Sin(0+\frac{\pi }{6})=2\times \frac{1}{2}=1cm$

Velocity at $t=0$,

$v=\frac{dx}{dt}=A\omega Cos(\omega t+\varphi )$

$v=2\times 100Cos(0+\frac{\pi }{6})=200\times Cos(0+\frac{\pi }{6})=200\times \frac{\sqrt{3}}{2}=100\sqrt{3}cm/sec=\sqrt{3}m/sec$

Acceleration at $t=0$,

$a=\frac{dv}{dt}=-A{\omega }^{2}Sin(\omega t+\varphi )=-{\omega }^{2}x$

$a=-{100}^2\times 1=-{10}^{4}cm/se{c}^2=-100m/se{c}^2$

$|a|=100m/se{c}^2$