Question

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

Answer 1

$x=\left(2.0cm\right)sin[\left(100s^{-1}\right)t+\frac{\pi }{8}]$

$m=10g$

(a) Amplitude = 2 cm

$\omega =100se{c}^{-1}$

$\therefore T=\frac{2\pi }{100}=\frac{\pi }{50}sec.$

$=0.063sec.$

We know that,

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

$\Rightarrow T^2=4{\pi }^2\frac{m}{k}$

$\Rightarrow k=\frac{4{\pi }^2}{T^2}=10.5dyne/cm$

$=100N/m$

[Because $0=\frac{2\pi }{T}=100se{c}^{-1}$]

$\left(b\right)Att=0,x=2cmsin\frac{\pi }{6}$

$=2\times \frac{1}{2}=1cm$ from the mean position

We know that,

$x=Asin(\omega t+\Phi ),$

$v=a\omega cos(\omega t+\Phi )$

$=2\times 100cos(0+\frac{\pi }{6})$

$=200\times \frac{\sqrt{3}}{2}$

$=100\sqrt{3}cmse{c}^{-1}$

$=1.73m{s}^{-1}$

$\left(c\right)a=-w^{2}x$

$={100}^2\times 1=1cm/s^2$