Question

A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.

Answer 1

Given, $r=10cm,ATt=0,n=5cm,T=6sec$

So, $\omega =\frac{2\pi }{T}=\frac{2\pi }{6}=\frac{\pi }{3}se{c}^{-1}$

$Att=0,x=5cm$

So, $5=10sin(w\times 0+\Phi )[Y=rsinwt]$

$=10sin\Phi$

$sin\Phi =\frac{1}{2}$

$\Rightarrow \Phi =\frac{\pi }{6}$

$\therefore$ Equation of displacement

$x=\left(10cm\right)sin(\frac{\pi }{3}t+\frac{\pi }{6})$

$\left(ii\right)Att=4$ second

$x=10sin[\frac{\pi }{3}4+\frac{\pi }{6}]$

$=10sin\left[\frac{8\pi +\pi }{6}\right]$

$=10sin\left(\frac{9\pi }{6}\right)$

$=10sin\left(\frac{3\pi }{2}\right)$

$=10sin\pi +\frac{\pi }{2}$

$=-10sin\frac{\pi }{2}=-10$

Acceleration,

$a=-w^{2}x$

$=\left(\frac{-{\pi }^2}{9}\right)\times (-10)$

$=10.9\approx 0.11cm/sec$