Question

A certain oscillator has equation of motion
x¨ + 4x = 0.
The particle is initially at the point x =
√
3 when it is projected towards the origin with a speed 2. Determine the displacement, x(t), of the particle

Answer 1

Answer

Given equation

$\frac{d^2x}{dt^2}=-4x$

Given equation is of the simple harmonic motion so

$\omega=\sqrt4=2rad/s$

Displacement in shm is given by

$x(t) =x_ocos(wt+\phi)$

At t=0sec

$\sqrt{3}=Acos(2\times0+\phi) \\\sqrt{3}=Acos\phi$ . . . ... Eq. 1

And velocity v=2m/s

So velocity can be written as for shm

$v=-x_o\omega sin(\omega t+\phi)$

At t=0sec

$-2=-2x_osin(2\times0+\phi) \\x_osin\phi=1$ .. Eq. 2

By solving equation 1 and 2

$tan\phi=\frac{1}{\sqrt{3}}$

$\phi=\frac{\pi}{6}$

And $x_o=2$

So equation can be Written as

$x(t) =2cos(2t+\frac{\pi}{6})$