Question

A 0.5kg object attached to a spring with a force constant of 8.00N/m vibrates in simple harmonic motion with amplitude of 10.0cm.Calculate the maximum value of its
(a) speed and (b) acceleration when the object is 6.00cm from tge equilibrium position and (c) the time interval required for the object to move from x = 0 to x = 8.00cm

Answer 1

$x=Acos(\sqrt{\frac km}t+\varphi),$

a)

$v_{max}=A\sqrt{\frac km}=0.1\cdot \sqrt{\frac {8}{0.5}}=0.4~\frac ms,$

b)

$a=A{\frac km}cos(\frac{\pi}{2}\cdot\frac{6}{10})=0.1\cdot {\frac {8}{0.5}}\cdot 0.59=0.94~\frac {m}{s^2},$

c)

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{\omega}}=\frac{\pi}{2}~s,$

$t=\frac 14arcsin\frac{8}{10}=0.24 ~s.$