Question

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 70 cm/s, and if there is no damping, determine the position u of the mass at any time t. (Use g

Answer 1

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$