Respuesta :
Answer:
Explanation:
Given
mass of spring [tex]m=100\ gm[/tex]
extension in spring [tex]x=5\ cm[/tex]
downward velocity [tex]v=70\ cm/s[/tex]
Position in undamped free vibration is given by
[tex]u(t)=A\cos \omega _0t+B\sin \omega _0t[/tex]
where [tex]\omega _0^2=\frac{k}{m}[/tex]
also [tex]\frac{k}{m}=\frac{g}{L}[/tex]
[tex]\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}[/tex]
[tex]\omega _0=14[/tex]
[tex]u(t)=A\cos(14t)+B\sin(14t)[/tex]
it is given
[tex]u(0)=0[/tex]
[tex]u'(0)=70\ cm/s[/tex]
substituting values we get
[tex]A=0[/tex]
[tex]u(t)=B\sin (14t)[/tex]
[tex]u'(t)=14B\cos (14t)[/tex]
[tex]70=14B[/tex]
[tex]B=\frac{10}{2}[/tex]
[tex]B=5[/tex]
[tex]u(t)=5\sin (14t)[/tex]
