Solution :
Given :
Weight = 4 lb
Stretched of a spring in equilibrium,Δ = 2 ft
a). We know that
F = kΔ
4 = k x 2
k = 2 lb/ft
∴ Stiffness of the spring is, k = 2 lb/ft
Motion is critically damped.
[tex]$\epsilon = 1$[/tex]
[tex]$\epsilon=\frac{C}{C_c} = 1$[/tex]
[tex]$C=C_c[/tex]
[tex]$C=C_c=\sqrt{4k_m}[/tex]
[tex]$C=\sqrt{4 \times 2 \times \left(\frac{4}{32.174}\right)}[/tex]
C = 0.99729 lb.s/ft.
b). Displacement
[tex]$y(t)=y_0 \sin (\omega t)$[/tex]
[tex]$\dot{y}(t)= y_0 \omega \cos (\omega t)$[/tex]
[tex]$[\dot{y}(t)]_{mon}= y_0 \omega $[/tex]
Initial displacement = 6 inches
[tex]$y_0 = 0.5 \ ft , \ \ \ \dot y (t) = 3 \ ft/sec$[/tex]
[tex]$\Rightarrow 3=0.5 \times \omega$[/tex]
[tex]$\Rightarrow \omega = 6$[/tex]
Therefore, displacement :
[tex]$y(t)=y_0 \sin (\omega t)$[/tex]
[tex]$y(t)=0.5 \sin (6 t)$[/tex]
