Determine the equilibirium position for the weight.
[tex]\begin{gathered} \frac{97+11}{2}=\frac{108}{2} \\ =54 \end{gathered}[/tex]Determine the amplitude for the Simple harmonic function.
[tex]\begin{gathered} A=97-54 \\ =43 \end{gathered}[/tex]Determine the time taken by weight for the one complete cycle.
[tex]\begin{gathered} T=2(5.6-0.6) \\ =2\cdot5 \\ =10 \end{gathered}[/tex]The formula for the angular frequency is,
[tex]\omega=\frac{2\pi}{T}[/tex]Substitute the values in the formula to determine the angular frequancy of the weight.
[tex]\begin{gathered} \omega=\frac{2\pi}{10} \\ =\frac{\pi}{5} \end{gathered}[/tex]Determine the general equation of motion of weight.
[tex]y=54+43\cos (\frac{\pi}{5}\cdot t+\phi)[/tex]For t = 0.6 y = 11.
Determine the phase difference angle for t = 0.6 and y = 11.
[tex]\begin{gathered} 11=54+43\cos (\frac{\pi}{5}\cdot0.6+\phi) \\ 43\cos (\frac{\pi}{5}\cdot0.6+\phi)=-43 \\ \frac{\pi}{5}\cdot0.6+\phi=\cos ^{-1}(-1) \\ \phi=\pi-\frac{6\pi}{50} \\ =\frac{44\pi}{50} \\ =\frac{22\pi}{25} \end{gathered}[/tex]So equation for the height above the floor as a function of time is,
[tex]y=54+43\cos (\frac{\pi}{5}t+\frac{22\pi}{25})[/tex]Plot the graph of the height function.
Substitute 0 for t in the equation to determine the weight height at time zero.
[tex]\begin{gathered} y=54+43\cos (\frac{\pi}{5}\cdot0+\frac{22}{25}\pi) \\ =54+43\cos (\frac{22\pi}{25}) \\ =54-39.9803 \\ =14.0196 \\ \approx14.02 \end{gathered}[/tex]So height above the floor at time zero is 14.02 cm.
