Questions: Video Part 1: Graph The position vs. time graph below is for an object that is exhibiting SHM. The position function is: x(t)=A cos(ωt+Φ). Use this graph to fill in the table. A (m) T (s) f (Hz) vmax (m / s) 1 Part 2: Block on a horizontal surface Part 3: Finding v Part 4: Finding x

The position vs. time graph below is for an object that is exhibiting SHM. The position function is: \(x(t)=A \cos (\omega t+\Phi)\). Use this graph to fill in the table.

Amplitude (A) The amplitude \(A\) is the maximum displacement from the equilibrium position. From the graph, the maximum value of \(x\) is 1 m, so \(A = 1\) m.

Period (T) The period \(T\) is the time it takes for one complete cycle of the motion. From the graph, one complete cycle takes 2 seconds, so \(T = 2\) s.

Frequency (f) The frequency \(f\) is the reciprocal of the period: \(f = \frac{1}{T}\). Since \(T = 2\) s, \(f = \frac{1}{2} = 0.5\) Hz.

Maximum Velocity (v_max) The maximum velocity is given by \(v_{max} = A\omega\), where \(\omega\) is the angular frequency. We know that \(\omega = \frac{2\pi}{T}\). Since \(A = 1\) m and \(T = 2\) s, we have \(v_{max} = (1)\frac{2\pi}{2} = \pi \approx 3.14\) m/s.

\begin{tabular}{|c|c|c|c|} \hline \begin{tabular}{c} A \\ $(\mathrm{m})$ \end{tabular} & \begin{tabular}{c} T \\ $(\mathrm{s})$ \end{tabular} & \begin{tabular}{c} $\mathbf{f}$ \\ $(\mathrm{Hz})$ \end{tabular} & \begin{tabular}{c} $\mathbf{v}_{\text {max }}$ \\ $(\mathrm{m} / \mathrm{s})$ \end{tabular} \\ \hline $\boxed{1}$ & $\boxed{2}$ & $\boxed{0.5}$ & $\boxed{\pi \approx 3.14}$ \\ \hline \end{tabular}

\(A = 1 \text{ m}\) \(T = 2 \text{ s}\) \(f = 0.5 \text{ Hz}\) \(v_{max} = \pi \approx 3.14 \text{ m/s}\)