1. Exploring Velocity.
(a)
Starting from the general one-dimensional position function for simple harmonic motion
\begin{equation*}
x(t) = x_o + x_{\text{max}} \cos{(\omega t + \phi_o)}
\end{equation*}
take the derivative to find the velocity function \(\vec{v}(t)\text{.}\)
Solution.
\begin{equation}
v(t) = - \omega x_{\text{max}} \sin(\omega t + \phi_o) \tag{15.5.1}
\end{equation}
(b)
Graph both the position and the velocity functions. Choose one or two instants in time and check that the two graphs agree with each other.
Solution.
As the object passes through the equilibrium position located at \(\vec{x} = 0 \text{,}\) the speed (magnitude of the velocity) is maximum. When the object is maximally displaced from equilibrium, it temporarily comes to rest before changing its direction. Recall that a maxima or minima in the position as a function of time graph corresponds to a turning point in the motion. Therefore, when the object is located at \(x= \pm A\) the instantaneous velocity of the object is zero.
(c)
What are the SI units of the quantity \(\omega x_{\text{max}}\text{?}\) What is the physical significance of this quantity?
Solution.
The equation as a whole is a velocity and has dimension of length/time with SI units of meters/second. Trigonometric functions are dimensionless. Therefore, the quantity \(\omega A\) must have dimension of length/time for the equation to be true. Note that the SI units of \(\omega\) are \(rad/s\text{.}\) While radians are an SI unit, they are dimensionless. The physical significance of this quantity is that it corresponds to the maximum speed \(v_{max} = \omega x_{\text{max}} \) of the object. This allows you to re-write equation the velocity equation of motion as:
\begin{equation}
\vec{v}(t) = - v_{max} \sin(\omega t + \phi_0) \hat{x} \tag{15.5.2}
\end{equation}
