Answer
a) The maximum displacement of the given simple harmonic motion is $6\text{ inches}\text{.}$.
b)
The frequency of the given simple harmonic motion is $0.75Hz\text{ or }\frac{3}{4}\text{ cycles per second}$.
c) The time required for one cycle is $\frac{4}{3}\text{sec}$.
Work Step by Step
(a)
The provided equation is $ d=6\cos \left( \frac{3\pi }{2}t \right)$.
Now, the maximum displacement of a simple harmonic function is given by its amplitude. Let the maximum displacement be ${{d}_{\max }}$.
Here, $ A=6$.
Therefore,
$\begin{align}
& {{d}_{\max }}=A \\
& {{d}_{\max }}=6 \\
\end{align}$
Thus, the maximum displacement in inches is $6$.
(b)
The given equation is $ d=6\cos \left( \frac{3\pi }{2}t \right)$.
The time period, T is given as
$\begin{align}
& T=\frac{2\pi }{\frac{3\pi }{2}} \\
& T=\frac{4}{3} \\
\end{align}$
Thus,
$\begin{align}
& f=\frac{1}{T} \\
& f=\frac{1}{\frac{4}{3}} \\
& f=\frac{3}{4} \\
& f=0.75 \\
\end{align}$
Therefore, the frequency of the simple harmonic motion is $0.75Hz\text{ or }\frac{3}{4}\text{ cycles per second}$.
(c)
The given equation is $ d=6\cos \left( \frac{3\pi }{2}t \right)$.
Comparing with $ y=a\cos \left( bt \right)$, we get:
$\begin{align}
& a=6 \\
& b=\frac{3\pi }{2}
\end{align}$
The time required to complete one cycle of a simple harmonic motion is called its time period. Thus, the time period of the given simple harmonic motion is
$\begin{align}
& T=\frac{2\pi }{\frac{3\pi }{2}} \\
& T=\frac{4}{3} \\
\end{align}$
Thus, the time required to complete one cycle is $\frac{4}{3}\text{sec}$.
