Answer:
If we express each function in the form \[x=a\cos \pi t\] where
\[\phi =+\pi .\] is the initial phase, i.e., \[\text{x=aco}s\left( \frac{2\pi
t}{T}+\phi \right)\text{=2cos}\left( \frac{\text{2 }\!\!\pi\!\!\text{
t}}{\text{4}}\text{+ }\!\!\pi\!\!\text{ } \right)\text{=-2cos}\left(
\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{t} \right)\] represents the
angle which the initial radius vector of the particle makes with the positive
direction of x-axis.
(a)\[\left(
\text{3t+ }\!\!\pi\!\!\text{ /3} \right)\]\[\left( \text{ }\!\!\pi\!\!\text{
/6-t} \right)\]
or
\[\left( \text{2 }\!\!\pi\!\!\text{ t+ }\!\!\pi\!\!\text{ /4} \right)\]
(i)
Comparing
it with equation (0, we note that \[a=2,\,\,\omega =3\] and = 5 71/6
Hence,
the reference circle will be as shown in Fig.
10(NCT).5(a).
(b)\[\pi
t\]
Comparing
it with equation (i), we note that
\[\phi
\]
The
reference circle will be as shown in Fig. 10(NCT).5(b).
(c)
x = 3 \[\phi \]
Comparing
it with equation (i), we note that \[a=3,\] \[\text{x=-2sin}\left( \text{3t+ }\!\!\pi\!\!\text{
/3} \right)\text{=2cos}\]
The reference circle will be as shown in Fig. 10 (NCT).
5(d).
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