Answer:
(a) The maximum extension of the spring in both the cases
will be = F/k, where k is the spring constant of the spring used.
(b)In
Fig. 10 (NCT). 6 (a), if; \[x\] is the extension in the spring, when mass m is
returning to its mean position after being released free, then restoring force
on the mass is
\[\left(
\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+3t+}\frac{\text{
}\!\!\pi\!\!\text{ }}{\text{3}} \right)\]
As,
this F is directed towards mean position of the mass, hence the mass attached
to the spring will execute SHM.
Here, spring factor = spring
constant = k.
inertia factor = mass of the given mass = m
As, time period,
\[\text{x=2cos}\left( \text{3t+}\frac{\text{5
}\!\!\pi\!\!\text{ }}{\text{6}} \right)\] \[\omega \] \[\text{x=cos}\left(
\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{-t} \right)\text{=cos}\left(
\text{t- }\!\!\pi\!\!\text{ /6} \right)\text{ }\!\!\}\!\!\text{ }\left[
\because \cos \left( -\theta \right)=\cos \theta \right]\]
In
Fig. 10 (NCT). 6 (b), we have a two body system of spring constant k and
reduced mass, \[\text{a=2, }\!\!\omega\!\!\text{ =1 and }\phi \text{= - }\pi
\text{/6}\text{.}\]
Here,
inertia factor = mil and spring factor = k.
\[\therefore \] time period, \[\text{ }\!\!\omega\!\!\text{
=2 }\!\!\pi\!\!\text{ and }\phi \text{=}\frac{\text{3 }\!\!\pi\!\!\text{
}}{\text{2}}\text{+}\frac{\text{ }\!\!\pi\!\!\text{
}}{\text{4}}\text{=}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{4}}\]
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