Answer:
Here, at \[t=0,\,x=1\,cm\] and \[V=\omega
\,cm\,{{s}^{-1}}\]; \[a=-200{{x}^{2}}\]=? ; A = ? ; \[a=100{{x}^{3}}\]
Given \[\left( \omega t+\phi \right)\]
\[\text{ }\!\!\omega\!\!\text{ cm
}{{\text{s}}^{\text{-1}}}\text{,}\] \[\text{ }\!\!\pi\!\!\text{
}{{\text{s}}^{\text{-1}}}\] (1)
Velocity, \[\left( \omega t+\alpha \right)\]
\[\text{ }\!\!\omega\!\!\text{ cm
}{{\text{s}}^{\text{-1}}}\] \[\phi \]or\[\text{ }\!\!\omega\!\!\text{ =
}\!\!\pi\!\!\text{ }{{\text{s}}^{\text{-1}}}\]
Or \[x=A\cos \left( \omega t+\phi \right)\]
(2)
Squaring and adding (i) and (ii)
\[\therefore \] or \[1=A\cos
\left( \pi \times 0+\phi \right)=A\cos \phi \]
Or \[\text{V=}\frac{\text{dx}}{\text{dt}}\text{=-A
}\!\!\omega\!\!\text{ sin}\left( \omega t+\phi \right)\]
Dividing (ii) by (i), we get
\[\therefore \] or \[\text{V=
}\!\!\omega\!\!\text{ =-A }\!\!\omega\!\!\text{ sin}\left( \pi \times
0+\phi \right)\] or \[\text{1=-Asin}\phi \]
For, \[\text{Asin}\phi =-1\] ..(3)
\[{{\text{A}}^{\text{2}}}\left(
{{\cos }^{2}}\phi +{{\sin }^{2}}\phi \right)=1+1=2\]
Differentiating (iii), w. r. t, we
have
Velocity, \[{{\text{A}}^{\text{2}}}\text{=2}\]
Applying initial conditions, i.e.,
at t= 0, V = \[\text{A=}\sqrt{\text{2}}\text{cm}\]
\[\tan \phi =-1\]
Squaring and adding (iv) and (v),
we get
\[\phi
=\frac{3\pi }{4}\]or\[\frac{7\pi }{4}\]or\[x=B\sin \left( \omega t+\alpha
\right)\]
Dividing
(iv) by (v), we have
\[\text{At t=0, x=1, so,1=B sin }\left( \text{
}\!\!\omega\!\!\text{ }\!\!\times\!\!\text{ 0+ }\!\!\alpha\!\!\text{ }
\right)\text{=Bsin }\!\!\alpha\!\!\text{ }\] or \[\text{V=}\frac{\text{dx}}{\text{dt}}\text{=B
}\!\!\omega\!\!\text{ cos }\left( \text{ }\!\!\omega\!\!\text{ t+
}\!\!\alpha\!\!\text{ } \right)\text{ }\] or \[\omega \]or \[\text{
}\!\!\omega\!\!\text{ =B }\!\!\omega\!\!\text{ cos}\left( \text{
}\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 0+ }\!\!\alpha\!\!\text{ }
\right)\text{or1=Bcos }\!\!\alpha\!\!\text{ }\]
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