question_answer 7)

The motion of a particle in S.H.M. is described by the displacement function, \[x=A\cos (\omega t+\phi )\]. If the initial \[(t=0)\] position of the particle is 1 cm and its initial velocity is \[\text{2 }\!\!\pi\!\!\text{ / }\!\!\omega\!\!\text{ }\] what are its amplitude and initial phase angle? The angular frequency of the particle is \[{{\text{e}}^{\text{-}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}{{\text{t}}^{\text{2}}}}}\]. If instead of the cosine function, we choose the sine function to describe the SHM : \[x=B\,\sin (\omega t+\alpha )\], what are the amplitude and initial phase of the particle with the above initial conditions?

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 \[\text{ }\!\!\omega\!\!\text{ =B }\!\!\omega\!\!\text{ cos}\left( \text{ }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 0+ }\!\!\alpha\!\!\text{ } \right)\text{or1=Bcos }\!\!\alpha\!\!\text{ }\]