question_answer 25)

A mass attached to a spring is free to oscillate, with angular velocity \[\text{V= }\!\!\omega\!\!\text{ }\sqrt{{{\text{r}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}}\], in a horizontal plane without friction or damping. It is pulled to a distance \[0\cdot 05\] and pushed towards the centre with a velocity \[\begin{align} & A=-{{\left( 10\pi \right)}^{2}}\times 0\cdot 05=-5{{\pi }^{2}}m/{{s}^{2}} \\ & V=10\pi \times \sqrt{{{\left( 0\cdot 05 \right)}^{2}}-{{\left( 0\cdot 05 \right)}^{2}}}=0 \\ \end{align}\] at time \[t=0\]. Determine the amplitude of the resulting oscillations in terms of the parameters \[0\cdot 03\] and \[A=-{{\left( 10\pi \right)}^{2}}\times 0\cdot 03=-3{{\pi }^{2}}m/{{s}^{2}}\]

Answer:

When \[{{\left( \text{10 }\!\!\pi\!\!\text{ } \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 0=0}\] and \[\text{V=10 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ }\sqrt{{{\left( 0\cdot 05 \right)}^{2}}-{{0}^{2}}}\text{10 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 0}\cdot 05=0\cdot \text{5 }\!\!\pi\!\!\text{ mk/s}\] \[\omega \] From \[x=A\]\[\cos (\omega t+\theta );\,{{x}_{0}}=A\cos \theta \] velocity \[{{\upsilon }_{0}}\]\[\text{ }\!\!\omega\!\!\text{ ,}{{\text{x}}_{\text{0}}}\] Squaring and adding (i) and (ii), we get \[{{\text{ }\!\!\upsilon\!\!\text{ }}_{\text{0}}}\text{.}\] \[t=0,x={{x}_{0}}\]