A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{3}{2}\]
Correct Answer: C
Solution :
\[K.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\omega }^{2}}\,\,\,({{A}^{2}}-{{x}^{2}})\] at \[x=0,\]\[K.E.=\frac{1}{2}m{{v}^{2}}_{\max }=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] at \[x=\frac{A}{2},\,K.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\omega }^{2}}\left[ {{A}^{2}}-{{\left( \frac{A}{2} \right)}^{2}} \right]\] \[=\frac{3}{4}\left[ \frac{1}{2}m{{\omega }^{2}}{{A}^{2}} \right]\] \[\Rightarrow \] \[\frac{K{{E}_{x=0}}}{K{{E}_{x=A/2}}}=\frac{4}{3}\]
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