A) \[4:1\]
B) \[8:1\]
C) \[3:1\]
D) \[2:1\]
Correct Answer: C
Solution :
: The kinetic energy of the particle executing SHM at a distance x from its equilibrium position is \[K=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] and the potential energy is \[U=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] where A is the amplitude, co is the angular frequency and M is the mass of the body. At \[x=\frac{A}{2}\] \[K=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{A}{2} \right)}^{2}} \right)\] and \[U=\frac{1}{2}m{{\omega }^{2}}{{\left( \frac{A}{2} \right)}^{2}}\] Their corresponding ratio is \[\frac{K}{U}=\frac{\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{A}{2} \right)}^{2}} \right)}{\frac{1}{2}m{{\omega }^{2}}{{\left( \frac{A}{2} \right)}^{2}}}=\frac{\left( {{A}^{2}}-\frac{{{A}^{2}}}{4} \right)}{\frac{{{A}^{2}}}{4}}\] \[=\frac{\frac{3}{4}{{A}^{2}}}{\frac{{{A}^{2}}}{4}}=\frac{3}{1}\]
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