A) zero
B) 1
C) 3/4
D) None of these
Correct Answer: C
Solution :
In SHM, if amplitude is a then, \[\operatorname{Total} energy =\frac{1}{2}m\,{{\omega }^{2}}{{a}^{2}}\] Then, kinetic energy at displacement x is \[=\,\,\,\frac{1}{2}m\,{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\] So, kinetic energy \[\left( at,\,\,x=\frac{a}{2} \right)=\frac{1}{2}m\,{{\omega }^{2}}\left[ {{a}^{2}}-{{\left( \frac{a}{2} \right)}^{2}} \right]\] \[=\,\,\,\frac{\frac{3}{4}\,\,\left( \frac{1}{2}m\,{{\omega }^{2}}\,{{a}^{2}} \right)}{\frac{1}{2}\,m\,\,{{\omega }^{2}}\,{{a}^{2}}}\] \[=\,\,\,\frac{3}{4}\]
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