A) \[2\sqrt{2}m\]
B) 2m
C) \[\sqrt{2}m\]
D) none of the above
Correct Answer: B
Solution :
: Total energy of particle executing SHM \[=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] Potential energy of particle\[=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[\frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\]or\[k=m{{\omega }^{2}}\] \[\therefore \]Total energy\[=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] Or, \[1=\frac{k{{a}^{2}}}{2}\] or \[{{a}^{2}}=\frac{2}{k}=\frac{2}{0.5}=\frac{4}{1}\] or \[a=2\text{ }m\].
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