A) \[12\times {{10}^{-3}}J\]
B) \[25\times {{10}^{-3}}J\]
C) \[0.48\times {{10}^{-3}}J\]
D) \[0.24\times {{10}^{-3}}J\]
Correct Answer: C
Solution :
Displacement of particle in the case of SHM \[y=A\sin (\omega t+\phi )\] ...(i) \[y=3\sin (0.2t)\] ... (ii) (given) Comparing Eqs. (i) and (ii), we get \[A=3,\] \[\omega =0.2\] Now, particle distance \[x=\frac{A}{3}=1\] Kinetic energy in SHM \[=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] \[=\frac{1}{2}\times 3\times {{10}^{-3}}{{(0.2)}^{2}}[{{3}^{2}}-{{1}^{2}}]\] \[=0.48\times {{10}^{-3}}J\]
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