A) \[\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
B) \[\text{4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
C) \[\text{8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
D) \[\text{16 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
Correct Answer: A
Solution :
Given, \[k=16N/m,\] \[m=1kg\] amplitude \[a=5cm=5\times {{10}^{-2}}m.\] Kinetic energy of a body executing SHM. \[=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] When displacement \[y=0,\] then KE is maximum. \[\therefore \] Maximum \[KE=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] \[=\frac{1}{2}m{{\left( \frac{2\pi }{T} \right)}^{2}}{{a}^{2}}\] \[=\frac{1}{2}m{{\left( \sqrt{\frac{k}{m}} \right)}^{2}}{{a}^{2}}\] \[=\frac{1}{2}m\times \frac{k}{m}\times {{a}^{2}}=\frac{1}{2}k{{a}^{2}}\] \[=\frac{1}{2}\times 16kt{{(5\times {{10}^{-2}})}^{2}}\] \[=200\times {{10}^{-4}}=2\times {{10}^{-2}}J\]
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