| (I) amplitude |
| (II) period |
| (III) displacement |
A) \[\text{I}\] and \[\text{II}\] are correct
B) \[\text{II}\] and \[\text{III}\] are correct
C) \[\text{I}\] and \[\text{III}\] are correct
D) \[\text{I, II }and\,\text{III}\,are\,correct\]
Correct Answer: A
Solution :
Key Idea: Total energy of a particle executing simple harmonic motion is obtained by summing its potential and kinetic energies. Potential energy of particle in SHM \[U=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] or \[U=\frac{1}{2}m{{(2\pi f)}^{2}}{{x}^{2}}\] or \[U=2{{\pi }^{2}}m{{f}^{2}}{{x}^{2}}\] ?(i) Kinetic energy of particle in SHM \[K=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] or \[K=2{{\pi }^{2}}m{{f}^{2}}({{A}^{2}}-{{x}^{2}})\] ?(ii) Hence, total energy \[E=K+U\] \[=2{{\pi }^{2}}m{{f}^{2}}{{x}^{2}}+2{{\pi }^{2}}m{{f}^{2}}({{A}^{2}}-{{x}^{2}})\] \[=2{{\pi }^{2}}m{{f}^{2}}{{A}^{2}}=\frac{2{{\pi }^{2}}m{{A}^{2}}}{{{T}^{2}}}\] \[\left( \because \,T=\frac{1}{f} \right)\] Thus, it is obvious that total energy of particle executing simple harmonic motion depends on amplitude (A) and period (T). \[\text{II}\] and \[\text{III}\] are correct
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