A) \[\frac{1}{4}m{{\omega }^{2}}{{A}^{2}}\]
B) \[\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\]
C) \[m{{\omega }^{2}}{{A}^{2}}\]
D) 0
Correct Answer: A
Solution :
Average energy \[=\frac{\int_{0}^{T}{U\,dt}}{\int_{0}^{T}{dt}}=\frac{1}{T}\int_{0}^{T}{U\,dt}\] \[=\frac{1}{2T}\int_{0}^{T}{m{{\omega }^{2}}{{A}^{2}}{{\cos }^{2}}(\omega t+\phi )dt}\] \[=\frac{1}{4}m{{\omega }^{2}}{{A}^{2}}\]
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