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The initial phase of a simple harmonic oscillator is zero. At what fraction of the period, the velocity is half of its maximum value?A. 1B. `1//2`C. `(2//3)`D. `1//6` |
Answer» Correct Answer - D `v=Aomegacosomegat` `therefore (1)/(2)Aomega=Aomegacosomegat` `therefore omegat=cos^(-1)((1)/(2))` `(2pi)/(T)t=(pi)/(3)` `therefore (t)/(T)=(1)/(6)` |
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