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Mechanical Properties Of Fluids

Question
CBSEENPH11019214

Time period of simple harmonic oscillator is T. In what time will it traverse a distance equal to half of amplitude from mean position?

Solution

The displacement of particle executing simple harmonic oscillation is given by, 
                      y = A sin ωt

Let the particle take to time to go from mean position to half of extreme position. 
Therefore, 
therefore space space space space space space space space space straight A over 2 equals straight A space s i n space omega t subscript 0
rightwards double arrow space space space space space space space space space space s i n space omega t subscript 0 equals 1 half
rightwards double arrow space space space space space space space space space space space omega t subscript straight omicron equals straight pi over 6
rightwards double arrow space space space space space space space space space straight t subscript 0 equals fraction numerator straight pi over denominator 6 straight omega end fraction equals fraction numerator straight pi cross times straight T over denominator 6 cross times 2 straight pi end fraction equals straight T over 12
So, at a time of straight T over 12 the simple harmonic oscillator is equal to half of amplitude from the mean position.