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A simple harmonic oscillator has a time period of 2s. What will be the change in the phase after 0.25s after leaving the mean position? |
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Answer» Solution :Time period T= 2 sec, time = 0.25 sec, PHASE DIFFERENCE after sec `=f xx t/T xx 2PI=(0.25)/(2) xx 2pi = pi/4 , pi/4 = 90^@` For a phase `pi/4` starting from MEAN position the body will be EXTREME position. (Phase difference between mean position and extreme position is `pi/4` Rad or `90^@`) |
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