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Two particles are executing simple harmonic of the same amplitude (A) and frequency `omega` along the x-axis . Their mean position is separated by distance `X_(0)(X_(0)gtA). If the maximum separation between them is (X_(0)+A), the phase difference between their motion is: |
Answer» `x_(1) = A sin omegat, x_(2) =Asin (omegat + theta) +x_(0)` `x_(2) -x_(1) = x_(0) +A (sin(omegat +theta)-sin(omegat))` `x_(2) -x_(1) =x_(0) +2A sin ((theta)/(2)) cos (omegat+(theta)/(2))` The distance between the two `SHM` is also oscillating simple harmonically with an amplitude of `x_(0) = 2A sin ((theta)/(2))`. maximum distance between two `SHM` is `x_(0) +A` from the above `x_(0) +2A sin ((theta)/(2)) = x_(0) +A` `sin((theta)/(2)) = (1)/(2), theta = (pi)/(3)` |
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