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Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. Their phase difference is |
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Answer» Let equations of motion of two particles be `x=Asinomegat` `x=Asin(omegat+phi)` (ii) When `x=A//2`, from (i), `sin omegat=1//2=sinpi//6` `omegat=pi//6` When `x=A/2`, from (ii), `sin (omegat+phi)=1/2=sinpi//6` or `sin ((5pi)/(6))` `omegat+phi=pi//6` or `(5pi)/(6)` (a) If `omegat+phi=pi//6impliespi//6+phi=pi//6impliesphi=0` (not possible) (b) If `omegat+phi=5pi//6impliespi//6+phi=(5pi)/(6)impliesphi=(2pi)/(3)=120^@` |
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