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Two waves are represented by the equations `y_(1)=a sin (omega t+kx+0.785)` and `y_(2)=a cos (omega t+kx)` where, x is in meter and t in second The phase difference between them and resultant amplitude due to their superposition areA. `45^(@)` and 1.84 aB. `30^(@)` and aC. `30^(@)` and 2aD. `45^(@)` and a |
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Answer» Correct Answer - A `y_(1)= a sin (omega t+kx+0.57)` `:.` Phase `phi_(1)=omegat+kx+0.57` `y_(2)=a cos (omega t+kx)` `=a sin (omega t+kx +pi/2)` `:.` Phase, `phi_(2)=omegat+kx+pi/2` Phase diferent, `Delta phi=phi_(2)-phi_(1)` `=[omega t+kx+pi/2]-(omegat+kx+0.57)` `=pi/2 -0.785` `=(1.57-0.785) rad =0.785` rad `=pi/4 rad = 45^(@)` Resultant amplitude, `a=2 a "cos" phi/2=2a " cos"pi/8=1.84 a` |
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