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Two sound waves of frequencies 100Hz and 102Hz and having same amplitude 'A' are interfering. At a stationary detector, which can detect resultant amplitude greater than or equal to A. So, ina given time interval of 12 seconds, finds the total duration in which detector is active. |
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Answer» `y_(2)=A sin omega_(2) t` `y_(1)=2A cos {((omega_(2)-omega_(1)))/2}{"sin"((omega_(2)+omega_(1)))/2 t}` Resultant amplitude `A_(r)=2A_(0)Icos(/_\omega)t//2I` `(/_\omega)t/2=(pi)/2impliest=1/4s` `(/_\omega)t/2=(pi)/3impliest=1/6s` In one cycle of INTENSITY of `1//2s` the detector remain idle for `2(1/4-1/2)s=1/6sec` `:.` In `1//2` sec cyclem active time is `(1/2-1/6)` `=1//3` sec `:.` In `12` sec INTERVAL, active time is `12xx((1//3))/((1//2))=8SEC` 
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