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What is your observation when two sources are emitting sound with frequency 499 Hz and 501 Hz ? |
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Answer» (A) Frequency of 500 Hz is HEARD with change in intensity take PLACE twice. `(f _(1) + f _(2))/(2) =(499 +501)/(2) =500 Hz` (Here intensity of resultant wave changes `f _(2) -f_(1)` times in `1S = (501 - 499 )=2 Hz)` |
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