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Two distant sources situated together emit sound each of frequency 300cycles persecond. Ifoneofthemweretoapproach and the other to recede from a stationary observer each with a velocity of 1//100^th the velocity ofsound, calculate the number of beats per second heard by the observer. |
Answer» Solution :The situation is shown in figure-6.87.  The first source is approaching the stationary OBSERVER. The frequencyof the source as heard by observeris given by `n_(1) = n((v)/(v - v_(s)))` Where n is the actual frequency of source `n_(1) = 300((v)/(v - v//100))` `(300 xx 100)/(99)` = 303.03 Hz The second source is receding from the observer in the direction opposite to the sound. Hence the APPARENT frequency `n_(2)` `n_(2) = ((v)/(v + 0.01v)) xx 300` `(300 xx 100)/(101)` = 297.03 Hz Hence BEAT frequency detected by observer is `Delta n = 303.03 - 297.03 = 6 Hz` |
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