A person hums in a well (closed) and finds strong consecutive resonance frequencies of 60 Hz, 100 Hz, and 140Hz...
1 answer:
Answer:
fundamental frequency= 20Hz
length= 4.3m
Explanation:
resonating frequencies = 60 Hz,100 Hz,140 Hz
speed of sound =344 m/s
as well behaves as organ pipe with are end closed
only odd harmonics are allowed
, so the vibration will corresponds to n=1,3,5,7....
(2n−1) λ
/4 = l (∵λ
= V/f)
(2n-1) V / 4f = l
(2m-1)/6= (2n-1)/10= (2p-1)/14 =A
The given frequencies happen to be 3rd,5th & 7th harmonics of the fundamental frequency n=20Hz.
Therefore, The fundamental frequency will be 20 Hz because 60 Hz is corresponding 3rd harmonic
In order to find length,
(2n-1) V / 4f = l
= 4.3m
therefore, the well is 4.3m deep
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