Question

A person hums in a well (closed) and finds strong consecutive resonance frequencies of 60 Hz, 100 Hz, and 140Hz...

explain why these can't be the first consecutive resonance frequencies?
What is the fundamental frequency of the well?
How deep is the well?

Answer 1

fundamental frequency= 20Hz

length= 4.3m

Step-by-step 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)344)/(4*60) = ((2n-1)344)/(4*100) = ((2p-1)344)/(4*140)`

​

(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

`(1*344)/(4 20)` = 4.3m

therefore, the well is 4.3m deep