Question

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

Answer 1

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

$\frac{(2m-1)344}{4*60} = \frac{(2n-1)344}{4*100} = \frac{(2p-1)344}{4*140}$

​  

 (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

$\frac{1*344}{4 20}$ = 4.3m

therefore, the well is 4.3m deep