Question

A well with vertical sides and water at the bottom resonates at 5.63 Hz and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of 1.10 kg/m3 and a bulk modulus of 1.33 × 10^5 Pa. How far down in the well is the water surface?

Answer 1

The value is s = 15.4 \ m

Step-by-step explanation:

From the question we are told that

The frequency is
`f =  5.63 \  Hz`

The density of air in the well is
`\sigma  =  1.33 *10^(5) \  Pa`

The Bulk modulus is
`B =  1.33*10^( 5) \  Pa`

Generally the distance of the water surface from the opening is mathematically represented as

`s =  (1)/(4 f )  *  \sqrt{(B)/(\rho ) }`

=>
`s =  (1)/( 4 *  (5.63 )) \sqrt{(1.33*10^(5))/( 1.10) }`

=>
`s =  15.4 \ m`