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

Answer:

The  value  is  s =  15.4 \ m

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

      $s =  \frac{1}{4 f }  *  \sqrt{\frac{B}{\rho } }$

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

=>   $s =  15.4 \ m$