Question

(8%) Problem 3: Sound in water travels at a velocity governed by the relation v = √(B/rho) where B is the bulk modulus and rho is the density. For salt water, take B = 2.38 × 109 Pa and rho = 1046 kg/m3. A whale sends out a high frequency (9 kHz) song to another whale 1.0 km away. How much time, in seconds, does it take for the sound to travel between the whales, t?

Answer 1

Answer:

the time required for the sound to travel between the whales 0.66 S.

Explanation:

As given in the problem, the velocity of sound wave ($v$) is governed by the equation

$v = \sqrt{\dfrac{B}{\rho}}$

Given, $B = 2.38 \times 10^{9} Pa$ and $\rho = 1046 Kg m^{-3}$

So for salt water, the velocity of sound wave ($v_{s}$) can be written as

$v_{s} = \sqrt{\dfrac{2.38 \times 10^{9}}{1046}} ms^{-1} = 1.508 \times 10^{3} ms^{-1}$

As the whales are d = 1 Km or 1000 m apart from each other, so the time ($t$) required for the sound wave to travel this distance is given by

$t = \dfrac{d}{v_{s}} = \dfrac{1000 m}{1.508 \times 10^{3}} = 0.66 s$