Question

Calculate the magnitude of the change in A that occurs. The density of air and water is 1.3x10−3g/cm3 and 1.0g/cm3, respectively, and the velocity of sound is 331 m/s in air and 1410 m/s in water. The frequency of the wave remains constant as the wave propagates. Please write your answer with one decimal place. (Hint: Begin with Equation for wave energy of air being similar to that of wave energy in water.to find Ratio between the amplitude of the sound wave in air to that of the amplitude of the sound wave in water..)

Answer 1

Answer:

The amplitude of the wave in air $A_{a}$ will be 57.2 times to the amplitude of the wave in water $A_{w}$.

Explanation:

We know that the power carried by a wave (P) while propagating through a medium is given by

$P = 2 \pi v \rho \nu^{2} A^{2}$

where 'v' is the velocity of the wave propagating through a medium, '$\rho$' is the density of the medium, '$\nu$' is the frequency of the wave and 'A' is the amplitude of the wave.

For air medium,

$P_{a} = 2 \pi v_{a} \rho_{a} \nu^{2} A_{a}^{2}$

and for water medium,

$P_{w} = 2 \pi v_{w} \rho_{w} \nu^{2} A_{w}^{2}$

According to the conservation of energy, we can equate the above two equations, as given the frequency of the wave remains constant. So,

$&& v_{a} \rho_{a} A_{a}^{2} = v_{w} \rho_{w} A_{w}^{2}\\&or,& \dfrac{A_{a}}{A_{w}} = \sqrt{\frac{v_{w}\rho_{w}}{v_{a}\rho_{a}}}\\&or,& \dfrac{A_{a}}{A_{w}} = \sqrt{\dfrac{1410 \times 10^{2} cm s^{-1} \times 1 g cm^{-3}}{331 \times 10^{2} cm s^{-1} \times 1.3 \times 10^{-3} g cm^{-3}}} \approx 57.2\\&or,& A_{a} = 57.2 A_{w}$