Question

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 = 2.28 × 109 Pa and rho = 1043 kg/m3. A whale sends out a high frequency (10 kHz) song to another whale 1.0 km away.

Answer 1

Answer:

t = 0.67635 s

n = 6764

Explanation:

Given:

- The velocity of sound in water v:

                                    v = √(B/rho)

Where, B: Bulk Modulus = 2.28*10^9 Pa

            rho: Density of salt water = 1043 kg/m^3

- The wale sends out a high frequency f = 10 kHz

- The distance between two wales s = 1.0 km

Find:

- Time taken for the sound to travel between whales t?

- How many wavelengths can fit between the two whales n?

Solution:

- The time taken for the sound to travel from one whale to another can be determined from:

                               t = s / v

                               t = s / √(B/rho)

                               t = s*√(rho/B)

- Plug in the values:

                               t = 1000*√(1043/2.28*10^9)

                              t = 0.67635 s

- The wavelength λ of the sound emitted can be calculated by the following formula:

                               λ = √(B/rho) / f

                               λ = √(2.28*10^9/1043) / 10^4

                               λ =  0.14785 m

- The number of wavelengths n that could fit in the distance s is:

                               n*λ = s

                               n = 1000 / 0.14785

                              n = 6764

                               

Answer 2

a. The time taken for the sound to travel between the two (2) whales is 0.68 seconds.

b. The number of wavelengths that can fit between the two (2) whales is 6764.

Given the following data:

• Bulk modulus of salt water = $2.28 \times 10^9 \;Pa$

• Density of salt water = 1,043 $kg/m^3$

• Frequency = 10 kHz to Hz = 10,000 Hz

• Distance = 1.0  km to m = 1,000 m

a. To determine the time taken for the sound to travel between the two (2) whales:

Mathematically, velocity of sound in water is given by the formula:

$V=\sqrt{\frac{B}{\rho} }$    .....equation 1.

Where:

• B is the bulk modulus.

• $\rho$ is the density.

• V is the velocity of sound.

Also, time is given by the formula:

$Time =\frac{distance}{velocity}$   .....equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$Time = \frac{d}{\sqrt{\frac{B}{\rho} }} \\\\Time =d(\sqrt{\frac{\rho}{B}} )$

Substituting the given parameters into the formula, we have;

$Time =1000 \times (\sqrt{\frac{1043}{2.28 \times 10^9 }} )\\\\Time =1000 \times\sqrt{4.58 \times 10^{-7}} \\\\Time =1000 \times6.76 \times 10^{-4}$

Time = 0.68 seconds.

b. To determine the number of wavelengths that can fit between the two (2) whales:

Mathematically, wavelength is calculated by using this formula;

$Wavelength = \frac{Velocity }{frequency}$   ...equation 3.

Substituting eqn. 1 into eqn. 3, we have:

$Wavelength = \frac{\sqrt{\frac{B}{\rho} }}{f} \\\\Wavelength = \frac{\sqrt{\frac{2.28 \times 10^9 }{1043} }}{10000} \\\\Wavelength =\frac{\sqrt{2.19 \times 10^6} }{10000} \\\\Wavelength = \frac{1478.5}{10000} \\\\Wavelength = 0.14785 \;meter$

$Number\;of\;wavelength = \frac{Distance}{Wavelength} \\\\Number\;of\;wavelength = \frac{1000}{0.14785 }$

Number of wavelength = 6,763.6 ≈ 6764.

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