Answer to Question #134212 in Physics for James Winnie

Question

An ocean-depth sounding device emits a signal 38kHz in water whose average temperature is 4 degree celcius. The impulse is reflected from the ocean bed and returned to the same location 1.62 seconds after the signal is emmited. Calculate:
(i) the depth of the water
(ii) the wavelength of the signal in water

Answer 1

An empirical equation for the speed of sound in sea water is the following (see https://en.wikipedia.org/wiki/Speed_of_sound#Water):

$$c(T) = a_1 + a_2 T + a_3 T^2 + a_4 T^3$$

where $a_1 = 1448.96, a_2 = 4.591, a_3 = -5.304 \times 10^{-2}, a_4 = 2.374 \times 10^{-4}$. Thus, for $T = 4\degree C$ :

$$c(T) = 1448.96 + 4.591\cdot 4 + -5.304 \times 10^{-2}\cdot 4^2 + 2.374 \times 10^{-4}\cdot 4^3 =\\ \approx 1467m/s$$

Then, in $t = 1.62s/2 = 0.81s$ the impulse will reach the ocean bed. Thus, it will travell the following distance:

$$h = ct = 1467m/s\cdot 0.81 = 1188.27m$$

The wavelength is given by the following expression:

$$\lambda = \dfrac{c}{\nu}$$

where $\nu = 38kHz = 38\times10^3Hz$ is the frequency of the wave. Obtain then:

$$\lambda = \dfrac{1467m/s}{38\times10^3Hz} = 38.6\times 10^{-3}m$$

Answer. (i) $1188.27m$, (ii) $38.6\times 10^{-3}m$.