Question

A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only if the sharks are separated by 3.50m, one being that much farther away than the other. (a) If the ultrasound has a frequency of 100 kHz, show this ability is not limited by its wavelength. (b) If this ability is due to the dolphin’s ability to detect the arrival times of echoes, what is the minimum time difference the dolphin can perceive?

Short answer

(a)This ability that sharks are separated by 3.50mis not limited by its wavelength.

(b) Minimum time for echo difference is $4.55\times {10}^{-3}s$.

Step-by-step solution

The frequency of ultrasound

Ultrasound frequency is defined as the number of ultrasound waves per second. OR The product of frequency and wavelength is the speed of the wave.

(a) Check the ability of the sharks

Write the equation for the wavelength as below.

$\lambda =\frac{v_w}{f}$ ….. (1)

Here,$\lambda$ is the wavelength of the ultrasound wave that would be equal to the small traveling distance of the wave, f is the frequency, and$v_w$ is the speed of sound in seawater.

Consider the given data as below.

The frequency,$\begin{array}{rcl}f& =& 100kHz\\ & =& {10}^{5}Hz\end{array}$

The speed of sound in seawater, $v_w=1540m{s}^{-1}$

Then wavelength in seawater is define by substituting the known values into equation (1).

$\begin{array}{rcl}\lambda & =& \frac{1540}{{10}^5}\\ & =& 0.0154m\end{array}$

And you can say that$3.50m>0.0154m$.

Therefore, this ability that sharks are separated by 3.50mis not limited by its wavelength.

(b) Minimum time for echoes difference

Calculate the arrival time of echoes that is equivalent to a round trip of the sound wave.

$\Delta t=2\times \left(\frac{\Delta d}{v_w}\right)$ ….. (2)

Multiplication of 2 here represents of the round trip of a sound wave.

As known that,

The speed of sound in tissue,$v=1540m/s$

The depth,$\Delta d=3.50m$

Substitute these values into equation (2), and you have

$\begin{array}{rcl}\Delta t& =& 2\times \left(\frac{3.5}{1540}\right)\\ & =& 4.55\times {10}^{-3}s\end{array}$

Hence, a minimum time for echo difference is $4.55\times {10}^{-3}s$.