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 3.50 m, 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?
Final Answer
- . Since this wavelength is much less than the minimum perceptible separation, it means that wavelength is not the limiting factor in the dolphin's perceptual abilities.
Solution video
OpenStax College Physics, Chapter 17, Problem 82 (Problems & Exercises)
vote with a rating of
votes with an average rating of
.
Prev
Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dychko. A dolphin can't tell if there are two sharks unless the sharks are separated by more than 3.50 meters and we are going to explore why that is in this question. So in part (a) we suppose that the wavelength of the ultrasound the dolphin is emitting that creates this limit; if this wavelength was bigger than 3.50 meters then that would account for their inability to distinguish details that are smaller than that. So the wave equation says that the speed of a sound wave equals its frequency multiplied by its wavelength so let's solve for λ by dividing both sides by frequency. And so λ is the speed of sound in saltwater divided by the frequency and the speed of sound in saltwater is 1540 meters per second according to table [17.1] and so we divide that speed by 100 times 10 to the 3 hertz and that's 0.0154 meters. So this is very good resolution in terms of the wavelength and so that's not the limiting factor. The limiting factor is their perception of the time difference between the reflections of the sound from the two different sharks. So the sound is emitted by the dolphin here and first it hits this shark that's closer and then that gets reflected and then the dolphin will detect the time at which this reflection occurs and then compare that with the time that this reflection occurs and they will be able to measure a difference when these sharks are separated by 3.50 meters in which case, the sound will have taken some time to go from the first shark to the second and then back again; that's the total time difference for these two sound waves to get back to the dolphin. So this total time difference will be the distance traveled by this second ray that's additional and divided by the speed of sound and this distance is 2 times the separation between the sharks. So that separation we are labeling as L and so we can substitute 2L in place of this d distance. So the extra time for the second echo is 2L over v so that's 2 times 3.50 meters divided by 1540 meters per second which is 4.55 milliseconds. So this is the smallest amount of time that the dolphin is able to perceive.