The Vespertilionidae family of bats detect the distance to an object by timing how long it takes for an emitted signal to reflect off the object and return. Typically they emit sound pulses 3 ms long and 70 ms apart while cruising. (a) If an echo is heard 60 ms later $\left(v_{\text {sound }}=331 \mathrm{m} / \mathrm{s}\right),\( how far away is the object? (b) When an object is only \)30 \mathrm{cm}$ away, how long will it be before the echo is heard? (c) Will the bat be able to detect this echo?

Short Answer

Expert verified
Answer: (c) The bat will be able to detect the echo at 30 cm away if the time it takes for the echo to be heard is less than the time between pulses (70 ms).

Step by step solution

01

(a) Finding the distance to the object

: To determine the distance of the object from the bat when it hears the echo after 60 ms (0.06 s), we can use the speed of sound (\(v_{sound} = 331 m/s\)) and the time it takes to hear the echo following the formula: total distance = speed × time. Since the sound needs to travel to the object and back, we will divide the total distance by 2 to get the distance to the object. The total distance traveled by the sound is: Total distance = \(v_{sound}\) × time Total distance = \(331 m/s\) × \(0.06 s\) Distance to the object = Total distance / 2
02

(b) Finding the time before the echo is heard

: We are given that the object is 30 cm (0.3 m) away from the bat. To find the time it takes for the echo to be heard by the bat, we can once again use the relationship between speed, distance, and time: time = distance / speed. Since the sound travels to the object and back, we need to multiply the distance by 2 to get the total distance: Total distance = distance to object × 2 Total distance = \(0.3 m\) × 2 Time before the echo is heard = Total distance / \(v_{sound}\)
03

(c) Determining if the bat can detect the echo

: Finally, we need to determine if the bat can detect the echo when the object is 30 cm away. From part (b), we found the time it takes for the echo to be heard at this distance. We have the given information that the sound pulses are 3 ms long and 70 ms apart, and we will use this to determine if the bat can detect the echo. If the time before the echo is heard is less than the time it takes between pulses (70 ms), then the bat will be able to detect the echo. Otherwise, the bat won't be able to detect the echo. Comparing the time between pulses and the time before the echo is heard (from part (b)), we can determine if the bat can detect the echo.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Your friend needs advice on her newest "acoustic sculpture." She attaches one end of a steel wire, of diameter \(4.00 \mathrm{mm}\) and density $7860 \mathrm{kg} / \mathrm{m}^{3},$ to a wall. After passing over a pulley, located \(1.00 \mathrm{m}\) from the wall, the other end of the wire is attached to a hanging weight. Below the horizontal length of wire she places a 1.50 -m-long hollow tube, open at one end and closed at the other. Once the sculpture is in place, air will blow through the tube, creating a sound. Your friend wants this sound to cause the steel wire to oscillate at the same resonant frequency as the tube. What weight (in newtons) should she hang from the wire if the temperature is \(18.0^{\circ} \mathrm{C} ?\)
A geological survey ship mapping the floor of the ocean sends sound pulses down from the surface and measures the time taken for the echo to return. How deep is the ocean at a point where the echo time (down and back) is $7.07 \mathrm{s} ?\( The temperature of the seawater is \)25^{\circ} \mathrm{C}$
A cello string has a fundamental frequency of \(65.40 \mathrm{Hz}\) What beat frequency is heard when this cello string is bowed at the same time as a violin string with frequency of \(196.0 \mathrm{Hz} ?\) [Hint: The beats occur between the third harmonic of the cello string and the fundamental of the violin. \(]\)
An auditorium has organ pipes at the front and at the rear of the hall. Two identical pipes, one at the front and one at the back, have fundamental frequencies of \(264.0 \mathrm{Hz}\) at \(20.0^{\circ} \mathrm{C} .\) During a performance, the organ pipes at the back of the hall are at $25.0^{\circ} \mathrm{C},\( while those at the front are still at \)20.0^{\circ} \mathrm{C} .$ What is the beat frequency when the two pipes sound simultaneously?
An organ pipe that is open at both ends has a fundamental frequency of $382 \mathrm{Hz}\( at \)0.0^{\circ} \mathrm{C}$. What is the fundamental frequency for this pipe at \(20.0^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free