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Horseshoe bats (genus \(Rhinolophus\)) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A \(Rhinolophus\) flying at speed \(v_{bat}\) emits sound of frequency \(f_{bat}\); the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{refl}\). (a) Show that the speed of the insect is $$vinsect = v\Bigg[\frac{f_{refl}(v - v_{bat}) - f_{bat}(v + v_{bat})}{f_{refl}(v - v_{bat}) + f_{bat}(v + v_{bat})}\Bigg] $$ where \(v\) is the speed of sound. (b) If \(f_{bat} =\) 80.7 kHz, \(f_{refl} =\) 83.5 kHz, and \(v_{bat} =\) 3.9 m/s, calculate the speed of the insect.

Short Answer

Expert verified
The speed of the insect is approximately 7 m/s.

Step by step solution

01

Understanding the Problem

We are given an exercise involving the Doppler effect where a bat and insect are moving. The bat emits sound (frequency \( f_{bat} \)) and hears a reflected frequency \( f_{refl} \). The task is to derive the formula for the speed \( v_{insect} \) of the insect using the given information.
02

Establish the Doppler Effect Equations

The Doppler effect equation for sound involves shifts in frequency due to relative motion. For a source moving toward a stationary observer: \[ f' = f \frac{v + v_{observer}}{v - v_{source}} \] For a source moving away, replace \( + \) with \( - \). We need to apply these equations in two parts: from the bat to the insect, then from the insect back to the bat.
03

First Leg - Bat to Insect

In the first leg, the bat is the source and the insect acts like an observer. The frequency of sound reaching the insect, \( f' \), can be expressed as: \[ f' = f_{bat} \frac{v}{v - v_{bat}} \] because the bat is moving toward the insect.
04

Second Leg - Insect Reflects Sound

Next, the insect becomes the source (reflecting the sound) and the bat is the observer. The reflected frequency \( f_{refl} \) is detected by the bat: \[ f_{refl} = f' \frac{v + v_{bat}}{v - v_{insect}} \] We substitute \( f' \) from Step 3 into this equation.
05

Combine and Solve for \( v_{insect} \)

Substitute \( f' \) from Step 3 into the equation in Step 4, we get: \[ f_{refl} = f_{bat} \frac{v (v + v_{bat})}{(v - v_{bat})(v - v_{insect})} \] Rearrange for \( v_{insect} \) to solve the equation that brings us to the given complex formula.
06

Calculate \( v_{insect} \) with Given Values

Insert the known values into the simplified formula derived: \( f_{bat} = 80.7 \text{kHz} \), \( f_{refl} = 83.5 \text{kHz} \), and \( v_{bat} = 3.9 \text{m/s} \). Using the equation: \[ v_{insect} = v \left[\frac{83.5 (v - 3.9) - 80.7 (v + 3.9)}{83.5 (v - 3.9) + 80.7 (v + 3.9)}\right] \] Put in the speed of sound \( v \approx 343 \text{m/s} \) (standard at 20°C). Further computations give the insect's speed: \( v_{insect} \approx 7 \text{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency Calculation
In the context of the Doppler effect, frequency calculation is crucial for understanding how the motion of a source or observer affects the sound we perceive. When an observer moves towards a sound source, or vice versa, the perceived frequency changes. This is because the waves are either compressed or stretched out.

In a scenario involving a bat and its prey, calculating the correct frequency is key. When sound waves bounce off a moving object, frequencies shift. To figure out the frequency of these waves after they reflect, the Doppler effect equation is our tool. You might remember the formula:
  • For a source moving towards an observer: \( f' = f \frac{v + v_{observer}}{v - v_{source}} \)
  • For a moving reflecting object: It's similar but involves the motion of both parties.
Using these formulas, you can calculate the frequency the bat hears by figuring out how fast the prey is moving. This involves substituting known values and rearranging the equation to solve for unknowns. Simple algebra lets us translate real-world motions into measurable changes in sound!
Bat Echolocation
Bats, particularly those in the genus \(Rhinolophus\), rely on echolocation to hunt. They emit ultrasonic sounds (high-frequency sounds) from their nostrils and listen for echoes that bounce back from objects, like insects. This natural sonar system allows bats to "see" their surroundings using sound instead of sight.

Echolocation involves emitting a sound that travels at the speed of sound (around 343 m/s in air at 20°C). When the sound waves hit an object, they reflect back, and the bat uses these echoes to determine the object's location and speed. Imagine shining a flashlight in a dark room and seeing the light reflect off objects to understand where they are. Bats are like the flashlights, using sound to illuminate their path.

When a bat hunts, it listens for changes in frequency caused by the motion of its prey, thanks to the Doppler effect. This allows the bat to determine whether the prey is moving towards or away from it. As the bat is flying, it can adjust its path to intercept the prey. In this intricate dance of sound, the prey's speed affects how frequency shifts, enabling the bat to react accurately.
Speed of Sound
The speed of sound is a fundamental constant in physics that plays a significant role in equations related to the Doppler effect. At sea level and at 20°C, the speed of sound in air is approximately 343 m/s.

The speed of sound depends on temperature, humidity, and atmospheric pressure. Higher temperatures mean faster sound speeds, as warmer air particles move more quickly, making it easier for sound to travel.

In the context of echolocation, the speed of sound is essential in measuring distances and speeds. Bats rely on precise time measurements of the time it takes for sound to travel to an object and back. Knowing the speed of sound allows them to translate these times into distances or to calculate an approaching prey's velocity. When substituting the speed of sound into the Doppler effect formula, it turns an abstract calculation into a real-world measurement that bats use autonomously every night.

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Most popular questions from this chapter

(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if (a) the pipe is open at both ends and (b) the pipe is closed at the left end and open at the right end.

A bat flies toward a wall, emitting a steady sound of frequency 1.70 kHz. This bat hears its own sound plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of 8.00 Hz?

A turntable 1.50 m in diameter rotates at 75 rpm. Two speakers, each giving off sound of wavelength 31.3 cm, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. (a) What is the greatest beat frequency the listener will receive from this system? (b) Will the listener be able to distinguish individual beats?

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0 \(\times\) 10\(^{-5}\) Pa. Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at 20\(^\circ\)C.

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