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Chapter 16: Problem 69

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掳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掳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掳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 stationary police car emits a sound of frequency 1200 Hz that bounces off a car on the highway and returns with a frequency of 1250 Hz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving toward or away from the police car? (b) What frequency would the police car have received if it had been traveling toward the other car at 20.0 m/s?

A baby's mouth is 30 cm from her father's ear and 1.50 m from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother?

\(\textbf{Longitudinal Waves on a Spring.}\) A long spring such as a Slinky\(^{\mathrm{TM}}\) is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke鈥檚 law has mass \(m\), length \(L\), and force constant \(k'\), the speed of longitudinal waves on the spring is \(v = L\sqrt{ k'/m}\) (see Section 16.2). (b) Evaluate \(v\) for a spring with \(m =\) 0.250 kg, \(L =\) 2.00 m, and \(k' =\) 1.50 N\(/\)m. \(\textbf{ULTRASOUND IMAGING}\). A typical ultrasound transducer used for medical diagnosis produces a beam of ultrasound with a frequency of 1.0 MHz. The beam travels from the transducer through tissue and partially reflects when it encounters different structures in the tissue. The same transducer that produces the ultrasound also detects the reflections. The transducer emits a short pulse of ultrasound and waits to receive the reflected echoes before emitting the next pulse. By measuring the time between the initial pulse and the arrival of the reflected signal, we can use the speed of ultrasound in tissue, 1540 m/s, to determine the distance from the transducer to the structure that produced the reflection. As the ultrasound beam passes through tissue, the beam is attenuated through absorption. Thus deeper structures return weaker echoes. A typical attenuation in tissue is \(-\)100 dB/m \(\cdot\) MHz; in bone it is \(-\)500 dB/m \(\cdot\) MHz. In determining attenuation, we take the reference intensity to be the intensity produced by the transducer.

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 10.0% higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

A police siren of frequency \(f_{siren}\) is attached to a vibrating platform. The platform and siren oscillate up and down in simple harmonic motion with amplitude \(A_p\) and frequency \(f_p\). (a) Find the maximum and minimum sound frequencies that you would hear at a position directly above the siren. (b) At what point in the motion of the platform is the maximum frequency heard? The minimum frequency? Explain.
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