Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A bat flying toward a wall at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) emits an ultrasound wave with a frequency of \(30.0 \mathrm{kHz}\). What frequency does the reflected wave have when it reaches the flying bat?

Short Answer

Expert verified
Answer: The frequency of the reflected wave when it reaches the flying bat is 30.0 kHz.

Step by step solution

01

Understand the problem

We are given the speed of the flying bat, the initial frequency of the ultrasound wave, and the wall as the "observer." First, we need to determine the frequency of the wave as it returns to the bat after reflecting off of the wall. To do this, we'll use the Doppler effect formula.
02

Identify velocities and Doppler effect formula for transmitted wave

Before we apply the Doppler effect formula, we have to identify the velocities involved in this problem. Since the bat is flying towards the wall, the bat's velocity, \(v_b\), will be \(7.0 \mathrm{~m} / \mathrm{s}\). The wall's velocity, \(v_w\), will be \(0 \mathrm{~m} / \mathrm{s}\) since it's stationary. The Doppler effect formula when the source is moving towards the stationary observer is: $$f_o = f_s\frac{v + v_o}{v + v_s}$$ Here, \(f_o\) is the observed frequency, \(f_s\) is the source frequency (in our case the frequency emitted by the bat), \(v\) is the speed of sound in air, \(v_o\) is the observer's velocity (the wall), and \(v_s\) is the source velocity (the bat). We don't know the exact speed of sound, but it doesn't matter in our case because later, when considering the reflected wave, we'll see that it will cancel out.
03

Calculate the frequency observed by the wall

Now we can plug the values into the Doppler effect formula to find the frequency observed by the wall: $$f_w = f_s\frac{v + v_w}{v + v_s} = 30.0 \mathrm{kHz} \cdot \frac{v + 0}{v - 7.0}$$
04

Identify velocities and Doppler effect formula for reflected wave

Now we need to find the frequency of the reflected wave received by the bat. In this part, the wall becomes the source, and the bat becomes the observer. We will use the Doppler effect formula for when the observer is moving toward the source: $$f_o = f_s\frac{v - v_o}{v - v_s}$$ Now \(f_s\) will be the frequency of the wave observed by the wall (from step 3), \(v_o\) will be the velocity of the bat, and \(v_s\) will be zero (since the wall is stationary).
05

Calculate the frequency of the reflected wave received by the bat

Using the values calculated before, we can now find the frequency of the reflected wave received by the bat: $$f_b = f_w\frac{v - v_b}{v - v_w} = 30.0 \mathrm{kHz} \cdot \frac{v + 0}{v - 7.0} \cdot \frac{v - 7.0}{v + 0} $$ As you can see, \(v\) cancels out in the equation, and we are left with: $$f_b = 30.0 \mathrm{kHz}$$
06

Conclusion

The frequency of the reflected wave when it reaches the flying bat is \(30.0\,\mathrm{kHz}\).

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!

Key Concepts

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

Doppler effect formula
The Doppler effect is a phenomenon observed when there is relative motion between a source of sound and an observer. This effect causes a shift in the frequency of the sound wave perceived by an observer. The Doppler effect formula is essential for calculating these frequency shifts and is expressed as:

\[\begin{equation}f_o = f_s * \frac{v + v_o}{v + v_s}\end{equation}\]
where:
  • f_o is the observed frequency,
  • f_s is the source frequency,
  • v is the speed of sound in the medium,
  • v_o is the observer's velocity relative to the medium, and
  • v_s is the source's velocity relative to the medium.
In the situation where an observer is moving towards the source, the formula is slightly adjusted:

\[\begin{equation}f_o = f_s * \frac{v - v_o}{v - v_s}\end{equation}\]
This formula captures the intuitive outcome that sounds seem higher-pitched as a source approaches and lower-pitched as it recedes, a familiar experience for many in everyday life such as hearing the changing pitch of a passing siren.
Sound wave reflection
Sound waves are much like light waves in that they can be reflected off of surfaces. When a sound wave hits a surface, it bounces back, which is what we term reflection. In the scenario of the bat and the wall, the ultrasound waves emitted by the bat are reflected off the wall and return to the bat with the same frequency if the wall is not moving. However, due to the bat's movement towards the wall and the principles of the Doppler effect, the frequency of the sound wave when it returns may be perceived differently by the bat.

Sound wave reflection is involved in numerous real-world applications, such as echolocation used by bats and marine animals, as well as human technologies like sonar and radar systems. Understanding how sound waves reflect off different surfaces is crucial in interpreting sonar readings in navigation, geological mapping, and even in medical ultrasound imaging.
Doppler effect application
The Doppler effect has a wide range of applications in various fields. In astronomy, it helps scientists determine whether a star or galaxy is moving towards or away from Earth based on the shift in frequency of the light source, which in turn contributes to the understanding of the expansion of the universe. Meteorology uses the Doppler effect in weather radar to track the velocity of precipitation, providing critical information for storm prediction and warnings.

Emergency vehicles leverage this effect to alert people to their approach, as the siren's pitch sounds higher when the vehicle approaches and lowers as it moves away. Medical diagnostics employ the Doppler effect in ultrasound to monitor blood flow in the body. These practical applications show how the Doppler effect is an essential concept in science and technology, having significant impact on safety, healthcare, and our comprehension of natural phenomena.
Sound frequency calculation
Calculating the frequency of a sound wave involves knowing the speed of sound in the medium through which the wave travels and the wavelength of the sound. The general formula for calculating the frequency (\[\begin{equation} f \end{equation}\]) of a wave is:\[\begin{equation}f = \frac{v}{\r} \end{equation}\]
where:
  • v is the speed of sound and
  • \r is the wavelength.
However, in the context of the Doppler effect, we calculate the perceived frequency based on the relative velocities of the observer and the source. The above formula takes on a specific form for the Doppler effect, incorporating these velocities into the calculation as seen in our bat and wall example. Within this framework, it's important to recognize that the frequency calculated is not the actual frequency emitted by the source, but the one perceived by an observer due to the motion of the source, observer, or both. The Doppler effect equations are thus key to sound frequency calculations in dynamic situations.

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

You are playing a note that has a fundamental frequency of \(400 .\) Hz on a guitar string of length \(50.0 \mathrm{~cm}\). At the same time, your friend plays a fundamental note on an open organ pipe, and 4 beats per seconds are heard. The mass per unit length of the string is \(2.00 \mathrm{~g} / \mathrm{m}\). Assume the velocity of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What are the possible frequencies of the open organ pipe? b) When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string. c) What is the length of the organ pipe?

A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?

A bugle can be represented by a cylindrical pipe of length \(L=1.35 \mathrm{~m} .\) Since the ends are open, the standing waves produced in the bugle have antinodes at the open ends, where the air molecules move back and forth the most. Calculate the longest three wavelengths of standing waves inside the bugle. Also calculate the three lowest frequencies and the three longest wavelengths of the sound that is produced in the air around the bugle.

A (somewhat risky) way of telling if a train that cannot be seen or heard is approaching is by placing your ear on the rail. Explain why this works.

Electromagnetic radiation (light) consists of waves. More than a century ago, scientists thought that light, like other waves, required a medium (called the ether) to support its transmission. Glass, having a typical mass density of \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3},\) also supports the transmission of light. What would the elastic modulus of glass have to be to support the transmission of light waves at a speed of \(v=2.0 \cdot 10^{8} \mathrm{~m} / \mathrm{s} ?\) Compare this to the actual elastic modulus of window glass, which is \(5 \cdot 10^{10} \mathrm{~N} / \mathrm{m}^{2}\).

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