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A bat emitting an ultrasonic wave of frequency \(4 \cdot 5 \times 10^{4} \mathrm{~Hz}\) flies at a speed of \(6 \mathrm{~m} / \mathrm{s}\) between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is \(330 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The bat hears frequencies of approximately 46,471 Hz and 44,153 Hz with a beat frequency of 2,318 Hz.

Step by step solution

01

Identify the frequencies as the bat approaches wall 1

As the bat flaps towards wall 1, the frequency it hears increases due to the Doppler Effect. Using the formula \( f' = f \left( \frac{v + v_0}{v} \right) \), where \( f \) is the original frequency, \( v \) is the speed of sound, and \( v_0 \) is the speed of the bat, we compute:\[f' = 4.5 \times 10^4 \left( \frac{330 + 6}{330} \right)\]Calculate this to find the frequency heard by the bat approaching wall 1.
02

Calculate the shifted frequency from wall 1

Since the ultrasonic wave hits wall 1 and reflects, the frequency shifts again. This is similar to a source now moving towards a stationary observer, so we use the formula:\[f'' = f' \left( \frac{v}{v - v_0} \right) = 4.5 \times 10^4 \left( \frac{330 + 6}{330} \right) \left( \frac{330}{330 - 6} \right)\]Compute \( f'' \), the actual frequency heard by the bat from wall 1.
03

Identify the frequencies as the bat approaches wall 2

Similarly, as the bat travels towards wall 2, calculate the frequency using:\[f' = 4.5 \times 10^4 \left( \frac{330 - 6}{330} \right)\]This accounts for the Doppler shift in frequency as the bat moves towards wall 2.
04

Calculate the reflected frequency from wall 2

The emitted wave strikes wall 2 and reflects back to the bat. Apply the Doppler formula again:\[f'' = f' \left( \frac{v}{v + v_0} \right) = 4.5 \times 10^4 \left( \frac{330 - 6}{330} \right) \left( \frac{330}{330 + 6} \right)\]Compute \( f'' \), the frequency from wall 2 as heard by the bat.
05

Calculate the beat frequency

The beat frequency is the absolute difference between the two frequencies heard by the bat from wall 1 and wall 2. Use:\[|f_{w1}'' - f_{w2}''|\]Calculate this to find out how often per second the bat hears the frequency differences between these two reflections.

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Key Concepts

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

Ultrasonic Wave
Ultrasonic waves are sound waves with frequencies higher than the upper audible limit of human hearing. Essentially, humans cannot hear these waves as they typically have frequencies above 20 kHz. Bats, however, rely on ultrasonic waves for navigation and hunting. Their echolocation ability allows them to emit ultrasonic waves, which then bounce off objects and return to them as echoes. By analyzing these echoes, bats can build a mental map of their surroundings and detect prey.
When we talk about ultrasonic waves, it's crucial to understand that objects reflect these high-frequency sounds much like a mirror reflects light. The frequency and wavelength of these waves play a key role in determining how they behave upon hitting obstacles.
*Key points about ultrasonic waves:*
  • They exceed 20 kHz, making them inaudible to humans.
  • Used in medical imaging like an ultrasound.
  • Bats use them for echolocation.
Beat Frequency
Beat frequency is the phenomenon that occurs when two sound waves of slightly different frequencies interfere with each other. The result is a sound that alternates between louder and softer volumes, which is called a "beat." In essence, the beat frequency is a new frequency that equals the absolute difference between the two original frequencies. Bats, when using echolocation, listen to the changes in these frequencies to understand their surroundings.
In the context of the bat flying between two walls, the bat is essentially perceiving two different frequencies due to the Doppler Effect, as it moves towards each wall at different rates. This results in two different frequencies after reflecting off the walls, and the beat frequency that the bat detects is the difference between these two reflected frequencies.
*Why is beat frequency important?*
  • Helps in detecting variations in sound frequency.
  • Used in music to tune instruments.
  • Serves as an effective echolocation tool for bats.
Sound Reflection
Sound reflection occurs when a sound wave strikes a surface and bounces back. Much like light reflecting off a mirror, sound waves can reflect off surfaces such as walls, water, and other obstacles. Reflected sound waves are the basis of echo and reverberation.
For bats, sound reflection is a vital component of echolocation. By emitting ultrasonic waves and measuring the time it takes for them to return after reflecting off surfaces, bats can estimate the distance, shape, and size of nearby objects.
*Important aspects of sound reflection:*
  • Reflective surfaces can be walls, trees, or any solid object.
  • Crucial for echolocation and sonar technologies.
  • The angle of reflection equals the angle of incidence.
Frequency Shift
The frequency shift is a critical consequence of the Doppler Effect, which occurs when there is a relative motion between the source of sound and an observer. As an object emitting a sound wave moves towards or away from an observer, the sound appears to change frequency.
In this exercise, the Doppler Effect causes the frequency shift in the sound waves that the bat emits and receives. As the bat moves towards wall 1, the frequency increases, and as it moves towards wall 2, the frequency decreases. This shift in frequency is essential for the bat's echolocation capabilities, enabling it to detect the direction and speed of its own motion relative to obstacles and prey.
*Key implications of frequency shift:*
  • Determines the perceived pitch of the sound.
  • Used in technologies like radar and medical ultrasound.
  • Enables the understanding of movements and speed in real-time.

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Most popular questions from this chapter

A man stands before a large wall at a distance of \(50 \cdot 0 \mathrm{~m}\) and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.

A source of sound \(S\) and a detector \(D\) are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line \(S D\) as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of \(20 \mathrm{~cm}\). Find the frequency of the sound emitted. Velocity of sound in air is \(336 \mathrm{~m} \mathrm{~s}^{-1}\).

A cylindrical metal tube has a length of \(50 \mathrm{~cm}\) and is open at both ends. Find the frequencies between \(1000 \mathrm{~Hz}\) and \(2000 \mathrm{~Hz}\) at which the air column in the tube can resonate. Speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\).

Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of \(32 \mathrm{~cm}\) and the other of \(32 \cdot 2 \mathrm{~cm} .\) The speed of sound in air is \(350 \mathrm{~m} / \mathrm{s}\).

Two coherent narrow slits emitting sound of wavelength \(\lambda\) in the same phase are placed parallel to each other at a small separation of \(2 \lambda\). The sound is detected by moving a detector on the screen \(\Sigma\) at a distance \(D(\gg>)\) from the slit \(S_{1}\) as shown in figure (16-E6). Find the distance \(x\) such that the intensity at \(P\) is equal to the intensity at \(O\).

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