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Two submarines are under water and approaching each other head-on. Sub A has a speed of \(12 \mathrm{m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{m} / \mathrm{s}\). Sub A sends out a \(1550-\mathrm{Hz}\) sonar wave that travels at a speed of \(1522 \mathrm{m} / \mathrm{s}\) (a) What is the frequency detected by sub B? (b) Part of the sonar wave is reflected from sub \(\mathrm{B}\) and returns to sub \(\mathrm{A}\). What frequency does sub \(\mathrm{A}\) detect for this reflected wave?

Short Answer

Expert verified
(a) Sub B detects 1622 Hz. (b) Sub A detects a reflected frequency of 1706 Hz.

Step by step solution

01

Understand the Problem

We have two submarines moving towards each other with specific speeds, and Sub A emits a sonar wave detected by Sub B. We're to find the frequency of the sonar wave detected by Sub B using the Doppler effect and then determine the frequency detected by Sub A after the wave is reflected from Sub B.
02

Doppler Effect Formula

The frequency detected by an observer when a source is moving towards or away from them can be calculated using the Doppler effect formula: \( f' = f \left( \frac{v + v_o}{v + v_s} \right) \), where \( f \) is the source frequency, \( v \) is the speed of sound, \( v_o \) is the speed of the observer, and \( v_s \) is the speed of the source.
03

Determine Variables for Part A

For Part A, the source (Sub A) moves towards the observer (Sub B). Here, \( f = 1550 \text{ Hz} \), \( v = 1522 \text{ m/s} \), \( v_o = 8 \text{ m/s} \) (since Sub B is the observer and moving towards the source), and \( v_s = 12 \text{ m/s} \) (since Sub A is moving towards the observer).
04

Calculate Frequency Detected by Sub B

Using the Doppler effect formula, we calculate the frequency detected by Sub B as follows: \[ f' = 1550 \left( \frac{1522 + 8}{1522 - 12} \right) \]Calculate the result of the expression inside the brackets and multiply it by 1550 Hz to find \( f' \).
05

Reflective Frequency Calculation for Sub A

Once Sub B detects the frequency \( f' \), the wave reflects back. Now Sub B acts as a source moving away at \( 8 \text{ m/s} \) and Sub A as an observer moving towards the source at \( 12 \text{ m/s} \). Calculate the returned frequency using the formula: \[ f'' = f' \left( \frac{1522 + 12}{1522 - 8} \right) \]Substitute the previously calculated \( f' \) into this formula to find the frequency detected by Sub A, which is \( f'' \).
06

Substitute and Calculate

Perform the calculations for both cases: the frequency a.) received by Sub B and b.) received by Sub A after reflection, using the relevant substitution in steps. Ensure careful attention to the sign of velocities, depending on their direction.

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Key Concepts

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

Exploring Sonar Waves
Sonar waves are a type of sound wave used by submarines to communicate or detect objects underwater. These waves are very similar to the sounds we hear, but they operate at a frequency that is often outside the realm of human hearing. Submarines emit these waves and, when the waves bounce back off an object, the time it takes for them to return can tell a lot about the distance and speed of the object in relation to the submarine.
Understanding sonar is crucial because the speed and frequency of the waves can change depending on the movement of the submarine or the object it is detecting. These sound waves travel through water at nearly constant speed, which, in this exercise, is provided as 1522 m/s. This is how submarines detect how fast or how close objects, like other submarines, are to them. When two submarines are approaching each other and a sonar wave is sent, the frequency might change due to their movement, and this is where the Doppler Effect becomes significant.
Frequency Calculation Using the Doppler Effect
The Doppler Effect explains how the frequency of a wave changes for an observer when there is relative motion between the source of the wave and the observer. In this exercise, we're working to find how Sub B detects the frequency of a sonar wave emitted by Sub A as they move towards each other. The formula used is:
  • \( f' = f \left( \frac{v + v_o}{v + v_s} \right) \)

To solve for the frequency detected by Sub B, we substitute the given values:
  • Source frequency \( f = 1550 \text{ Hz} \)
  • Speed of the sonar wave \( v = 1522 \text{ m/s} \)
  • Speed of observer (Sub B) \( v_o = 8 \text{ m/s} \)
  • Speed of source (Sub A) \( v_s = 12 \text{ m/s} \)
By inserting these numbers in the formula, you can calculate the frequency \( f' \) that Sub B will detect.
Furthermore, once Sub B detects this altered frequency and the wave reflects back, Sub A, now an observer, will detect a different frequency for the returning wave. You use a similar Doppler calculation, inserting \( f' \) in place of \( f \) and adjusting the velocities for the new scenario where Sub B is moving away.
Understanding Submarine Acoustics
Submarine acoustics is the study of how sound propagates under water, which is critical for determining how submarines use sonar. The movement of submarines affects how these sonar waves are perceived. In scenarios like this exercise, the water acts as a medium that can slightly distort the sonar waves due to its density.
Submarines rely heavily on the principles of acoustics because the sound travels well underwater, often better than light, which cannot penetrate the depths for communication or location detection.
  • Sound waves can bend around obstacles, providing an advantage in finding targets or navigating the ocean environment.
  • Using the Doppler Effect, submarines not only detect the presence of another object but can also calculate its speed and direction, which is vital for navigation and strategy.
In essence, acoustics allow submarines to interpret changes in sonar wave frequency, determining if another vessel is approaching or receding, making real-time decisions based on precise calculations.

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Most popular questions from this chapter

An amplified guitar has a sound intensity level that is \(14 \mathrm{dB}\) greater than the same unamplified sound. What is the ratio of the amplified intensity to the unamplified intensity?

The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\). Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{m} / \mathrm{s}\), find the wavelength and frequency of the sound you hear.

A wave has the following properties: amplitude \(=0.37 \mathrm{m},\) period \(=0.77 \mathrm{s},\) wave speed \(=12 \mathrm{m} / \mathrm{s} .\) The wave is traveling in the \(-x\) direction. What is the mathematical expression (similar to Equation 16.3 or 16.4 ) for the wave?

One of the most important concepts we encountered in this chapter is the transverse wave. For instance, transverse waves travel along a guitar string when it is plucked or along a violin string when it is bowed. Problem 112 reviews how the travel speed depends on the properties of the string and on the tension in it. Problem 113 illustrates how the Doppler effect arises when an observer is moving away from or toward a stationary source of sound. In fact, we will see that it's possible for both situations to occur at the same time. A siren, mounted on a tower, emits a sound whose frequency is \(2140 \mathrm{Hz}\). A person is driving a car away from the tower at a speed of \(27.0 \mathrm{m} / \mathrm{s}\). As the figure illustrates, the sound reaches the person by two paths: the sound reflected from the building in front of the car, and the sound coming directly from the siren. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). Concepts: (i) One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or reflected sound, does the wavelength change? (ii) Why does the driver hear a frequency for the reflected sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz} ?\) (iii) Why does the driver hear a frequency for the direct sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz}\) ? Calculations: What frequency does the person hear for the (a) reflected and (b) direct sound?

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C},\) how many times greater is the speed of sound in the rail than in the air?

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