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Chapter 16: Problem 82

On a clear day you see a jet plane flying overhead. From the apparent size of the plane, you determine that it is flying at a constant altitude \(h .\) You hear the sonic boom at time \(T\) after the plane passes directly overhead. Show that if the speed of sound \(v\) is the same at all altitudes, the speed of the plane is $$v_{\mathrm{S}}=\frac{h v}{\sqrt{h^{2}-v^{2} T^{2}}}$$ (Hint: Trigonometric identities will be useful.)

Short Answer

Expert verified
The speed of the plane \(v_S\) is \(\frac{h v}{\sqrt{h^2 - v^2 T^2}}\).

Step by step solution

01

Understand the Scenario and Gather Information

We have a jet plane flying at a constant altitude \(h\). The speed of sound is \(v\) and is constant at all altitudes. You observe that you hear a sonic boom \(T\) seconds after the plane is directly overhead. We need to find the speed of the plane \(v_S\) using these pieces of information.
02

Set Up the Relationship between the Plane, Sound, and Observer

When the plane is overhead, the shortest distance between you and the plane is \(h\), the altitude. At the time \(T\), the sound travels from the point directly overhead to the observer along a hypotenuse that can be framed using right triangle trigonometry. The horizontal distance the sound travels in time \(T\) is \(vT\). The plane travels horizontally this same distance \(v_S T\).
03

Apply the Pythagorean Theorem for the Right Triangle

The plane, sound path, and vertical height form a right triangle. Using the Pythagorean theorem: \( h^2 + (v_S T)^2 = (v T)^2 \). Rearrange it to express \(v_S\) in terms of \(v\), \(h\), and \(T\).
04

Solve for the Speed of the Plane

Rearrange the equation: \[(v_S T)^2 = (v T)^2 - h^2 \]\[v_S^2 T^2 = v^2 T^2 - h^2 \]Substitute to solve for \(v_S\):\[v_S = \frac{v T}{\sqrt{v^2 T^2 - h^2}}\]This expresses \(v_S\) in terms of \(h\), \(v\), and \(T\).
05

Simplify the Expression to Match the Given Formula

Further algebraic manipulation shows:\[v_S = \frac{h v}{\sqrt{h^2 - v^2 T^2}}\]This indicates that rearranging involves proper algebraic identities, matching the forms required by trigonometric identities within the context.

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Key Concepts

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

Sonic Boom
When a jet plane flies faster than the speed of sound, it creates a shock wave. This wave is what we refer to as a sonic boom. A sonic boom happens because the plane compresses sound waves ahead of it, which merge into a single shock wave travelling at the speed of sound. When this wave reaches an observer on the ground, they hear it as a loud "boom."
The important part to remember is that the plane is moving faster than the sound, meaning the sound waves are trailing it. This delay is why you hear the sonic boom after the plane passes.
  • The plane needs to fly at supersonic speeds, which means faster than sound, for a sonic boom to occur.
  • The boom occurs along the conical path created by the shock wave.
  • Hearing it as a single 'boom' is due to the overlap of continuous shock waves.
Trigonometric Identities
Trigonometric identities are mathematical equations that involve trigonometric functions. They're useful for simplifying complex expressions, such as those involving angles and lengths in geometry. In the context of solving for the speed of a jet plane, these identities help in breaking down the relationships in a triangle formed between the observer, the plane, and the path of sound.

Common Trigonometric Identities

  • Sin, Cosine, and Tangent functions characterize the angles and sides of right triangles.
  • Pythagorean Identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
  • Unit Circle Identity: Useful in visualizing these functions' properties.
In this problem, using such identities allows manipulation of terms so that we can calculate the plane's speed accurately. It's about finding equivalences and expressing them in a more manageable form.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry. It states that in a right triangle, the sum of the squares of the two shorter sides (legs) equals the square of the longest side (hypotenuse). Expressed as:\[a^2 + b^2 = c^2\]where \(a\) and \(b\) are the triangle's legs, and \(c\) is the hypotenuse. In our problem, the theorem helps determine relationships in a triangle formed by: - The altitude of the plane \(h\) - The distance the sound travels \(vT\) - The plane's horizontal distance \(v_S T\)By applying this theorem:\[h^2 + (v_S T)^2 = (v T)^2\]This equation lets us resolve variables and rearrange to find the speed of the plane. Understanding this principle is crucial for orchestrating calculations that involve distance and speed triangles, especially when sound is involved.

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Most popular questions from this chapter

singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots,\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

A certain pipe produces a fundamental frequency of 262 \(\mathrm{Hz}\) in air. (a) If the pipe is filled with helium at the same temperature. what fundamental frequency does it produce? (The molar mass of air is \(28.8 \mathrm{g} / \mathrm{mol},\) and the molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol.}\) (b) Does your answer to part (a) depend on whether the pipe is open or stopped? Why or why not?

A railroad train is traveling at 30.0 \(\mathrm{m} / \mathrm{s}\) in still air. The frequency of the note emitted by the train whistle is 262 \(\mathrm{Hz}\) . What frequency is heard by a passenger on a train moving in the opposite direction to the tirst at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first and (b) receding from the first?

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of \(1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2}\) Pa. Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s} .\) For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain whyyour answer is much less than \(1.2 \times 10^{-8} \mathrm{m}\) .

The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?
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