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Chapter 17: Problem 124

The shock wave off the front of a fighter jet has an angle of \(\theta=70.00^{\circ} .\) The jet is flying at \(1200 \mathrm{km} / \mathrm{h}\). What is the speed of sound?

Short Answer

Expert verified
The speed of sound is approximately 314.74 m/s.

Step by step solution

01

Convert km/h to m/s

The jet's speed is given in km/h and needs to be converted into m/s because the speed of sound is typically measured in m/s. This can be done by multiplying the speed by \(1000/3600\). The resultant speed in m/s is \(v = 1200 \times 1000 / 3600 = 333.33 m/s\).
02

Convert Degrees to Radians

This exercise assumes that we're working in radians, therefore the Mach angle \(\theta\) needs to be converted from degrees to radians. This is done by multiplying the degrees by \(\pi /180\). Therefore, \(\theta = 70.00 \times \(\pi /180 = 1.2217\) radians.
03

Solve for Speed of Sound

We have now converted all necessary units and can solve for \(v_s\) using \(\sin(\theta) = v_s/v\). Algebraically solved for \(v_s\) gives \(v_s = v \times \sin(\theta)\). Substituting in the known values gives the speed of sound as \(v_s = 333.33 \times \sin(1.2217) = 314.74 m/s \).

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Key Concepts

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

Speed Conversion
When working with physics problems that involve speed, it鈥檚 crucial to use consistent units. Speeds are often given in kilometers per hour (km/h), but many calculations require meters per second (m/s). To convert km/h to m/s, you need to multiply the speed by \( \frac{1000}{3600} \) because there are 1000 meters in a kilometer and 3600 seconds in an hour.
This conversion factor simplifies to \( \frac{5}{18} \), allowing us to convert speeds quickly:
  • Example: 1200 km/h \( \times \frac{5}{18} = 333.33 \) m/s
This unit conversion is a foundational skill in physics for ensuring that calculations align with the units of constants like the speed of sound.
Mach Angle
The Mach angle is an essential concept when dealing with objects moving faster than sound, creating shock waves. It is defined as the angle between the direction of the plane and the shock wave.
The angle of the shock wave, known as the Mach angle \( \theta \), relates to the speed of the object and the speed of sound using trigonometry:
  • \( \sin(\theta) = \frac{v_s}{v} \)
Where \( v_s \) is the speed of sound and \( v \) is the object's speed.
By solving this relation, we can determine how sound waves behave around high-speed objects like fighter jets.
Speed of Sound
The speed of sound is a critical value in physics as it determines how quickly sound waves travel through a medium. In air at sea level, it is typically around 343 m/s, but this can vary with temperature, humidity, and altitude.
In our problem, the speed of sound was solved using the relationship from the Mach angle:
  • \( v_s = v \times \sin(\theta) \)
  • Using the angle and jet speed, we find \( v_s = 314.74 \) m/s.
This demonstrates the direct application of trigonometry in calculating real-world physics problems like the flight of a jet.
Trigonometry in Physics
Trigonometry plays a vital role in physics, particularly when analyzing wave behaviors and angles. In problems involving speeds and angles, trigonometric functions like sine \( \sin \), cosine \( \cos \), and tangent \( \tan \) become indispensable.
For the Mach angle, the sine function is used to relate the speed of an object to the speed of sound:
  • \( \sin(\theta) = \frac{v_s}{v} \)
This requires angles to be in radians for proper calculation in physics formulas. Understanding how to convert angles and correctly apply trigonometric functions is key to solving such problems efficiently.

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Most popular questions from this chapter

Two sound speakers are separated by a distance \(d\),each sounding a frequency \(f .\) An observer stands at one speaker and walks in a straight line a distance \(x\) perpendicular to the the two speakers, until he comes to the first maximum intensity of sound. The speed of sound is \(v\) How far is he from the speaker?

(a) If a submarine's sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.

A string \(\left(\mu=0.006 \frac{\mathrm{kg}}{\mathrm{m}}, L=1.50 \mathrm{m}\right)\) is fixed at both ends and is under a tension of 155 N. It oscillates in the \(n=10\) mode and produces sound. A tuning fork is ringing nearby, producing a beat frequency of \(23.76 \mathrm{Hz}\). (a) What is the frequency of the sound from the string? (b) What is the frequency of the tuning fork if the tuning fork frequency is lower? (c) What should be the tension of the string for the beat frequency to be zero?

A flute plays a note with a frequency of \(600 \mathrm{Hz}\). The flute can be modeled as a pipe open at both ends, where the flute player changes the length with his finger positions. What is the length of the tube if this is the fundamental frequency?

A 512-Hz tuning fork is struck and placed next to a tube with a movable piston, creating a tube with a variable length. The piston is slid down the pipe and resonance is reached when the piston is \(115.50 \mathrm{cm}\) from the open end. The next resonance is reached when the piston is \(82.50 \mathrm{cm}\) from the open end. (a) What is the speed of sound in the tube? (b) How far from the open end will the piston cause the next mode of resonance?
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