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As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{m} .\) A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The sound reaches microphone 1 first, and then, \(1.46 \mathrm{ms}\) later, it reaches microphone \(2 .\) Find the distances \(L_{1}\) and \(L_{2}\).

Short Answer

Expert verified
\(L_1 \approx 3\) meters and \(L_2 \approx 3.50\) meters.

Step by step solution

01

Define relationship between time and distance

The time taken for sound to travel a certain distance between two points is given by the equation: \( t = \frac{d}{v} \), where \( d \) is the distance and \( v \) is the speed of sound. The problem provides the time delay between the sound reaching microphones 1 and 2 as \( t = 1.46 \times 10^{-3} \) seconds.
02

Express the distance difference in terms of speed and time

Based on the relationship between speed, distance, and time, the difference in distances \( L_2 - L_1 \) that the sound traveled can be expressed as: \( L_2 - L_1 = v \times t = 343 \times 1.46 \times 10^{-3} \). Calculate this to find \( L_2 - L_1 = 0.50078 \) meters.
03

Set up equations for the two distances

We have two equations forming a system:1. \( L_2 - L_1 = 0.50078 \) meters.2. From the geometry of the problem, use the Pythagorean theorem: \( L_2 = \sqrt{L_1^2 + D^2} \).
04

Solve the system of equations

Substitute \( L_2 = L_1 + 0.50078 \) into \( L_2 = \sqrt{L_1^2 + 1.5^2} \) and solve for \( L_1 \):\( (L_1 + 0.50078)^2 = L_1^2 + 1.5^2 \).Expand the equation:\( L_1^2 + 2 \times L_1 \times 0.50078 + 0.50078^2 = L_1^2 + 2.25 \).Simplify and solve the quadratic equation:\( 2 \times L_1 \times 0.50078 + 0.25078 = 2.25 \).\( L_1 \approx 3 \) meters.
05

Find L_2 using L_1 value

Now that we have \( L_1 \approx 3 \) meters, we use the equation for \( L_2 \):\( L_2 = L_1 + 0.50078 = 3 + 0.50078 \).Calculate \( L_2 \):\( L_2 \approx 3.50078 \) meters.

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Key Concepts

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

Speed of Sound
The speed of sound is an important concept in physics, especially when dealing with problems related to acoustics. It measures how fast sound waves propagate through a medium, like air. On average, the speed of sound in air at room temperature is about 343 meters per second (m/s). However, several factors can influence this speed.

  • Temperature: As the temperature increases, the speed of sound tends to increase. This is because warmer molecules move faster, helping sound waves travel quicker.
  • Medium: Sound can travel through gases, liquids, and solids, but it travels fastest in solids and slowest in gases. This is due to the closer packed molecules in solids.
  • Pressure and Density: Surprisingly, in gases like air, changes in pressure and density have less effect on sound speed compared to temperature changes.
Understanding the speed of sound helps us accurately calculate distances in problems involving sound propagation, as it allows us to relate the time sound takes to travel a given distance to that distance itself.
Time and Distance Calculations
Time and distance calculations are fundamental in physics. They allow us to determine how long it takes for an object to travel a certain distance or how far it will go in a specific time. The basic relationship is expressed in the formula:\[ t = \frac{d}{v} \]where:
  • \( t \) is the time.
  • \( d \) is the distance.
  • \( v \) is the velocity or speed of the object.
In problems where speed is constant, like the speed of sound, once you know two of these variables, you can easily find the third. For example, if you know how fast sound travels (343 m/s in air) and how long it takes to reach a point, you can calculate how far it traveled using the formula above.

This concept comes into play when there's a time delay, as in the problem where sound reaches two microphones at different times. The delay can be used to find the difference in distances the sound traveled to reach each microphone.
Pythagorean Theorem
The Pythagorean theorem is a key principle in geometry, used to relate the lengths of sides in a right triangle. It's fundamental for solving problems involving distances in coordinate systems. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. This is expressed as:\[ c^2 = a^2 + b^2 \]where:
  • \( c \) is the hypotenuse.
  • \( a \) and \( b \) are the other two sides.
In the context of our problem, the location of one microphone at the origin and the other along the y-axis forms one leg of a right-angled triangle, and the distance from the sound source forms the hypotenuse. By applying the Pythagorean theorem, we can relate these distances and solve for unknowns.

Using the relationships from the theorem, we can set up equations to manipulate and substitute values, which is essential for finding solutions to problems mixing linear distance and geometry.

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