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Chapter 15: Problem 18

Hiking in the mountains, you shout "hey," wait \(2.00 \mathrm{~s}\) and shout again. What is the distance between the sound waves you cause? If you hear the first echo after \(5.00 \mathrm{~s}\), what is the distance between you and the point where your voice hit a mountain?

Short Answer

Expert verified
Answer: The distance between the person and the point where their voice hits the mountain is 857.5 meters.

Step by step solution

01

Find the speed of sound in air

The typical speed of sound in air at room temperature (20 degrees Celsius) is approximately 343 meters per second (m/s). We'll use this speed in our calculations.
02

Calculate the distance between the sound waves

To find the distance between the two sound waves caused by the shouts, we can use the formula distance = speed * time. We know the time between the shouts is 2 seconds, and we already know the speed of sound in air. Therefore, we can calculate the distance like this: Distance = 343 m/s * 2 s Distance = 686 meters So, the distance between the two sound waves is 686 meters.
03

Calculate the round-trip time for the echo

The problem states that the first echo is heard after 5 seconds. This means that the sound has traveled to the mountain and then back to the person. To find the time it takes for the sound to reach the mountain, we need to divide the total time by 2: Time to reach the mountain = 5 s / 2 = 2.5 s
04

Calculate the distance between the person and the mountain

Now that we know the time it takes for the sound to travel to the mountain, we can use the same formula, distance = speed * time, to find the distance between the person and the mountain: Distance = 343 m/s * 2.5 s Distance = 857.5 meters So, the distance between the person and the mountain is 857.5 meters.

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Most popular questions from this chapter

Consider a monochromatic wave on a string, with amplitude \(A\) and wavelength \(\lambda\), traveling in one direction. Find the relationship between the maximum speed of any portion of string, \(v_{\max },\) and the wave speed, \(v\)

Write the equation for a sinusoidal wave propagating in the negative \(x\) -direction with a speed of \(120 . \mathrm{m} / \mathrm{s}\), if a particle in the medium in which the wave is moving is observed to swing back and forth through a \(6.00-\mathrm{cm}\) range in \(4.00 \mathrm{~s}\). Assume that \(t=0\) is taken to be the instant when the particle is at \(y=0\) and that the particle moves in the positive \(y\) -direction immediately after \(t=0\).

Which of the following transverse waves has the greatest power? a) a wave with velocity \(v\), amplitude \(A\), and frequency \(f\) b) a wave of velocity \(v\), amplitude \(2 A\), and frequency \(f / 2\) c) a wave of velocity \(2 v\), amplitude \(A / 2\), and frequency \(f\) d) a wave of velocity \(2 v\), amplitude \(A\), and frequency \(f / 2\) e) a wave of velocity \(v\), amplitude \(A / 2\), and frequency \(2 f\)

a) Starting from the general wave equation (equation 15.9 ), prove through direct derivation that the Gaussian wave packet described by the equation \(y(x, t)=(5.00 m) e^{-0.1(x-5 t)^{2}}\) is indeed a traveling wave (that it satisfies the differential wave equation). b) If \(x\) is specified in meters and \(t\) in seconds, determine the speed of this wave. On a single graph, plot this wave as a function of \(x\) at \(t=0, t=1.00 \mathrm{~s}, t=2.00 \mathrm{~s},\) and \(t=3.00 \mathrm{~s}\) c) More generally, prove that any function \(f(x, t)\) that depends on \(x\) and \(t\) through a combined variable \(x \pm v t\) is a solution of the wave equation, irrespective of the specific form of the function \(f\)

A particular steel guitar string has mass per unit length of \(1.93 \mathrm{~g} / \mathrm{m}\). a) If the tension on this string is \(62.2 \mathrm{~N},\) what is the wave speed on the string? b) For the wave speed to be increased by \(1.0 \%\), how much should the tension be changed?
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