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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

A guitar string with a mass of \(10.0 \mathrm{~g}\) is \(1.00 \mathrm{~m}\) long and attached to the guitar at two points separated by \(65.0 \mathrm{~cm} .\) a) What is the frequency of the first harmonic of this string when it is placed under a tension of \(81.0 \mathrm{~N} ?\) b) If the guitar string is replaced by a heavier one that has a mass of \(16.0 \mathrm{~g}\) and is \(1.00 \mathrm{~m}\) long, what is the frequency of the replacement string's first harmonic?

A wave on a string has a wave function given by $$ y(x, t)=(0.0200 \mathrm{~m}) \sin \left[\left(6.35 \mathrm{~m}^{-1}\right) x+\left(2.63 \mathrm{~s}^{-1}\right) t\right] $$ a) What is the amplitude of the wave? b) What is the period of the wave? c) What is the wavelength of the wave? d) What is the speed of the wave? e) In which direction does the wave travel?

You and a friend are holding the two ends of a Slinky stretched out between you. How would you move your end of the Slinky to create (a) transverse waves or (b) longitudinal waves?

If transverse waves on a string travel with a velocity of \(50 \mathrm{~m} / \mathrm{s}\) when the string is under a tension of \(20 \mathrm{~N},\) what tension on the string is required for the waves to travel with a velocity of \(30 \mathrm{~m} / \mathrm{s} ?\) a) \(7.2 \mathrm{~N}\) c) \(33 \mathrm{~N}\) e) \(45 \mathrm{~N}\) b) \(12 \mathrm{~N}\) d) \(40 \mathrm{~N}\) f) \(56 \mathrm{~N}\)

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.75 x-40 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10 \mathrm{~g} / \mathrm{m}\), determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\) (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.
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