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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 sound waves caused by the two shouts is 686 m, while the distance between the person and the point where their voice hit the mountain is 857.5 m.

Step by step solution

01

Calculate the distance between the sound waves caused by the two shouts

To determine the distance between the sound waves, we first need to find out how far the sound traveled during the 2.00 s interval between the shouts. We can do this using the speed of sound and the time interval given. The formula to calculate the distance traveled by the sound is: Distance = Speed × Time We know that the speed of sound is 343 m/s, and the time interval is 2.00 s. Plugging in these values, we get: Distance = (343 m/s) × (2.00 s) Distance = 686 m The distance between the sound waves caused by the two shouts is 686 m.
02

Calculate the distance between the person and the point where their voice hit the mountain

To find the distance between the person and the point where their voice hit the mountain, we need to use the echo formula. An echo is the reflection of sound waves, so the total distance traveled by the sound waves is double the distance between the person and the point where the voice hit the mountain. We are given that the first echo is heard after 5.00 s. The formula to calculate the total distance traveled by the sound waves is: Total distance = Speed × Time We know that the speed of sound is 343 m/s, and the time is 5.00 s. Plugging in these values, we get: Total distance = (343 m/s) × (5.00 s) Total distance = 1715 m Now, to find the distance between the person and the point where their voice hit the mountain, we need to divide this total distance by 2: Distance to the point where the voice hit the mountain = (Total distance) / 2 Distance to the point where the voice hit the mountain = (1715 m) / 2 Distance to the point where the voice hit the mountain = 857.5 m The distance between the person and the point where their voice hit the mountain is 857.5 m.

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Key Concepts

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

Speed of Sound
Understanding how fast sound travels is crucial when calculating distances related to sound waves. The speed of sound is not a constant value; it varies with the medium through which it travels as well as environmental conditions like temperature, humidity, and air pressure. For most calculations, we use the average speed of sound in air at sea level, which is about 343 meters per second (m/s) at 20 degrees Celsius.

Naturally, if you're hiking in cooler mountain air, this value might be slightly lower, as sound travels slower in colder air. For high school physics problems or standard estimations, using the 343 m/s value is usually sufficient. However, for precise calculations, one should take into account the specific conditions at the time and location of interest. Students should remember that the medium is key: in water, sound travels around 1482 m/s, and in steel, it goes up to 5960 m/s!
Echo Calculation
An echo occurs when a sound wave reflects off a surface and returns to the listener. Calculating the distance of an echo involves measuring the time it takes for the sound to travel to the reflecting surface and back. Since the sound makes a round trip, the total distance covered is twice the distance between the source and the reflecting surface.

For clarity, when you shout and hear the echo after a certain time, that time includes the sound journeying to the mountain and back. If it takes 5 seconds for the echo to return, and knowing the speed of sound is 343 m/s, you can calculate the total round-trip distance the sound wave traveled before halfing it to pinpoint the distance to the reflective surface. This is a great practical application of physics in the real world, and echoes are a useful phenomenon for various technological applications like sonar and echolocation.
Distance-Time Relationship
The distance-time relationship is a fundamental concept in physics that relates speed, distance, and time. It's expressed simply as: Distance = Speed × Time. This relationship tells us that the distance traveled by an object (or sound wave, in our case) is directly proportional to its speed and the time it moves at that speed.

When you're calculating the distance between sound waves, you're utilizing this relationship. It's a linear relationship, meaning that as time increases, distance increases proportionately, assuming speed remains constant. It's worth noting, as you study these relationships, that a clear understanding can help in a variety of real-world situations, from planning travel to designing soundproofing solutions where you might want to minimize the distance sound can travel.

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Most popular questions from this chapter

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(F=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different angular frequencies \((\omega=3000 . \mathrm{rad} / \mathrm{s}\) and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s})\) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point where both waves have a node, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

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A cowboy walks at a pace of about two steps per second, holding a glass of diameter \(10.0 \mathrm{~cm}\) that contains milk. The milk sloshes higher and higher in the glass until it eventually starts to spill over the top. Determine the maximum speed of the waves in the milk.

A ping-pong ball is floating in the middle of a lake and waves begin to propagate on the surface. Can you think of a situation in which the ball remains stationary? Can you think of a situation involving a single wave on the lake in which the ball remains stationary?

A sinusoidal wave traveling on a string is moving in the positive \(x\) -direction. The wave has a wavelength of \(4 \mathrm{~m}\), a frequency of \(50.0 \mathrm{~Hz}\), and an amplitude of \(3.00 \mathrm{~cm}\). What is the wave function for this wave?

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