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Semmy [17]
3 years ago
10

You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?

Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

2\cdot h = v\cdot t (1)

Where:

h - Height, measured in meters.

v - Velocity of sound, measured in meters per second.

t - Time, measured in seconds.

If we know that  v = 344\,\frac{m}{s} and t = 3\,s, then the height of the canyon is:

h = \frac{v\cdot t}{2}

h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}

h = 516\,m

The sound travelled 516 meters before bouncing off a cliff.

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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

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\vartheta = observed frequency of sound

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(b) The observed frequency is given by:

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\vartheta = \frac{334}{334 - 80}\vartheta_{o}

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Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

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