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Q11E

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University Physics with Modern Physics
Found in: Page 539
University Physics with Modern Physics

University Physics with Modern Physics

Book edition 14th edition
Author(s) Hugh D. Young, Roger A. Freedman
Pages 1596 pages
ISBN 9780321973610

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Short Answer

Question: A 60.0-m-long brass rod is struck at one end. A person at the other end hears two sounds as a result of two longitudinal waves, one traveling in the metal rod and the other traveling in air. What is the time interval between the two sounds? (The speed of sound in air is 344 m/s; see Tables 11.1 and 12.1 for relevant information about brass.)

The time interval between two sounds is 0.1559 s.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data

The given data can be listed below as,

  • The length of the brass rod is,\(l = 60\;{\rm{m}}\).
  • The sounds are heard by the person; one is traveling in metal, and the second is traveling in the air.

Step 2: Concept

When a sound wave is moving in a medium, the type of medium will affect the properties or characteristics of the respective sound wave. Two sound waves traveling in two different mediums will have different values of physical quantities associated with them.

Step 3: Determination of the time interval of two sounds

For metal brass:

The wave is traveling in brass.

The velocity of the wave can be calculated as,

\(v = \sqrt {\frac{Y}{\rho }} \)

Here \(Y\)is the young modulus of the elasticity of brass metal and \(\rho \)is the density of brass.

From tables 11.1 and 12.1, we can obtain,

\(Y = 9 \times {10^{10}}\;{\rm{Pa}}\) and \(\rho = 8600\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

Substitute these values in the above expression, and we get,

\(\begin{array}{c}{v_{{\rm{brass}}}} = \sqrt {\frac{{9 \times {{10}^{10}}\;{\rm{Pa}}}}{{8600\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}}} \\ = \sqrt {\frac{{9 \times {{10}^{10}}}}{{8600}} \cdot \left( {\frac{{1\;{\rm{Pa}}}}{{1\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}} \times \frac{{1\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - {\rm{2}}}}}}{{1\;{\rm{N}}}} \times \frac{{1\;{\rm{N}} \cdot {{\rm{m}}^{ - 2}}}}{{1\;{\rm{Pa}}}}} \right)} \\ = \sqrt {10465116.28 \cdot {{\left( {1\;{\rm{m/s}}} \right)}^2}} \\ = 3234.98\;{\rm{m/s}}\end{array}\)

The time taken by the sound wave to reach the distance of 60 m can be calculated as,

\({t_1} = 60\;{\rm{m}}/{v_{{\rm{brass}}}}\)

Substitute the value in the above expression, and we get,

\(\begin{array}{c}{t_1} = 60\;{\rm{m}}/3234.98\;{\rm{m/s}}\\{t_1} = {\rm{0}}{\rm{.0185}}\;{\rm{s}}\end{array}\) (1)

For Air:

The velocity of the sound wave in the air is 344 m/s.

The time taken by the sound wave to reach the distance of 60 m can be calculated as,

\(\begin{array}{c}{t_2} = 60\;{\rm{m}}/344\;{\rm{m/s}}\\{t_2} = 0.1744\;{\rm{s}}\end{array}\) (2)

The time interval between two sounds can be calculated as,

\(\Delta t = {t_2} - {t_1}\)

Substitute the values from equations 1 and 2 in the above expression, and we get,

\(\begin{array}{c}\Delta t = 0.1744\;{\rm{s}} - {\rm{0}}{\rm{.0185}}\;{\rm{s}}\\ = 0.1559\;{\rm{s}}\end{array}\)

Thus, the time interval between two waves is 0.1559 s.

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