A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end… — JAMB UTME

A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end records two sound signals—one traveling through the aluminum pipeline and the other through the surrounding air—separated by a time interval of $2.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ and the speed of sound in aluminum is $5100\text{ m s}^{-1}$ , what is the length of the pipeline in meters?

Answer: 1020 m

Answer

The length of the pipeline is

1020\text{ m}

Sound travels significantly faster through solids like aluminum (

5100\text{ m s}^{-1}

) than through gases like air (

340\text{ m s}^{-1}

). The time taken for sound to travel a distance

L

through air is

t_{\text{air}} = \frac{L}{340}

, while through aluminum it is

t_{\text{metal}} = \frac{L}{5100}

. Setting their difference equal to

2.8\text{ s}

gives

\frac{L}{340} - \frac{L}{5100} = 2.8

, which solves to

L = 1020\text{ m}

Step-by-Step Solution

Formulate transit time expressions for both media

t_{\text{air}} = \frac{L}{340}

and

t_{\text{metal}} = \frac{L}{5100}

Time taken by a wave to travel distance

L

at constant speed

v

t = \frac{L}{v}

Set up the time difference equation

t_{\text{air}} - t_{\text{metal}} = 2.8\text{ s}

The sound wave travels faster through aluminum than air, so the air pulse arrives later by

2.8\text{ s}

Solve the algebraic equation for distance

L

L = 1020\text{ m}

Combining terms yields

\frac{14L}{5100} = 2.8

, which simplifies to

L = \frac{2.8 \times 5100}{14} = 1020\text{ m}

Key Concept

Propagation speed of sound waves in different physical media

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