Question

A ship is cruising slowly through a channel with steep canyon walls on each side. The captain blows his foghorn and an echo from one wall is heard 4.2 s after the horn was sounded. The echo from the other wall is heard 6.5 s after the first echo. How wide is the channel? Assume the speed of sound in air is 343 m/s.

Answer 1

The width of the channel with steep canyon walls is 1,835.05 m.

What is echo?

This is a phenomenon that occurs due to the reelection of sound waves.

The distance of the sound wave from the first wall is calculated as follows;

$d_1 = \frac{vt_1}{2} \\\\ d_1 = \frac{343 \times 4.2}{2} \\\\ d_1 = 720.3 \ m$

The distance of the sound wave from the second wall is calculated as follows;

$d_2 = \frac{vt_2}{2} \\\\ d_2 = \frac{343 \times 6.5}{2} \\\\ d_2 = 1,114.75 \ m$

The width of the channel with steep canyon walls is calculated as follows;

$W= d_1 + d_2\\\\ W = 720.3 \ m \ + \ 1,114.75 \ m\\\\ W = 1,835.05 \ m$

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