A steel tube of length $1.00 m$ is struck at one end. A person with his ear close to the other end hears the sound of the bow twice, one traveling through the body of the tube and the other the air in the tube. Find the time gap between the two hearings.

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Solution

The speed of sound in air is $V_{a i r} = 230 m / s$
The speed of the sound in steel is $V_{s} = 5200 m / s$ .
Here the distance covered by the sound is $S = 1 m$

So, $t = t_{1} - t_{2}$
$= (\frac{1}{330} - \frac{1}{5200})$

$= 2.75 \times 10^{- 3} s e c$
$= 2.75 m s$ .

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A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound of the bow twice, one traveling through the body of the tube and the other the air in the tube. Find the time gap between the two hearings.

The speed of sound in air is Vair=230 m/sThe speed of the sound in steel is Vs=5200 m/s. Here the distance covered by the sound is S=1 mSo, t=t1−t2=(1330−15200)=2.75×10−3 sec=2.75 ms.

A steel tube of length $1.00 m$ is struck at one end. A person with his ear close to the other end hears the sound of the bow twice, one traveling through the body of the tube and the other the air in the tube. Find the time gap between the two hearings.

The speed of sound in air is $V_{a i r} = 230 m / s$
The speed of the sound in steel is $V_{s} = 5200 m / s$ .
Here the distance covered by the sound is $S = 1 m$

So, $t = t_{1} - t_{2}$
$= (\frac{1}{330} - \frac{1}{5200})$

$= 2.75 \times 10^{- 3} s e c$
$= 2.75 m s$ .