A steel tube of length 1.00 m is struck at one end. A person with his ear closed to the other end hears the sound of the blow twice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sound in various substances.

#### Solution

Given:

Velocity of sound in air_{ }*v *= 330 m/s

Velocity of sound through the steel tube *v*_{s} = 5200 m/s

Here, Length of the steel tube *S* = 1 m

As we know,

\[t = \frac{S}{v}\]

\[t_1 = \frac{1}{330}\] and \[t_2 = \frac{1}{5200}\]

Where, *t*_{1}_{ }is the time taken by the sound in air.

*t*_{2}_{ }is the time taken by the sound in steel tube.

Therefore,

\[\text { Required time gap } t = t_1 - t_2\]

\[ \Rightarrow t = \left( \frac{1}{330} - \frac{1}{5200} \right)\]

\[ \Rightarrow t = 2 . 75 \times {10}^{- 3} s\]

\[ \Rightarrow t = 2 . 75 \text { ms }\]

Hence, the time gap between two hearings is 2.75 ms.