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.
