Question

Give the appropriate speed of sound in the given medium.
A. Iron
B. Water
C. Vacuum
D. Dry air

Answer 1

Hint: Sound is an elastic longitudinal wave.The speed of longitudinal wave in a medium depends on the properties of medium such as elasticity and inertia. We have to have an idea about the values of elastic constants and densities of different media to get the speed of sound in different media.
Formula used: $v=\sqrt{\dfrac{Y}{\rho }},v=\sqrt{\dfrac{B}{\rho }}$ and $v=\sqrt{\dfrac{\gamma P}{\rho }}$

Complete answer:
A. The velocity of sound in a solid is given by the formula
\[v=\sqrt{\dfrac{Y}{\rho }}\] , where $Y$ is the Young’s modulus and $\rho $ is the density of the solid. Now for iron,
$Y=2.0\times {{10}^{11}}N{{m}^{-2}}$ and $\rho =7.7\times {{10}^{3}}kg{{m}^{-3}}$ . Putting these values we get
\[v=\sqrt{\dfrac{2\times {{10}^{11}}}{7.7\times {{10}^{3}}}}m{{s}^{-1}}=\sqrt{0.2597\times {{10}^{8}}}m{{s}^{-1}}=0.5096\times {{10}^{4}}m{{s}^{-1}}=5096m{{s}^{-1}}\] .
B. The velocity of sound in a liquid is given by
$v=\sqrt{\dfrac{B}{\rho }}$ , where$B$ is the bulk modulus and $\rho $ is the density of the liquid. Now for water,
$B=2\times {{10}^{9}}N{{m}^{-2}},\rho =1\times {{10}^{3}}kg{{m}^{-3}}$ . Putting these values we get
\[v=\sqrt{\dfrac{2\times {{10}^{9}}}{1\times {{10}^{3}}}}m{{s}^{-1}}=1414m{{s}^{-1}}.\]
C. As we know that sound is an elastic wave, so it needs a medium to travel. Thus it cannot propagate in the vacuum. Its velocity in the vacuum will be zero.
D. The velocity of sound in a gas is given by the formula
$v=\sqrt{\dfrac{\gamma P}{\rho }}$ , where $\gamma $ is the ratio of the specific heats of the gas at constant pressure and at constant volume, $P$ is the pressure and $\rho $ is the density of the gas. Now for dry air, \[\gamma =\dfrac{7}{4}=1.4,P=1.01\times {{10}^{5}}N{{m}^{-2}}\] and$\rho =1.29kg{{m}^{-3}}.$ Putting these values we get
$v=\sqrt{\dfrac{1.4\times 1.01\times {{10}^{5}}}{1.29}}m{{s}^{-1}}=331m{{s}^{-1}}.$

Note:
The formulae for speed of sound in solids, liquids and gases are different. We have to use the appropriate formula for specific mediums. In case of gases, the formula with Laplace’s correction should be used and not Newton's formula. Again while calculating we have to keep in mind that square roots are needed to be evaluated.