Velocity Of Sound In Different Media MCQ - Practice Questions with Answers

Speed of sound wave in a material medium - 

For deriving the equation of speed let us consider a section AB of medium as shown in figure of cross-sectional area S. Let A and B be two cross-sections as shown. Let in this medium sound wave propagation be from left to right. If wave source is located at origin O and when it oscillates, the oscillations at that point propagate along the rod.       

The stress at any cross section can be written as -     . . . . .. . . . (i)

Let us consider a section AB of the material as shown in the figure, of medium at a general instant of time t. The end A is at a distance 'x' from O and point B is at a distance 'x+d x' from O. Let in time duration 'dt' due to oscillations, medium particles at A be displaced along the length of a medium by 'y' and those at B by 'y+d y'. The resulting positions of section are A' and B' as shown in figure. By this, we can say that section AB is elongated by a length 'dy'. Thus strain produced in it is - 

$$
E=\frac{d y}{d x} \ldots \ldots .(i i)
$$

If Young's modulus of the material of medium is $Y$, we have

$$
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\delta_1}{E}
$$

By using Hooke's law -
From Eqs. (i) and (ii), we have

$$
\begin{aligned}
& Y=\frac{F / S}{d y / d x} \\
& F=Y S \frac{d y}{d x} \ldots .(i i i)
\end{aligned}
$$
Here, $F=$ Force

if $d m$ is the mass of section $A B$ and $a$ is its acceleration, which can be given as for a medium of density $\rho$ as

$$
\begin{array}{r}
d m=\rho S d x \\
a=\frac{d^2 y}{d t^2}
\end{array}
$$

From Eq. (iv), we have

$$
\begin{aligned}
& d F=(\rho S d x) \frac{d^2 y}{d t^2} \\
& \frac{d F}{d x}=\rho S \frac{d^2 y}{d t^2}
\end{aligned}
$$

From Eq. (iii) on differentiating w.r.t. to $x$, we can write

$$
\frac{d F}{d x}=Y S \frac{d^2 y}{d x^2}
$$

From Eqs. (v) and (vi), we get

$$
\frac{d^2 y}{d t^2}=\left(\frac{Y}{\rho}\right) \frac{d^2 y}{d x^2}
$$

Equation (vii) is a different form of wave equation.

$$
\begin{aligned}
\quad \frac{\partial^2 y}{\partial t^2}=v^2 \frac{\partial^2 y}{\partial x^2} \\
v=\sqrt{\frac{Y}{\rho}}
\end{aligned}
$$

This above equation shows wave velocity

In the case of gas or liquid, which shows only volume elasticity, E = B, where B = Bulk modulus of elasticity.

For longitudinal waves for liquid or gas - 

$$
v=\sqrt{\frac{B}{\rho}}
$$

where $\rho=$ Density of the medium