3.0 Speed of a longitudinal Wave

Consider a fluid element $ab$ confined to a tube of cross sectional area $S$ as shown in figure.

The element has a thickness $\Delta x$.

Let us assume that the equilibrium pressure of the fluid is ${P_{_0}}$.

Because of the disturbance, the section $a$ of the element moves distance $y$ from its mean position to a new position $a'$ and section $b$ moves a distance $\left( {y + \Delta y} \right)$ to a new position $b'$.

The pressure on the left and right side of the element becomes $\left( {{P_0} + \Delta {P_1}} \right)$ and $\left( {{P_0} + \Delta {P_2}} \right)$respectively.

If $\rho $ is the equilibrium density, the mass of the element is $\Delta m$, $$\Delta m=\rho S\Delta x$$ When the element moves, its mass does not change. Even though its volume and density changes.

The net force acting on the element is,$$\begin{equation} \begin{aligned} F = \Delta P.\;S \\ F = (\Delta {P_1} - \Delta {P_2}).\;S\quad ...(i) \\\end{aligned} \end{equation} $$Acceleration of wave is given by, $$a = \frac{{{\delta ^2}y}}{{\delta {t^2}}}$$Net force associated with the wave, $$\begin{equation} \begin{aligned} F = \Delta m.\;a \\ F = \rho S\Delta x.\;\frac{{{\delta ^2}y}}{{\delta {x^2}}}\quad ...(ii) \\\end{aligned} \end{equation} $$From equation $(i)$ and $(ii)$ we get, $$(\Delta {P_1} - \Delta {P_2})S = \rho S\Delta x\frac{{{\delta ^2}y}}{{\delta {t^2}}}$$ Dividing both sides by $\Delta x$, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{(\Delta {P_1} - \Delta {P_2})S}}{{\Delta x}} = \rho S\frac{{{\delta ^2}y}}{{\delta {t^2}}} \\ \Rightarrow - \frac{{\Delta \left( {\Delta P} \right)}}{{\Delta x}}S = \rho S\frac{{{\delta ^2}y}}{{\delta {t^2}}} \\\end{aligned} \end{equation} $$ As, ${\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta \left( {\Delta P} \right)}}{{\Delta x}} = \frac{{d\left( {\Delta P} \right)}}{{dx}}}$,$$\frac{{\delta \left( {\Delta P} \right)}}{{\delta x}} = - \rho \frac{{{\delta ^2}y}}{{\delta {t^2}}}\quad ...(iii)$$The excess pressure $\Delta P$ may be written as,$$\begin{equation} \begin{aligned} \Delta P = - B\frac{{\delta y}}{{\delta x}} \\ \frac{{\delta \left( {\Delta P} \right)}}{{\delta x}} = - B\frac{{{\delta ^2}y}}{{\delta {x^2}}}\quad ...(iv) \\\end{aligned} \end{equation} $$

From equation $(iii)$ and $(iv)$ we get, $$\frac{{{\delta ^2}y}}{{\delta {x^2}}} = \frac{\rho }{B}.\frac{{{\delta ^2}y}}{{\delta {t^2}}}$$or $$\frac{{{\delta ^2}y}}{{\delta {t^2}}} = \frac{B}{\rho }.\frac{{{\delta ^2}y}}{{\delta {x^2}}}$$Comparing the above equation with the wave equation, $$\frac{{{\delta ^2}y}}{{\delta {t^2}}} = {v^2}\frac{{{\delta ^2}y}}{{\delta {x^2}}}$$

So, the velocity of the sound wave is given as, $$v = \sqrt {\frac{B}{\rho }} $$

$v$ is the speed of longitudinal waves within a gas or a liquid.

Question 2. Calculate the speed of longitudinal waves in the following gases at ${0^0}$ C and 1 atm ($ = {10^5}$).

(a) oxygen for which the bulk modulus is $1.41 \times {10^5}Pa$ and density is $1.43kg/{m^3}$.

(b) helium for which the bulk modulus is $1.7 \times {10^5}Pa$ and density is $0.18kg/{m^3}$.

(a) $${v_{{o_2}}} = \sqrt {\frac{B}{\rho }} = \sqrt {\frac{{1.41 \times {{10}^5}}}{{1.43}}} = 314m/s$$

(b)$${v_{He}} = \sqrt {\frac{B}{\rho }} = \sqrt {\frac{{1.7 \times {{10}^5}}}{{0.18}}} = 927m/s$$