2.2 Relation between pressure wave and density wave

  Sound Waves

According to the definition of bulk modulus $(B)$, $$B = \left( { - \frac{{dP}}{{dV/V}}} \right)$$ or $$\frac{{dV}}{V} = - \frac{{dP}}{B}\quad ...(i)$$ Also, $${\text{Volume = }}\frac{{{\text{Mass}}}}{{{\text{Density}}}}$$ So, $$V = \frac{m}{\rho }$$ Differentiating the above equation we get, $$\begin{equation} \begin{aligned} \frac{{dV}}{{d\rho }} = - \frac{m}{{{\rho ^2}}} \\ \frac{{dV}}{{d\rho }} = - \frac{1}{\rho }\left( {\frac{m}{\rho }} \right) \\ \frac{{dV}}{{d\rho }} = - \frac{V}{\rho }\quad \quad \left( {As,V = \frac{m}{\rho }} \right) \\ \frac{{dV}}{V} = - \frac{{d\rho }}{\rho }\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} \frac{{d\rho }}{\rho } = \frac{{dP}}{B} \\ d\rho = \frac{\rho }{B}\left( {dP} \right) \\\end{aligned} \end{equation} $$ The above equation can also be written as, $$\Delta \rho = \frac{\rho }{B}\left( {\Delta P} \right)$$ or $$\Delta \rho = \frac{{\left( {\Delta P} \right)}}{{{v^2}}}\quad \quad \left( {As,v = \sqrt {\frac{B}{\rho }} } \right)$$ As we know, $$\Delta P = {\left( {\Delta P} \right)_m}\sin \left( {kx - \omega t} \right)$$ So, density equation can be written as, $$\Delta \rho = {\left( {\Delta \rho } \right)_m}\sin \left( {kx - \omega t} \right)$$ where, $${\left( {\Delta \rho } \right)_m} = \frac{\rho }{B}{\left( {\Delta P} \right)_m}\quad or\quad {\left( {\Delta \rho } \right)_m} = \frac{{{{\left( {\Delta P} \right)}_m}}}{{{v^2}}}\quad $$

Note:

• Density equation is in phase with the pressure equation
• Density equation is $90^\circ $ out of phase with the displacement equation

Question 1. (a)What is the displacement amplitude for a sound wave having a frequency of $100Hz$ and a pressure amplitude of $10Pa$?

(b) The displacement amplitude of a sound wave of frequency $300Hz$ is ${10^{ - 7}}m$. What is the pressure amplitude of this wave? Speed of sound in air is $340m/s$ and density of air is $1.29kg/{m^3}$.

Solution:

(a)$${(\Delta P)_m} = BAk$$Here,$$k = \frac{\omega }{v} = \frac{{2\pi f}}{v}$$and$$B = \rho {v^2}$$$$\therefore {(\Delta P)_m} = (\rho {v^2})(A)\left( {\frac{{2\pi f}}{v}} \right)$$$$\therefore A = \frac{{{{(\Delta P)}_m}}}{{2\pi f\rho v}}....(i)$$

Substituting the values, $$A = \frac{{10}}{{2 \times 3.14 \times 100 \times 1.29 \times 340}}$$$$ = 3.63 \times {10^{ - 5}}m$$

(b) From equation $(i)$,$${\left( {\Delta P} \right)_m} = 2\pi f\rho vA$$Substituting the values,$${\left( {\Delta P} \right)_m} = 2 \times 3.14 \times 300 \times 1.29 \times 340 \times {10^{ - 7}}$$$$ = 8.26 \times {10^{ - 2}}N/{m^2}$$