Question

A 3 m long organ pipe open at both ends is driven to third harmonic standing wave. If the amplitude of pressure oscillations is 1 per cent of mean atmospheric pressure ($P_0={10}^{5}N/m^2$). Find the amplitude of particle displacement and density oscillations. Speed of sound $v=332m/s$ and density of air $\rho =1.03kg/m^3$.

• A. $1.28cm,9.0\times {10}^{-3}kg/m^3$
  
• B. $28cm,9.0\times {10}^{-3}kg/m^3$
• C. $0.28cm,9.0\times {10}^{-3}kg/m^3$
• D. $8cm,9.0\times {10}^{-3}kg/m^3$

Answer 1

The correct option is C $0.28cm,9.0\times {10}^{-3}kg/m^3$

$\frac{3\lambda }{2}=3\therefore \lambda =2m{P}_m=100N/m^{2}V=330m/s{\rho }_0=1kg/m^3{P}_m=B{s}_{0}k={\rho }_0{v}^2{s}_0\frac{2\pi }{\lambda }\Rightarrow \frac{\lambda P_m}{{\rho }_0{v}^{2}2\pi }\therefore s_0=\frac{2\times 100}{1\times 330\times 330x2\pi }\Rightarrow 0.28cmB=\frac{-dp}{dv/v}=\frac{dp}{d\rho /\rho }[\because m=\rho v=0=\frac{dp}{\rho }+\frac{dv}{v}]d\rho =\frac{\rho .dp}{B}\Rightarrow {\left(d\rho \right)}_{max}=\frac{\rho }{B}={\left(dp\right)}_{max}\Rightarrow \frac{\rho P_m}{B}{\left(d\rho \right)}_{max}\therefore {\left(d\rho \right)}_{max}=\frac{\rho P_m}{\rho v^2}=\frac{1}{1089}=9\times {10}^{-3}kg/m^3$