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Speed of Longitudinal Waves

Speed of Longitudinal Waves :- Consider a fluid (gas/liquid) element of thickness Δx confined in a tube of cross-sectional area S as shown in figure below :

Let P_o be the equilibrium pressure and ρ be the equilibrium density of the fluid. The mass of the element under consideration is ρSΔx. After the production of wave, let the fluid element changes its position such that pressure on its left side becomes P_o + ΔP₁ and pressure on its right side becomes P_o + ΔP₂. When the element changes its position, its volume and density changes but mass remains the same.

The net force acting on the element :

$\displaystyle F=\left( \Delta {{P}_{1}}-\Delta {{P}_{2}} \right)S$

The net acceleration of the element :

$\displaystyle a=\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$

Applying Newton’s second law to the element,

$\displaystyle F=ma$

$\displaystyle \Rightarrow \left( \Delta {{P}_{1}}-\Delta {{P}_{2}} \right)S=\rho S\Delta x\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$

$\displaystyle \Rightarrow \left( \Delta {{P}_{1}}-\Delta {{P}_{2}} \right)=\rho \Delta x\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$

Dividing both sides by Δx and taking limit Δx→0,

$\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\left( \Delta {{P}_{1}}-\Delta {{P}_{2}} \right)}{\Delta x}=\rho \frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$

$\displaystyle \Rightarrow -\frac{\partial P}{\partial x}=\rho \frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$ …..(1)

The negative sign indicates that as Δx decreases, pressure (P) increases.

Now we also know that excess pressure is given by,

$\displaystyle \Delta P=-B\frac{\partial y}{\partial x}=-B\frac{\partial }{\partial x}\left[ A\sin (\omega t-kx) \right]=BAk\left[ \cos (\omega t-kx) \right]$

Differentiating above equation partially w.r.t. position(x),

$\displaystyle \frac{\partial P}{\partial x}=BA{{k}^{2}}\sin (\omega t-kx)$ …..(2)

Now,

$\displaystyle \frac{\partial y}{\partial t}=A\omega \cos (\omega t-kx)$

$\displaystyle \Rightarrow \frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-A{{\omega }^{2}}sin(\omega t-kx)$ …..(3)

Using equations (2) and (3) in equation (1), we get

$\displaystyle -\left[ BA{{k}^{2}}\sin (\omega t-kx) \right]=\rho \left[ -A{{\omega }^{2}}sin(\omega t-kx) \right]$

$\displaystyle \Rightarrow \left[ B{{k}^{2}} \right]=\rho \left[ {{\omega }^{2}} \right]$

$\displaystyle \Rightarrow {{\left( \frac{\omega }{k} \right)}^{2}}=\frac{B}{\rho }$

$\displaystyle \Rightarrow {{v}^{2}}=\frac{B}{\rho }$

$\displaystyle v=\sqrt{\frac{B}{\rho }}$ …..(4)

Equation (4) gives speed of longitudinal waves in a fluid (gasses and liquids).

If a longitudinal wave propagates in a solid then speed of longitudinal waves is given by :

$\displaystyle v=\sqrt{\frac{Y}{\rho }}$ …..(5)

Here ρ is the density and Y is Young’s Modulus of the solid.

In general, speed of a longitudinal waves in a medium of elasticity E is given by :

$\displaystyle v=\sqrt{\frac{E}{\rho }}$ …..(6)

For fluids, E is replaced by B (Bulk Modulus)

$\displaystyle {{v}_{liquid/gas}}=\sqrt{\frac{B}{\rho }}$

Similarly for solids, E is replaced by Y (Young’s Modulus)

$\displaystyle {{v}_{solid}}=\sqrt{\frac{Y}{\rho }}$

Note :-

(1). Although the density of solids is much larger than the density of fluids but the Young’s Modulus (Y) is larger by a greater factor. Hence speed of longitudinal waves is much larger in solids than in fluids.

(2). The speed in different media :

$\displaystyle {{v}_{solid}}>{{v}_{liquid}}>{{v}_{gas}}$

Speed of Longitudinal Waves

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