Displacement Wave And Pressure Wave MCQ - Practice Questions with Answers

Relation between displacement wave and pressure wave -

As we have studied that, when a longitudinal wave propagates in a gaseous medium, it produces rarefaction and compression in the medium, periodically. The region where compression occurs, the pressure is more as compare to the normal pressure of the medium and the region where rarefaction occurs, the pressure is lesser as compare to the normal pressure of the medium. Thus we can also describe any longitudinal waves in a gaseous medium as pressure waves and these are also termed as compressional waves.

Let us consider a longitudinal wave propagating in a positive 'x' direction as shown in the given figure. This figure shows segment AB
of the medium of width 'dx'. Let a longitudinal wave propagate in this medium whose equation is given as - 

$$
y=A \sin (k x-\omega t)
$$

In this equation, ' $y$ ' is the displacement of a medium particle situated at a distance ' $x$ ' from the origin along the direction of propagation of the wave. From the figure, $A B$ is the medium segment such that $A$ is located at position $x=x$ and $B$ is at $x=x+d$ $x$ at an instant. If after some time $t$ medium particle at $A$ reaches to a point $A^{\prime}$ which is displaced by $y$ and the medium particle at $B$ reaches point $B^{\prime}$ which is at a displacement $y+d y$ from $B$. Here dy is given by equation as

$$
\begin{aligned}
& d y=A k \cos (k x-\omega t) d x \\
& d V=S d y=-S A k \cos (k x-\omega t) d x
\end{aligned}
$$

Where, $\mathrm{S}=$ Area of cross-section and $\mathrm{V}=$ Volume of section AB

$$
\begin{gathered}
\frac{d V}{V}=\frac{d y}{d x}=\frac{S A k \cos (k x-\omega t) d x}{S d x} \\
\frac{d V}{V}=A k \cos (k x-\omega t)
\end{gathered}
$$
 

If B is the bulk modulus of the medium, then the excess pressure in section AB can be given as - 

$\begin{aligned} & \Delta P=-B\left(\frac{d V}{V}\right)=-B\left(\frac{d y}{d x}\right) \\ & \Delta P=-B A k \cos (k x-\omega t) \\ & \Delta P=-\Delta P_{\max } \cos (k x-\omega t)\end{aligned}$

Here $\Delta P_{\max }$ is the pressure amplitude at a medium particle at position $x$ from origin and $\Delta P$ is the excess pressure at that point So,

$$
\Delta P_{\max }=B A k=\frac{2 \pi}{\lambda} A B
$$
 

In the compression zone, more particles stay in a unit volume of the medium. So, the density and pressure of the region will be higher. In the refracted zone, lesser particles stay in any unit volume. Let a sound wave propagate in a medium of Bulk modulus B and density .

                                                            So,

$$
\begin{gathered}
B=\left(-\frac{d p}{d V / V}\right) \\
\frac{d V}{V}=-\frac{d p}{p}
\end{gathered}
$$

From both Equation, we get, $d \rho=\frac{\rho}{B} d p$
The speed of sound is given by, $v=\sqrt{\frac{B}{\rho}} \Rightarrow \frac{\rho}{B}=\frac{1}{v^2}$ Hence, $d \rho=\frac{\rho}{B} \Delta p=\frac{1}{v^2} \Delta p$

So, this relation gives a relation between pressure with density. So the variation of density is like the variation of pressure -

$$
\Delta \rho=(\Delta \rho)_m \sin (k x-\omega t)
$$

where, $(\Delta \rho)_m=\frac{\rho}{B}(\Delta p)_m=\frac{(\Delta p)_m}{v^2}$
Note - The density equation is in phase with the pressure equation and this is $\frac{\pi}{2}$ out of phase with the displacement equation