Question

Two sound waves move in the same direction. If the average power transmitted across a cross-section by them are equal while their wavelengths are in the ratio of 1:2. Their pressure amplitudes would be in the ratio of ___.

Answer 1

Given,

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{2}$

$\therefore \frac{f_1}{f_2}=\frac{{\omega }_1}{{\omega }_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{2}{1}$........(i)

$(\because f\propto \frac{1}{\lambda })$

The average power s given by,

$P=\frac{1}{2}\rho {\omega }^2{A}^{2}Sv$

As the average power transmitted across a cross-section by the two waves are equal, therefore we have,

$\frac{1}{2}\rho {\omega }_1^2{A}_1^{2}Sv=\frac{1}{2}\rho {\omega }_2^2{A}_2^{2}Sv$

$\Rightarrow {\omega }_1{A}_1={\omega }_2{A}_2$

or $\frac{A_1}{A_2}=\frac{{\omega }_2}{{\omega }_1}=\frac{1}{2}$........(From equation (i)).

Now pressure amplitude, $P_0=B_{0}AK$

$\therefore \frac{(P_0{)}_1}{(P_0{)}_2}=\frac{A_1}{A_2}.\frac{K_1}{K_2}=\frac{A_1}{A_2}.\frac{{\lambda }_2}{{\lambda }_1}$

[Since $K=\frac{2\pi }{\lambda }$]

$\Rightarrow \frac{(P_0{)}_1}{(P_0{)}_2}=\left(\frac{1}{2}\right)\left(\frac{2}{1}\right)=1$.