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 amplitude would be in the ratio of

• A. 1
• B. 2
• C. 4
• D. $\frac{1}{2}$

Answer 1

The correct option is A 1

$P=\frac{1}{2}\rho {\omega }^2{A}^{2}sv\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{2}\therefore \frac{f_1}{f_2}=\frac{{\omega }_1}{{\omega }_2}=\frac{2}{1}(asf\propto \frac{1}{\lambda })$
$P_1=P_2$
$\therefore {\omega }_1{A}_1={\omega }_2{A}_{2}or\frac{A_1}{A_2}=\frac{{\omega }_2}{{\omega }_1}=\frac{1}{2}$
Now pressure amplitude
$P_0=B_{0}Ak\therefore \frac{\left(P_0\right)1}{\left(P_0\right)2}=\frac{A_1}{A_2}.\frac{k_1}{k_2}=\frac{A_1}{A_2}.\frac{{\lambda }_2}{{\lambda }_1}=\left(\frac{1}{2}\right)\left(\frac{2}{1}\right)=1$