Question

If the pressure variation in the medium is represented by equation $P=\Delta P_o\sin (kx-\omega t)$ then the resulting density variation in medium is suitably represented by equation

• A. $\rho =\Delta {\rho }_o\cos (kx-\omega t)$
• B. $\rho =-\Delta {\rho }_o\sin (kx-\omega t)$
• C. $\rho =-\Delta {\rho }_o\cos (kx-\omega t)$
• D. $\rho =\Delta {\rho }_o\sin (kx-\omega t)$

Answer 1

The correct option is D $\rho =\Delta {\rho }_o\sin (kx-\omega t)$

We know that when density is high in the medium pressure is also high and if density is low then pressure is low. This means density and pressure in the medium are in phase.
$\Rightarrow \Delta \varphi =0$
Therefore, for pressure equation, $P=\Delta P_o\sin (kx-\omega t)$, the density equation will be $\rho =\Delta {\rho }_o\sin (kx-\omega t)$.