Question

For energy density, power and intensity of any wave choose the correct option-
A) $u = $ energy density $ = \dfrac{1}{2}\rho {\omega ^2}{A^2}$
B) $P = $ power $ = \dfrac{1}{2}\rho {\omega ^2}{A^2}v$
C) $I = $ intensity $ = \dfrac{1}{2}\rho {\omega ^2}{A^2}Sv$
D) $I = \dfrac{P}{S}$

Answer 1

Hint
Energy density is the ratio of energy to volume and also intensity to velocity. Energy is written as $P \times t$. Volume is written as $S \times L$. Velocity is written as $\dfrac{L}{t}$. So by cancelling the distance and time we get intensity as the ratio of power to the area.

Complete step-by-step answer
$L$ is length, $S$ is area, $P$ is power, $t$ is time, $v$ is velocity and $V$ is volume. We will find the relation between the quantities of Intensity, Power, Area, velocity etc.
The energy associated with unit volume of the medium is defined as energy density. It is given by,
$u = \dfrac{{Energy}}{{Volume}}$
It is also given by,
$u = \dfrac{{Intensity}}{{Velocity}}$
$\dfrac{{Energy}}{{Volume}} = \dfrac{{Intensity}}{{Velocity}}$
Since energy is product of power and time-
$E = P \times t$,
$\dfrac{{P \times t}}{V} = \dfrac{I}{v}$
Volume can be written as a product of area and length-
$V = S \times L$.
$\dfrac{{P \times t}}{{S \times L}} = \dfrac{I}{v}$
Cross multiply by keeping intensity on the left hand side.
$I = \dfrac{{P \times t \times v}}{{S \times L}}$
As velocity is ratio of distance to time
$v = \dfrac{l}{t}$,
$I = \dfrac{{P \times t \times l}}{{S \times l \times t}}$
Hence, intensity is the ratio of power to area
$I = \dfrac{P}{S}$.
The correct relation in the given options is (D)

Note
The wave intensity is defined as the average amount of energy flow in medium per unit time and per unit of its cross-sectional area. Its unit is $W/{m^2}$. It is given by,
$I = \dfrac{E}{{S \times t}}$
At a distance $r$ from a point source of power$\;P$ the intensity is given by,
$I = \dfrac{P}{{4{\pi ^2}}}$
$I$ is inversely proportional to the square of the distance $r$,
$I \propto\ \dfrac{1}{{{r^2}}}$.