Question

The ratio of intensities of two waves are given by $4:1$ . Then the ratio of the amplitude of the two waves is :
(A) $2:1$
(B) $1:2$
(C) $4:1$
(D) $1:4$

Answer 1

Hint the intensity and amplitude are related to each other by square law. We will use that formula to calculate the ratio of amplitudes.

Complete Step-by-step Solution
The formula of intensity is given by energy per unit area per unit time.
$\Rightarrow$ $I = \dfrac{E}{{Ar \times t}}$ ……(i)
Where $I = $intensity of wave
$\Rightarrow$ $E = $energy of wave
$\Rightarrow$ $Ar = $area on which it is falling
$\Rightarrow$ $t = $time in seconds.
Eq (i) can be rewritten as –
$\Rightarrow$ $I = \dfrac{P}{{Ar}}$ …….(ii)
We will now calculate power and area to see the relation between amplitude and intensity.

Let the equation of the wave be given as:
$\Rightarrow$ $y = A\sin (\omega t - kx)$
The energy per unit length of this wave is determined experimentally and is given by:
$\Rightarrow$ $\dfrac{{dE}}{{dx}} = \mu {A^2}{\omega ^2}{\cos ^2}(\omega t - kx)$
The power of the wave is given by-
$
  P = \dfrac{{energy}}{{time}} \\
  P = \dfrac{{dE}}{{dt}} \times \dfrac{{dx}}{{dx}} \\
  P = v \times \dfrac{{dE}}{{dx}} \\
  P = v{A^2}{\omega ^2}\mu {\cos ^2}(\omega t - kx) \\
    \\
 $
In one frequency , the power travelled is average power.
$\Rightarrow$ ${P_{avg}} = \dfrac{{1{A^2}{\omega ^2}\mu v}}{2}$
In eq (ii) , we will use average power as the value of power.
Now coming to eq (ii)-
$
  I = \dfrac{P}{{Ar}} \\
  I = \dfrac{{{A^2}{\omega ^2}\mu v}}{{2Ar}} \\
    \\
 $
Where
$A = $ amplitude of wave
$\Rightarrow$ $\mu = $ mass per unit length
$\Rightarrow$ $\omega = $angular frequency
$\Rightarrow$ $v = $velocity of wave
$\mu = \dfrac{m}{L}$ putting this in the formula of intensity .
$m = $ mass of the string in which the wave is travelling
$L = $length of the string in which the wave is travelling.
$\Rightarrow$ $
  I = \dfrac{1}{2} \times {A^2}{\omega ^2}\dfrac{m}{V}v \\
  I = \dfrac{1}{2}{A^2}{\omega ^2}v\rho \\
  I = \dfrac{1}{2}{A^2}{(2\pi f)^2}v\rho \\
  I = 2{A^2}{\pi ^2}{f^2}\rho v \\
 $
This is the derived formula of intensity . We can clearly see that-
$I \propto {A^2}$
$\Rightarrow$ $
  \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{A_1^2}}{{A_2^2}} \\
  \dfrac{{{A_1}}}{{{A_2}}} = \sqrt {\dfrac{{{I_1}}}{{{I_2}}}} \\
  \dfrac{{{A_1}}}{{{A_2}}} = \sqrt {\dfrac{4}{1}} \\
  \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{2}{1} \\
 $

Hence the correct option is A.

Note We have considered a wave travelling in a string here. The string has certain mass and length and that is what $m$ and $L$ are.