Question

Two harmonic waves travelling in same medium have frequency ratio 1:2 and intensity ratio 1:36 the amplitude ratio is

Answer 1

we know, intensity of harmonic wave is given by, $I=2\pi^2A^2\nu^2\rho C$

where, A is amplitude, v is frequency of wave, $\rho$ is density of wave and C is speed of wave.

here, $\rho$ and C are constants
so, $\frac{I_1}{I_2}=\frac{A_1^2\nu_1^2}{A_2^2\nu_2^2}$

given, $\frac{\nu_1}{\nu_2}$ = 1 : 2
$\frac{I_1}{I_2}$ = 1 : 36

so, 1/36 = $\frac{A_1^2}{A_2^2}$ × 1/4

hence, $\frac{A_1}{A_2}$ = 1 : 3

Answer 2

Given :

Ration of frequency is 1 : 2 i.e  $\dfrac{\nu_2}{\nu_1}=\dfrac{1}{2}$ .

Ration of intensity is 1 : 36 i.e $\dfrac{A_2}{A_1}=\dfrac{1}{36}$ .

To Find :

The amplitude ratio .

Solution :

We know , intensity of harmonic waves is given by :

$I=2\pi^2 A^2\nu^2\rho C$

So , ratio is given by :

$\dfrac{I_2}{I_1}=\dfrac{2\pi^2 A_2^2\nu_2^2\rho C}{2\pi^2 A_1^2\nu_1^2\rho C}$

$\dfrac{I_2}{I_1}=(\dfrac{A_2}{A_1})^2\times (\dfrac{\nu_2}{\nu_1})^2\\\\\dfrac{A_2}{A_1}=\sqrt{\dfrac{I_2}{I_1}}\times \dfrac{\nu_1}{\nu_2}\\\\\dfrac{A_2}{A_1}=\sqrt{\dfrac{1}{36}}\times \dfrac{2}{1}\\\\\dfrac{A_2}{A_1}=\dfrac{1}{3}$

The ratio of amplitude is 1 : 3 .

Learn More :

Waves

https://brainly.in/question/8539665