The dependence of disturbances due to two wav | vedclass

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(A) The intensity $I$ of a wave is proportional to the square of its amplitude $A$ and the square of its angular frequency $\omega$ $(I \propto \omega

Vedclass · Accepted Answer

$1 : 1$

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(A) The intensity $I$ of a wave is proportional to the square of its amplitude $A$ and the square of its angular frequency $\omega$ $(I \propto \omega^2 A^2)$.From the figure,for the first wave (solid line),the amplitude $A_1 = 1$ and it completes two cycles in the same time that the second wave completes one cycle. Thus,the frequency $\omega_1 = 2\omega_2$.For the second wave (dotted line),the amplitude $A_2 = 2$ and frequency $\omega_2$.The ratio of intensities is given by:$\frac{I_1}{I_2} = \left(\frac{\omega_1}{\omega_2}ight)^2 \left(\frac{A_1}{A_2}ight)^2$Substituting the values:$\frac{I_1}{I_2} = \left(\frac{2\omega_2}{\omega_2}ight)^2 \left(\frac{1}{2}ight)^2 = (2)^2 	imes \left(\frac{1}{2}ight)^2 = 4 	imes \frac{1}{4} = 1$Therefore,the ratio is $1 : 1$.

The dependence of disturbances due to two waves on time is shown in the figure. The ratio of their intensities $I_1 / I_2$ will be

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