The displacement-time graphs for two sound wa | vedclass

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(D) The intensity $I$ of a sound wave is proportional to the square of its amplitude $a$ and the square of its angular frequency $\omega$,i.e.,$I \pro

Vedclass · Accepted Answer

$1:1$

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(D) The intensity $I$ of a sound wave is proportional to the square of its amplitude $a$ and the square of its angular frequency $\omega$,i.e.,$I \propto a^2 \omega^2$.From the given graph:For wave $A$,the amplitude $a_A = 2$ units and the time period $T_A = 2$ units (as it completes one cycle in $2$ units of time).For wave $B$,the amplitude $a_B = 1$ unit and the time period $T_B = 1$ unit (as it completes one cycle in $1$ unit of time).Since angular frequency $\omega = 2\pi/T$,we have $\omega_A / \omega_B = T_B / T_A = 1/2$.Now,the ratio of intensities is:$\frac{I_A}{I_B} = \left( \frac{a_A}{a_B} ight)^2 	imes \left( \frac{\omega_A}{\omega_B} ight)^2$$\frac{I_A}{I_B} = \left( \frac{2}{1} ight)^2 	imes \left( \frac{1}{2} ight)^2 = 4 	imes \frac{1}{4} = 1$Thus,the ratio $I_A/I_B$ is $1:1$.

The displacement-time graphs for two sound waves $A$ and $B$ are shown in the figure. The ratio of their intensities $I_A/I_B$ is equal to:

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