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Question

(A) The intensity $I$ of a sound wave is given by the relation $I = \frac{1}{2} \rho v \omega^2 A^2$,where $\rho$ is the density of the medium,$v$ is

Vedclass · Accepted Answer

$A = 10 	imes 10^{-4} \, m, \omega = 500 \, s^{-1}$

Vedclass · Answer

(A) The intensity $I$ of a sound wave is given by the relation $I = \frac{1}{2} ho v \omega^2 A^2$,where $ho$ is the density of the medium,$v$ is the wave velocity,$\omega$ is the angular frequency,and $A$ is the displacement amplitude.Since $ho$ and $v$ are constant for all waves traveling in the same medium,the intensity is proportional to the square of the product of angular frequency and amplitude: $I \propto (\omega A)^2$.Let us calculate the value of $(\omega A)$ for each option:$A) \omega A = 500 	imes 10 	imes 10^{-4} = 0.5$$B) \omega A = 2000 	imes 2 	imes 10^{-4} = 0.4$$C) \omega A = 115 	imes 2 	imes 10^{-4} = 0.023$$D) \omega A = 200 	imes 20 	imes 10^{-4} = 0.4$Comparing the values,option $A$ has the highest value of $(\omega A)$,and therefore,the highest intensity.

Waves of displacement amplitude $A$ and angular frequency $\omega$ travel in air with the same velocity. Which of the following waves has the highest intensity?

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