Two waves are represented byy_1 = 5 sin 2 pi( | vedclass

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(D) The intensity of a wave is given by $I = 2 \pi^2 \rho v f^2 A^2$,where $\rho$ is density,$v$ is wave speed,$f$ is frequency,and $A$ is amplitude.F

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$1: 16$

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(D) The intensity of a wave is given by $I = 2 \pi^2 ho v f^2 A^2$,where $ho$ is density,$v$ is wave speed,$f$ is frequency,and $A$ is amplitude.For the given waves:Wave $1$: $y_1 = 5 \sin(150 \pi t - 0.5 \pi x)$. Amplitude $A_1 = 5$,angular frequency $\omega_1 = 150 \pi$,wave number $k_1 = 0.5 \pi$. Wave speed $v_1 = \frac{\omega_1}{k_1} = \frac{150 \pi}{0.5 \pi} = 300 \ m/s$.Wave $2$: $y_2 = 10 \sin(300 \pi t - 1.0 \pi x)$. Amplitude $A_2 = 10$,angular frequency $\omega_2 = 300 \pi$,wave number $k_2 = 1.0 \pi$. Wave speed $v_2 = \frac{\omega_2}{k_2} = \frac{300 \pi}{1.0 \pi} = 300 \ m/s$.Since $v_1 = v_2$ and assuming the medium is the same $(ho_1 = ho_2)$,the intensity ratio is $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}ight)^2 	imes \left(\frac{f_1}{f_2}ight)^2 = \left(\frac{A_1}{A_2}ight)^2 	imes \left(\frac{\omega_1}{\omega_2}ight)^2$.Substituting the values: $\frac{I_1}{I_2} = \left(\frac{5}{10}ight)^2 	imes \left(\frac{150 \pi}{300 \pi}ight)^2 = \left(\frac{1}{2}ight)^2 	imes \left(\frac{1}{2}ight)^2 = \frac{1}{4} 	imes \frac{1}{4} = \frac{1}{16}$.

Two waves are represented by
$y_1 = 5 \sin 2 \pi(75 t - 0.25 x)$
$y_2 = 10 \sin 2 \pi(150 t - 0.50 x)$
The intensity ratio $\frac{I_1}{I_2}$ of the two waves is

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Two waves are represented by$y_1 = 5 \sin 2 \pi(75 t - 0.25 x)$$y_2 = 10 \sin 2 \pi(150 t - 0.50 x)$The intensity ratio $\frac{I_1}{I_2}$ of the two waves is