Two men are walking along a horizontal straig | vedclass

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(A) Let the stationary man be at point $C$ and the horizontal line be $AB$. Let $O$ be the point on the line $AB$ directly below $C$. Given $CO = 12 \

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the stationary man be at point $C$ and the horizontal line be $AB$. Let $O$ be the point on the line $AB$ directly below $C$. Given $CO = 12 \,m$. The men are at $A$ and $B$ such that $AO = OB = 5 \,m$. The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \,m$. The angle $	heta$ between the line $AC$ (or $BC$) and the vertical $CO$ is given by $\cos 	heta = \frac{12}{13}$ and $\sin 	heta = \frac{5}{13}$.The man at $A$ (behind) is moving towards $O$ with velocity $v_A = 2.0 \,m \,s^{-1}$. The component of his velocity along the line $AC$ is $v_{A, 	ext{eff}} = v_A \sin 	heta = 2.0 	imes \frac{5}{13} \,m \,s^{-1}$. Since he is moving towards $C$,the observed frequency is $f_A = f \left( \frac{v}{v - v_A \sin 	heta} ight) = 1430 \left( \frac{330}{330 - 2 	imes \frac{5}{13}} ight) \approx 1430 \left( 1 + \frac{10}{13 	imes 330} ight)$.The man at $B$ (front) is moving away from $O$ with velocity $v_B = 1.0 \,m \,s^{-1}$. The component of his velocity along the line $BC$ is $v_{B, 	ext{eff}} = v_B \sin 	heta = 1.0 	imes \frac{5}{13} \,m \,s^{-1}$. Since he is moving away from $C$,the observed frequency is $f_B = f \left( \frac{v}{v + v_B \sin 	heta} ight) = 1430 \left( \frac{330}{330 + 1 	imes \frac{5}{13}} ight) \approx 1430 \left( 1 - \frac{5}{13 	imes 330} ight)$.The beat frequency is $\Delta f = |f_A - f_B| = 1430 \left( \frac{10}{13 	imes 330} + \frac{5}{13 	imes 330} ight) = 1430 \left( \frac{15}{13 	imes 330} ight) = \frac{1430}{330} 	imes \frac{15}{13} = \frac{13}{3} 	imes \frac{15}{13} = 5 \,Hz$.

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