Two men are walking along a horizontal straig | vedclass

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(B) Let the stationary man be at point $C$ and the two moving men be at $A$ and $B$. The distance $CO = 12 \ m$. Since the men are $10 \ m$ apart and

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let the stationary man be at point $C$ and the two moving men be at $A$ and $B$. The distance $CO = 12 \ m$. Since the men are $10 \ m$ apart and the stationary man is equidistant,$AO = OB = 5 \ m$.The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \ m$.The angle $	heta$ between the line of motion and the line joining the observer to the source is given by $\cos 	heta = \frac{5}{13}$.For the man behind $(A)$ moving towards $O$ with speed $v_s = 2.0 \ m \ s^{-1}$,the component of velocity along the line $AC$ is $v_s \cos 	heta$ towards $C$. The observed frequency is $f_A = f \left( \frac{v}{v - v_s \cos 	heta} ight) = 1430 \left( \frac{330}{330 - 2 \cos 	heta} ight) \approx 1430 \left( 1 + \frac{2 \cos 	heta}{330} ight)$.For the man in front $(B)$ moving away from $O$ with speed $v_s = 1.0 \ m \ s^{-1}$,the component of velocity along the line $BC$ is $v_s \cos 	heta$ away from $C$. The observed frequency is $f_B = f \left( \frac{v}{v + v_s \cos 	heta} ight) = 1430 \left( \frac{330}{330 + 1 \cos 	heta} ight) \approx 1430 \left( 1 - \frac{\cos 	heta}{330} ight)$.The beat frequency is $\Delta f = f_A - f_B = 1430 \left( \frac{2 \cos 	heta + \cos 	heta}{330} ight) = 1430 \left( \frac{3 \cos 	heta}{330} ight) = 13 \cos 	heta$.Substituting $\cos 	heta = \frac{5}{13}$,we get $\Delta f = 13 	imes \frac{5}{13} = 5 \ Hz$.

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