Two cars are moving on two perpendicular road | vedclass

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(B) Given:Speed of car $A$ $(v_A)$ = $72\; km/hr = 72 	imes \frac{5}{18} = 20\; m/s$.Speed of car $B$ $(v_B)$ = $36\; km/hr = 36 	imes \frac{5}{18}

Vedclass · Accepted Answer

$298$

Vedclass · Answer

(B) Given:Speed of car $A$ $(v_A)$ = $72\; km/hr = 72 	imes \frac{5}{18} = 20\; m/s$.Speed of car $B$ $(v_B)$ = $36\; km/hr = 36 	imes \frac{5}{18} = 10\; m/s$.Frequency of source $(n)$ = $280\; Hz$.Speed of sound $(v)$ = $340\; m/s$ (assumed).The component of velocity of source $(A)$ along the line joining the cars is $v_A \cos 45^o$ (moving towards the observer).The component of velocity of observer $(B)$ along the line joining the cars is $v_B \cos 45^o$ (moving towards the source).Using the Doppler effect formula:$n' = n \left( \frac{v + v_B \cos 45^o}{v - v_A \cos 45^o} ight)$$n' = 280 \left( \frac{340 + 10 \cos 45^o}{340 - 20 \cos 45^o} ight)$$n' = 280 \left( \frac{340 + 10/\sqrt{2}}{340 - 20/\sqrt{2}} ight)$$n' = 280 \left( \frac{340 + 7.07}{340 - 14.14} ight) = 280 \left( \frac{347.07}{325.86} ight) \approx 298.2\; Hz$.Thus,the observed frequency is approximately $298\; Hz$.

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