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(D) Let $f_0$ be the actual frequency of the horn,$C$ be the speed of sound,and $V_s$ be the speed of the car.When the car approaches the observer: $f

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) Let $f_0$ be the actual frequency of the horn,$C$ be the speed of sound,and $V_s$ be the speed of the car.When the car approaches the observer: $f_1 = f_0 \left( \frac{C}{C - V_s} ight) = 100 \, Hz$.When the car moves away from the observer: $f_2 = f_0 \left( \frac{C}{C + V_s} ight) = 50 \, Hz$.Dividing the two equations: $\frac{f_1}{f_2} = \frac{100}{50} = 2 = \frac{C + V_s}{C - V_s}$.Solving for $V_s$: $2(C - V_s) = C + V_s \implies 2C - 2V_s = C + V_s \implies C = 3V_s \implies V_s = \frac{C}{3}$.Substituting $V_s$ into the first equation: $100 = f_0 \left( \frac{C}{C - C/3} ight) = f_0 \left( \frac{C}{2C/3} ight) = f_0 \left( \frac{3}{2} ight)$.Thus,$f_0 = 100 	imes \frac{2}{3} = \frac{200}{3} \, Hz$.If the observer moves with the car,the relative velocity between the source and the observer is zero,so the observed frequency is equal to the actual frequency $f_0$.Therefore,$f = f_0 = \frac{200}{3} \, Hz$.Comparing this with $\frac{x}{3} \, Hz$,we get $x = 200$.

When a car is approaching the observer,the frequency of the horn is $100 \, Hz$. After passing the observer,it is $50 \, Hz$. If the observer moves with the car,the frequency will be $\frac{x}{3} \, Hz$ where $x = .....$

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