A car P approaching a crossing at a speed of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The sound from the source $P$ reaches the observer at $Q$ along the path $PQ$. The source $P$ is approaching the crossing with a velocity $v_{s} =

Vedclass · Accepted Answer

$717$

Vedclass · Answer

(B) The sound from the source $P$ reaches the observer at $Q$ along the path $PQ$. The source $P$ is approaching the crossing with a velocity $v_{s} = 10\, m/s$.When the observer in car $Q$ hears the sound of the horn,the effective velocity of approach of the car $P$ towards the observer is $v_{s} \cos 	heta$.From the geometry of the problem,the distance $PQ = \sqrt{40^2 + 30^2} = 50\, m$. Thus,$\cos 	heta = \frac{40}{50} = \frac{4}{5}$.The apparent frequency $f'$ heard by the observer is given by the Doppler effect formula:$f' = \left( \frac{v}{v - v_{s} \cos 	heta} ight) f$Substituting the given values:$f' = \left( \frac{340}{340 - 10 	imes \frac{4}{5}} ight) 	imes 700$$f' = \left( \frac{340}{340 - 8} ight) 	imes 700 = \frac{340}{332} 	imes 700$$f' \approx 1.024 	imes 700 \approx 716.86\, Hz$Rounding to the nearest integer,we get $f' \approx 717\, Hz$.

Similar Questions

If $\alpha \neq \beta$ but $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is:

The value of $k$,if $(1,2)$ and $(k,-1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$ is:

The difference between the fundamental frequencies of an open pipe and a closed pipe of the same length is $100 \ Hz$. The difference between the frequencies of the second harmonic of the open pipe and the third harmonic of the closed pipe is: (in $Hz$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module