Question

The driver of a car travelling at $100km/h$ toward a vertical cliff briefly sound the horn. Exactly one second latter she hears the echo and notes that its frequency is $840Hz$. How far from the cliff was the car when the driver sounded the horn and what is the frequency of the horn?

Answer 1

Let the distance between the car and the cliff be $x$

Speed of sound =343 m/s

Speed of car= $100km/hr=100\times \frac{5}{18}m/s=\frac{500}{18}m/s$

Distance traveled by the car in 1s= $\frac{500}{18}m$

Time taken by sound from car to cliff= $T_1=\frac{x}{343}$

When the sound is returning back to the car, the car has already traveled a distance of 500/18 m.

Hence, time taken from cliff to car= $T_2=\frac{x-\frac{500}{18}}{343}$

Total time taken= 1s=$T_1+T_2=\frac{x}{343}-\frac{500}{18\times 343}+\frac{x}{343}$

On solving for $x$ we get $x=185.39m$

Let $v$= velocity of sound in air=343m/s

and $v_c$ be the velocity of car= $\frac{500}{18}=27.7m/s$

Let the frequency of sound of the horn be $f_o$

Frequency of sound at the cliff:

$f_1=\frac{v}{v-v_c}\times f_o=\frac{343}{315.3}\times f_o$ ....(1)

Frequency of sound heard by the person in car after reflection from the cliff:

$f_2=\frac{v+v_c}{v}\times f_1=\frac{370.7}{343}\times f_1$

Substituting the value of $f_1$ from Eq. 1 and $f_2=840$, we get

$840=\frac{370.7}{343}\times \frac{343}{315.3}\times f_0$

We get,

$f_o=714.46Hz$