Question

A car moving towards a cliff emits sound of frequency 'n'. If the difference in frequencies of the horn and its echo heard by the driver of the car is 10% of 'n', then the speed of the car is nearly (Speed of sound in air is 336ms$^{-1}$)

• A. 16 ms$^{-1}$
• B. 18 ms$^{-1}$
• C. 30 ms$^{-1}$
• D. 33 ms$^{-1}$

Hint:

Problems involving echoes and the Doppler effect can be treated as a two-step process. First, the moving object is the source and the stationary reflector is the observer. Second, the reflector becomes a stationary source emitting the received frequency, and the original object is now a moving observer.

Answer 1

The Correct Option is A

Step 1: Model the problem in two stages.
Stage 1: The car is a moving source, and the cliff is a stationary observer. The car's horn emits frequency $n$. Stage 2: The cliff reflects the sound, acting as a stationary source. The car's driver is a moving observer.

Step 2: Calculate the frequency detected by the cliff.
The formula for the Doppler effect when the source moves towards a stationary observer is $f' = f \left(\frac{v}{v-v_s}\right)$, where $v$ is the speed of sound, and $v_s$ is the speed of the source. Let $v_s = v_{car}$ and $v = v_{sound}$. The frequency received by the cliff is $n' = n \left(\frac{v_{sound}}{v_{sound}-v_{car}}\right)$.

Step 3: Calculate the frequency of the echo heard by the driver.
The cliff reflects the frequency $n'$. Now, the cliff is a stationary source emitting $n'$, and the driver is an observer moving towards the source. The formula for the Doppler effect when the observer moves towards a stationary source is $f'' = f' \left(\frac{v+v_o}{v}\right)$, where $v_o$ is the speed of the observer. Here, $f' = n'$ and $v_o = v_{car}$. The frequency of the echo heard by the driver is $n_{echo} = n' \left(\frac{v_{sound}+v_{car}}{v_{sound}}\right)$. Substitute the expression for $n'$: \[ n_{echo} = \left(n \frac{v_{sound}}{v_{sound}-v_{car}}\right) \left(\frac{v_{sound}+v_{car}}{v_{sound}}\right) = n \left(\frac{v_{sound}+v_{car}}{v_{sound}-v_{car}}\right). \]

Step 4: Use the given information about the frequency difference.
The difference between the echo frequency and the original horn frequency is given as 10% of $n$. \[ \Delta n = n_{echo} - n = 0.10n. \] \[ n \left(\frac{v_{sound}+v_{car}}{v_{sound}-v_{car}}\right) - n = 0.1n. \] Divide by $n$: \[ \frac{v_{sound}+v_{car}}{v_{sound}-v_{car}} - 1 = 0.1. \] \[ \frac{(v_{sound}+v_{car}) - (v_{sound}-v_{car})}{v_{sound}-v_{car}} = 0.1. \] \[ \frac{2v_{car}}{v_{sound}-v_{car}} = 0.1. \]

Step 5: Solve for the speed of the car, $v_{car$.}
\[ 2v_{car} = 0.1(v_{sound}-v_{car}). \] \[ 2v_{car} = 0.1v_{sound} - 0.1v_{car}. \] \[ 2.1v_{car} = 0.1v_{sound}. \] \[ v_{car} = \frac{0.1}{2.1} v_{sound} = \frac{1}{21} v_{sound}. \] Substitute the given speed of sound, $v_{sound} = 336$ m/s. \[ v_{car} = \frac{336}{21} = 16 \text{ m/s}. \] \[ \boxed{v_{car} = 16 \text{ m/s}}. \]