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This is one of our many quizzes on Doppler effect. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Doppler effect Quiz 2

physics

 A cooker whistles with $f=5\,Hz$. If the chef moving with ${{v}_{c}}=5\,{m}/{s}\;$away from the cooker, what frequency does the chef hear? $(v=330\,{m}/{s}\;)$ 

 For an observer moving away from a stationary source, use the Doppler effect equation, as the source, in this case the cooker whistle, is stationary and the observer, in this case a chef, is moving away from the whistle. The frequency of the cooker whistle is the fundamental frequency. 

  $f'=\left( \dfrac{v-{{v}_{c}}}{v} \right)\times f \\ \Rightarrow f'=\left( \dfrac{330-5}{330} \right)\times 5 \\ \Rightarrow f'=\dfrac{325}{330}\times 5 \\ \therefore f'=4.9\,Hz$

 The trash vehicle while playing music with a frequency of $60\,Hz$ crosses a house with a speed of $60\,{m}/{s}\;$. At the same time, a person jogging from the opposite direction crosses the same house with a speed of $v\,{m}/{s}\;$ hears the music to be of frequency $77\,Hz$. What is the speed with which the person is jogging? $(v=330\,\,{m}/{s}\;)$

 For the source and observer heading towards each other, use the Doppler Effect equation. The source is a garbage truck, and the observer is a human. The fundamental frequency corresponds to the frequency of music played in a garbage truck.

  $ f'=\dfrac{v+{{v}_{o}}}{v-{{v}_{s}}}f \\ \Rightarrow 77=\dfrac{330+v}{330-60}\times 60 \\ \Rightarrow \dfrac{77}{60}=\dfrac{330+v}{270} \\ \Rightarrow v=\dfrac{77\times 270}{60}-330 \\ \therefore v=16.5\,{m}/{s}$

 The dog chasing a cat, barks with a frequency of 80 Hz. The speed of the dog is $60\,{m}/{s}\;$ and the speed of the cat is $40\,{m}/{s}\;$. The frequency of barking as per heard by the cat is, $($ The speed of sound in air, $v=360\,\,{m}/{s}\;)$

 Use the Doppler Effect equation for the source moving towards the observer and the observer moving away from the source. Here the source is a dog and the observer is a cat. The fundamental frequency is the frequency of bark.

 $f'=\dfrac{v-{{v}_{o}}}{v-{{v}_{s}}}f \\ \Rightarrow f'=\dfrac{360-40}{360-60}\times 80 \\ \Rightarrow f'=\dfrac{320}{300}\times 80 \\ \therefore f'=85\,Hz$

 The teacher is reading a book with a sound of a frequency of 50 Hz and the principal is situated at some distance. The instructor is moving with a speed of $10\,\,{m}/{s}\;$ at an angle of $30{}^\circ $ with the S-O $($Source - Object$)$ axis. The wavelength of the sound wave of reading heard by the principal will be, $((v=330\,\,{m}/{s}\;))$

 For a source moving towards an observer at rest, use the Doppler Effect equation. The analyst is the principal, and the source is a teacher. The basic frequency is the teacher's reading frequency.

 $\text{The speed of source is,} \\ {{v}_{s}}=v\cos \theta  \\ \Rightarrow {{v}_{s}}=10\cos 30{}^\circ  \\ \therefore {{v}_{s}}=8.66\,{m}/{s} \\ \text{The apparent frequency is, } \\ f'=\dfrac{v}{v-{{v}_{s}}}f \\ \Rightarrow f'=\dfrac{330}{330-8.66}\times 50 \\ \Rightarrow f'=\dfrac{330}{321.34}\times 50 \\ \therefore f'=51.34\,Hz \\ \text{The equation of velocity,} \\ v={f}'\lambda \\ \text{Rearranging the equation,} \\ \lambda =\dfrac{v}{{{f}'}} \\ \text{Substituting the values in the equation,} \\ \lambda =\dfrac{51.34}{330} \\ \therefore \lambda =0.15\text{ }m$

 The bus and scooter are moving away from each other at a rate of $60\,\,{m}/{s}\;$ and $40\,\,{m}/{s}\;$ respectively. The frequency of the mouth organ blown by the passenger is $50 Hz$. The power of sound of mouth organ heard by the girl riding a scooter at $t=8\text{ }s$ is, $($The speed of sound in air, $v=360\,\,{m}/{s}\;)$

 For the source and observer moving away from each other, use the Doppler Effect equation, where the bus is the source and the scooter is the observer. The fundamental frequency is equal to the frequency of the mouth organ's vibration.

 $f'=\left( \dfrac{v-{{v}_{o}}}{v+{{v}_{s}}} \right)\times f \\ \Rightarrow f'=\left( \dfrac{360-40}{360+60} \right)\times 50 \\ \Rightarrow f'=\dfrac{320}{420}\times 50 \\ \therefore f'=38\,Hz  \\ \text{The energy of the sound from mouth organ,} \\ E=h{f}' \\ \text{Substituting the values in it,} \\ E=6.626\times {{10}^{-34}}\times 38 \\ \therefore E=251.788\text{ }J  \\ \text{The equation of power,} \\ P=\dfrac{E}{t} \\ \text{Substituting the values in it,} \\ P=\dfrac{251.788}{8} \\ \therefore P\approx 32\text{ }W$