
A car moves towards a hill with speed \[{v_c}\] . It blows a horn of frequency f which is heard by an observer following the car with speed \[{v_0}\] . The speed of sound in air is \[v\] . Sound is reflected from the hill. Then:
A. The length of sound reaching the hill is \[\dfrac{v}{f}\]
B. The wavelength of sound reaching the hill is $\dfrac{{v - {v_c}}}{f}$
C. The wavelength of sound of horn directly reaching the observer is $\dfrac{{v + {v_0}}}{f}$
D. The beat frequency observed by the observer is $\dfrac{{{v_c}\left( {v + {v_0}} \right)f}}{{{v^2} - v_c^2}}$
Answer
483.3k+ views
Hint: The Doppler Effect in sound is the basis for this problem. So, before we go into the question, let's review the basics of the Doppler Effect. When a wave source travels relative to its observer, the Doppler Effect occurs, which causes a change in wave frequency.
Complete answer:
We get the following Doppler Effect equation when a source is travelling towards a stationary observer: $f' = \dfrac{v}{{\left( {v - {v_s}} \right)}}f$
In this case the frequency of sound is the source that reaches the top of the hill, which is a stationary observer:
$f' = \dfrac{v}{{v - {v_c}}}f$
As a result, the wavelength of sound that reaches the hill is $ = \dfrac{{v - {v_c}}}{f}$
Sound frequency directly heard by the observer: $f''' = \dfrac{{v + {v_0}}}{{v + {v_c}}}f$
Sound wave length that immediately reaches the observer: $\dfrac{{v + {v_c}}}{{v + {v_0}}}\dfrac{v}{f}$
After reflection from the hill, the frequency of sound was observed: \[\;f'' = \;\dfrac{{v + {v_0}}}{v}f' = \dfrac{{v + {v_0}}}{{v - {v_c}}}f\]
As a result, beat frequency $ = \left| {f''' - f''} \right| = \dfrac{{2{v_c}f\left( {{v_0} + v} \right)}}{{{v^2} - v_c^2}}$
So, the correct option is: B. The wavelength of sound reaching the hill is $\dfrac{{v - {v_c}}}{f}$
Note:
One thing students should know about the Doppler Effect is that, unlike sound, light, and radio waves, matter waves' particles have momentum as well. Because of the source's velocity, this momentum is gained. The frequency of matter waves with increased momentum cannot be determined using Doppler shift.
Complete answer:
We get the following Doppler Effect equation when a source is travelling towards a stationary observer: $f' = \dfrac{v}{{\left( {v - {v_s}} \right)}}f$
In this case the frequency of sound is the source that reaches the top of the hill, which is a stationary observer:
$f' = \dfrac{v}{{v - {v_c}}}f$
As a result, the wavelength of sound that reaches the hill is $ = \dfrac{{v - {v_c}}}{f}$
Sound frequency directly heard by the observer: $f''' = \dfrac{{v + {v_0}}}{{v + {v_c}}}f$
Sound wave length that immediately reaches the observer: $\dfrac{{v + {v_c}}}{{v + {v_0}}}\dfrac{v}{f}$
After reflection from the hill, the frequency of sound was observed: \[\;f'' = \;\dfrac{{v + {v_0}}}{v}f' = \dfrac{{v + {v_0}}}{{v - {v_c}}}f\]
As a result, beat frequency $ = \left| {f''' - f''} \right| = \dfrac{{2{v_c}f\left( {{v_0} + v} \right)}}{{{v^2} - v_c^2}}$
So, the correct option is: B. The wavelength of sound reaching the hill is $\dfrac{{v - {v_c}}}{f}$
Note:
One thing students should know about the Doppler Effect is that, unlike sound, light, and radio waves, matter waves' particles have momentum as well. Because of the source's velocity, this momentum is gained. The frequency of matter waves with increased momentum cannot be determined using Doppler shift.
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