
A person is riding a motorcycle at a speed \[v\] towards a stationary car operating its siren at \[165\,{\text{Hz}}\]. A police car is chasing the person at a speed \[22\,{\text{m/s}}\] with its siren operating at \[176\,{\text{Hz}}\]. If the motorcyclist does not hear beats, the speed \[v\left( {{\text{m/s}}} \right)\] is (speed of sound in air=\[330\,{\text{m/s}}\])
A. 33
B. 22
C. 11
D. 0
Answer
572.1k+ views
Hint: Use the formula for the frequency of the source heard by the receiver for the different conditions of the state of motion of the source and receiver. Also consider the condition for the beat difference heard by the receiver if there are two sources of the sound.
Formulae used:
The expression for the frequency \[f\] of the stationary source heard by the moving receiver is
\[f = \dfrac{{c + {v_r}}}{c}{f_0}\] …… (1)
Here, \[c\] is the speed of the air, \[{v_r}\] is the speed of the moving receiver and \[{f_0}\] is the frequency of the source.
The expression for the frequency \[f\] of the moving source heard by the moving receiver moving towards the source is
\[f = \dfrac{{c + {v_r}}}{{c + {v_s}}}{f_0}\] …… (2)
Here, \[c\] is the speed of the air, \[{v_r}\] is the speed of the moving receiver, \[{v_s}\] is the speed of the moving source and \[{f_0}\] is the frequency of the source.
Complete step by step answer:
The person is riding on a motorcycle with a speed \[v\,{\text{m/s}}\] towards a stationary car operating his siren at \[165\,{\text{Hz}}\]. A police car is chasing the person at a speed \[22\,{\text{m/s}}\] with its siren operating at \[176\,{\text{Hz}}\].
Calculate the frequency of the stationary car siren heard by the person riding the motorcycle.
Rewrite equation (1) for the frequency \[{f_1}\] of the stationary car siren heard by the person riding the motorcycle.
\[{f_1} = \dfrac{{c + v}}{c}{f_{02}}\]
Here, \[{f_{02}}\] is the frequency of the siren in the stationary car.
Substitute \[330\,{\text{m/s}}\] for \[c\] and \[165\,{\text{Hz}}\] for \[{f_{02}}\] in the above equation.
\[{f_1} = \dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right)\]
Rewrite equation (1) for the frequency \[{f_2}\] of the police car siren heard by the person riding the motorcycle.
\[{f_2} = \dfrac{{c - v}}{{c - {v_3}}}{f_{03}}\]
Here, \[{f_{03}}\] is the frequency of the siren in the police car and \[{v_3}\] is the speed of the police car.
The negative sign indicates that the police car siren is moving towards the person and the person is moving away from the police car.
Substitute \[330\,{\text{m/s}}\] for \[c\], \[22\,{\text{m/s}}\] for \[{v_3}\] and \[176\,{\text{Hz}}\] for \[{f_{03}}\] in the above equation.
\[{f_2} = \dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\]
The person riding the motorcycle hears no beats. This shows that the beat difference of the stationary car siren and the police car siren heard by the person is zero.
\[ \Rightarrow {f_1} - {f_2} = 0\]
\[ \Rightarrow {f_1} = {f_2}\]
Substitute \[\dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right)\] for \[{f_1}\] and \[\dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\] for \[{f_2}\] in the above equation.
\[\dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right) = \dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\]
Solve the above equation for the velocity \[v\] of the person.
\[v = 22\,{\text{m/s}}\]
Therefore, the speed \[v\,\left( {{\text{m/s}}} \right)\] of the person is \[22\,{\text{m/s}}\].
Hence, the correct option is B.
Note:Make sure that the sign of the velocity of the receiver or the source is reversed if they are not moving towards each other because this the point where students do mistakes.
Formulae used:
The expression for the frequency \[f\] of the stationary source heard by the moving receiver is
\[f = \dfrac{{c + {v_r}}}{c}{f_0}\] …… (1)
Here, \[c\] is the speed of the air, \[{v_r}\] is the speed of the moving receiver and \[{f_0}\] is the frequency of the source.
The expression for the frequency \[f\] of the moving source heard by the moving receiver moving towards the source is
\[f = \dfrac{{c + {v_r}}}{{c + {v_s}}}{f_0}\] …… (2)
Here, \[c\] is the speed of the air, \[{v_r}\] is the speed of the moving receiver, \[{v_s}\] is the speed of the moving source and \[{f_0}\] is the frequency of the source.
Complete step by step answer:
The person is riding on a motorcycle with a speed \[v\,{\text{m/s}}\] towards a stationary car operating his siren at \[165\,{\text{Hz}}\]. A police car is chasing the person at a speed \[22\,{\text{m/s}}\] with its siren operating at \[176\,{\text{Hz}}\].
Calculate the frequency of the stationary car siren heard by the person riding the motorcycle.
Rewrite equation (1) for the frequency \[{f_1}\] of the stationary car siren heard by the person riding the motorcycle.
\[{f_1} = \dfrac{{c + v}}{c}{f_{02}}\]
Here, \[{f_{02}}\] is the frequency of the siren in the stationary car.
Substitute \[330\,{\text{m/s}}\] for \[c\] and \[165\,{\text{Hz}}\] for \[{f_{02}}\] in the above equation.
\[{f_1} = \dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right)\]
Rewrite equation (1) for the frequency \[{f_2}\] of the police car siren heard by the person riding the motorcycle.
\[{f_2} = \dfrac{{c - v}}{{c - {v_3}}}{f_{03}}\]
Here, \[{f_{03}}\] is the frequency of the siren in the police car and \[{v_3}\] is the speed of the police car.
The negative sign indicates that the police car siren is moving towards the person and the person is moving away from the police car.
Substitute \[330\,{\text{m/s}}\] for \[c\], \[22\,{\text{m/s}}\] for \[{v_3}\] and \[176\,{\text{Hz}}\] for \[{f_{03}}\] in the above equation.
\[{f_2} = \dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\]
The person riding the motorcycle hears no beats. This shows that the beat difference of the stationary car siren and the police car siren heard by the person is zero.
\[ \Rightarrow {f_1} - {f_2} = 0\]
\[ \Rightarrow {f_1} = {f_2}\]
Substitute \[\dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right)\] for \[{f_1}\] and \[\dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\] for \[{f_2}\] in the above equation.
\[\dfrac{{\left( {330\,{\text{m/s}}} \right) + v}}{{330\,{\text{m/s}}}}\left( {165\,{\text{Hz}}} \right) = \dfrac{{\left( {330\,{\text{m/s}}} \right) - v}}{{\left( {330\,{\text{m/s}}} \right) - \left( {22\,{\text{m/s}}} \right)}}\left( {176\,{\text{Hz}}} \right)\]
Solve the above equation for the velocity \[v\] of the person.
\[v = 22\,{\text{m/s}}\]
Therefore, the speed \[v\,\left( {{\text{m/s}}} \right)\] of the person is \[22\,{\text{m/s}}\].
Hence, the correct option is B.
Note:Make sure that the sign of the velocity of the receiver or the source is reversed if they are not moving towards each other because this the point where students do mistakes.
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