
A siren placed at a railway platform is emitting a sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz when the train approaches the siren. During his journey in a different train B he records the frequency of 6 kHz while approaching the same siren. The ratio of velocity of train B to train A is:
A. \[\dfrac{{242}}{{252}}\]
B. \[\dfrac{5}{6}\]
C. \[2\]
D. \[\dfrac{{11}}{6}\]
Answer
509.4k+ views
Hint: The concept of apparent frequency is used to resolve the problem. The apparent frequency heard is calculated by undertaking the original frequency along with velocity of sound and the velocity of the observer.
Complete Step by Step Answer:Given:
Original frequency of the siren is, \[{f_0} = 5\;{\rm{kHz}}\].
Apparent frequency heard for train A is, \[{f_1} = 5.5\;{\rm{kHz}}\]
Apparent frequency heard for train B is, \[{f_2} = 6\;{\rm{kHz}}\]
Let the speed of train A and B be \[{v_A}\] and \[{v_B}\].
The formula for the apparent speed for train A is,
\[f = {f_0}\left[ {\dfrac{{{v_s} + {v_A}}}{{{v_s}}}} \right]\]
Here, \[{v_s}\] is the speed of sound and its standard value in air is 330 m/s.
Solving for train A as,
\[\begin{array}{l}
{f_1} = {f_0}\left[ {\dfrac{{{v_s} + {v_A}}}{{{v_s}}}} \right]\\
5.5\;{\rm{Hz}} = 5\;{\rm{Hz}}\left[ {\dfrac{{330\;{\rm{m/s}} + {v_A}}}{{330\;{\rm{m/s}}}}} \right]\\
{v_A} = 33\;{\rm{m/s}}...................................................\left( 1 \right)
\end{array}\]
Solving for train B as,
\[\begin{array}{l}
{f_2} = {f_0}\left[ {\dfrac{{{v_s} + {v_B}}}{{{v_s}}}} \right]\\
6\;{\rm{Hz}} = 5\;{\rm{Hz}}\left[ {\dfrac{{330\;{\rm{m/s}} + {v_B}}}{{330\;{\rm{m/s}}}}} \right]\\
{v_B} = 66\;{\rm{m/s}}...................................................\left( 2 \right)
\end{array}\]
Taking ratio of equation 1 and 2 as,
\[\begin{array}{l}
\dfrac{{{v_B}}}{{{v_A}}} = \dfrac{{66\;{\rm{m/s}}}}{{33\;{\rm{m/s}}}}\\
\dfrac{{{v_B}}}{{{v_A}}} = 2
\end{array}\]
Therefore, the required ratio of velocity of train B to train A is 2 and the correct option is C.
Note:The correct mathematical formula for the apparent frequency is to be remembered and the concept of apparent frequency is to be known along with the direction of source of sound. And the value of speed of sound in air should be known.
Complete Step by Step Answer:Given:
Original frequency of the siren is, \[{f_0} = 5\;{\rm{kHz}}\].
Apparent frequency heard for train A is, \[{f_1} = 5.5\;{\rm{kHz}}\]
Apparent frequency heard for train B is, \[{f_2} = 6\;{\rm{kHz}}\]
Let the speed of train A and B be \[{v_A}\] and \[{v_B}\].
The formula for the apparent speed for train A is,
\[f = {f_0}\left[ {\dfrac{{{v_s} + {v_A}}}{{{v_s}}}} \right]\]
Here, \[{v_s}\] is the speed of sound and its standard value in air is 330 m/s.
Solving for train A as,
\[\begin{array}{l}
{f_1} = {f_0}\left[ {\dfrac{{{v_s} + {v_A}}}{{{v_s}}}} \right]\\
5.5\;{\rm{Hz}} = 5\;{\rm{Hz}}\left[ {\dfrac{{330\;{\rm{m/s}} + {v_A}}}{{330\;{\rm{m/s}}}}} \right]\\
{v_A} = 33\;{\rm{m/s}}...................................................\left( 1 \right)
\end{array}\]
Solving for train B as,
\[\begin{array}{l}
{f_2} = {f_0}\left[ {\dfrac{{{v_s} + {v_B}}}{{{v_s}}}} \right]\\
6\;{\rm{Hz}} = 5\;{\rm{Hz}}\left[ {\dfrac{{330\;{\rm{m/s}} + {v_B}}}{{330\;{\rm{m/s}}}}} \right]\\
{v_B} = 66\;{\rm{m/s}}...................................................\left( 2 \right)
\end{array}\]
Taking ratio of equation 1 and 2 as,
\[\begin{array}{l}
\dfrac{{{v_B}}}{{{v_A}}} = \dfrac{{66\;{\rm{m/s}}}}{{33\;{\rm{m/s}}}}\\
\dfrac{{{v_B}}}{{{v_A}}} = 2
\end{array}\]
Therefore, the required ratio of velocity of train B to train A is 2 and the correct option is C.
Note:The correct mathematical formula for the apparent frequency is to be remembered and the concept of apparent frequency is to be known along with the direction of source of sound. And the value of speed of sound in air should be known.
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