Question

A girl is sitting near the open window of a train that is moving at a velocity \[10.00\,{\text{m/s}}\] to the east. The girl’s uncle stands near the track and watches the train move away. The locomotive whistle emits sound at frequency \[500.0\,{\text{Hz}}\]. The air is still. What frequency does the girl hear? A wind begins to blow from the east at \[10.00\,{\text{m/s}}\].

Answer 1

Hint:We can use the formula for the frequency of the sound heard by the observer when the observer is approaching the source and source of sound is moving away from the observer. First we should calculate the relative velocity of the girl and locomotive whistle with respect to the wind. Then use this formula and calculate the frequency of the sound heard by the girl.

Formula used:
The expression for the frequency \[f'\] of the sound heard by the observer when the observer is approaching the source and source of sound is moving away from the observer is given by
\[f' = \left( {\dfrac{{v + {v_o}}}{{v + {v_s}}}} \right)f\] …… (1)
Here, \[v\] is the velocity of the sound, \[{v_o}\] is velocity of the observer, \[{v_s}\] is velocity of the source of sound and \[f\] is frequency of the sound from the source of sound.

Complete step by step answer:
We have given that the velocity of the train towards the east is \[10.00\,{\text{m/s}}\].
\[{v_t} = 10.00\,{\text{m/s}}\]
The frequency of the locomotive whistle emitting sound is \[500.0\,{\text{Hz}}\].
\[f = 500.0\,{\text{Hz}}\]
The velocity of the wind from east to west direction is \[10.00\,{\text{m/s}}\].
\[{v_w} = - 10.00\,{\text{m/s}}\]
Since the girl is sitting in the train, the velocity of the girl is the same as that of the velocity of the train. Also the velocity of the whistle is the same as that of the train.
\[{v_t} = {v_g} = {v_s}\]
The velocity \[{v_{gw}}\] of the girl with respect to the wind is given by
\[{v_{gw}} = {v_g} - {v_w}\]
Substitute \[10.00\,{\text{m/s}}\] for \[{v_t}\] and \[ - 10.00\,{\text{m/s}}\] for \[{v_w}\] in the above equation.
\[{v_{gw}} = \left( {10.00\,{\text{m/s}}} \right) - \left( { - 10.00\,{\text{m/s}}} \right)\]
\[ \Rightarrow {v_{gw}} = 20.00\,{\text{m/s}}\]
Hence, the velocity of the girl relative to the air is \[20.00\,{\text{m/s}}\].
The velocity \[{v_{sw}}\] of the source of sound with respect to the wind is given by
\[{v_{sw}} = {v_s} - {v_w}\]
Substitute \[10.00\,{\text{m/s}}\] for \[{v_s}\] and \[ - 10.00\,{\text{m/s}}\] for \[{v_w}\] in the above equation.
\[{v_{sw}} = \left( {10.00\,{\text{m/s}}} \right) - \left( { - 10.00\,{\text{m/s}}} \right)\]
\[ \Rightarrow {v_{sw}} = 20.00\,{\text{m/s}}\]
Hence, the velocity of the whistle relative to the air is \[20.00\,{\text{m/s}}\].
Let us now calculate the frequency of the sound heard by the girl. Rearrange equation (1) for the frequency of the sound heard by the girl.
\[f' = \left( {\dfrac{{v + {v_{gw}}}}{{v + {v_{sw}}}}} \right)f\]
Substitute \[20.00\,{\text{m/s}}\] for \[{v_{gw}}\], \[20.00\,{\text{m/s}}\] for \[{v_{sw}}\] and \[500.0\,{\text{Hz}}\] for \[f\] in the above equation.
\[f' = \left( {\dfrac{{v + \left( {20.00\,{\text{m/s}}} \right)}}{{v + \left( {20.00\,{\text{m/s}}} \right)}}} \right)\left( {500.0\,{\text{Hz}}} \right)\]
\[ \therefore f' = 500.0\,{\text{Hz}}\]

Hence, the frequency of the sound heard by the girl is \[500.0\,{\text{Hz}}\].

Note: The students should not forget to calculate the relative velocity of the girl and locomotive whistle with respect to the velocity of wind. In general, we consider only the velocity of the source of sound and observer. But in such cases, the wind is considered still. In the present case, the wind is not still. Hence, we should calculate the relative velocities of the observer and the source of sound.