question_answer

A source of sound emits sound waves at frequency \[{{f}_{0}}.\]It is moving towards an observer with fixed speed \[{{\upsilon }_{s}}({{\upsilon }_{s}}<\upsilon ,\] where \[\upsilon \] is the speed of sound in air). If the observer were to move towards the source with speed \[{{\upsilon }_{0}},\] one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as \[{{\upsilon }_{0}}\]is changed. The variation of f with \[{{\upsilon }_{0}}\]is given correctly by: [JEE Main Online Paper (Held On 11 April 2015)]  

A)  graph A with slope \[=\frac{{{f}_{0}}}{(\upsilon -{{\upsilon }_{s}})}\]

B)    graph A with slope\[=\frac{{{f}_{0}}}{(\upsilon +{{\upsilon }_{s}})}\]

C)    graph B with slope \[=\frac{{{f}_{0}}}{(\upsilon -{{\upsilon }_{s}})}\]

D)    graph B with slope\[=\frac{{{f}_{0}}}{(\upsilon +{{\upsilon }_{s}})}\]

Correct Answer: A

Solution :

For the given situationfrequency listened by an observer is f. So, \[f={{f}_{0}}\left[ \frac{V+{{V}_{0}}}{V-{{V}_{s}}} \right]\] \[f=\frac{{{f}_{0}}V}{V-{{V}_{s}}}+\frac{{{f}_{0}}}{V-{{V}_{s}}}={{V}_{0}}\] equating the equation \[y=mx+C\] \[m=\frac{{{f}_{0}}}{V-{{V}_{s}}}\]So choice is (A) .