JEE Mains · Physics · STD 11 - 14. waves and sound
A source of sound emits sound waves at frequency \(f_0\). It is moving towards an observer with fixed speed \(v_s\) (\(v_s < v\), where \(v\) is the speed of sound in air). If the observer were to move towards the source with speed \(v_0\), one of the following two graphs (\(A\) and \(B\)) will given the correct variation of the frequency \(f\) heard by the observer as \(v_0\) is changed The variation of \(f\) with \(v_0\) is given correctly by

- A graph \(A\) with slope \( = \,\frac{{{f_0}}}{{\left( {v + {v_s}} \right)}}\)
- B graph \(B\) with slope \( = \,\frac{{{f_0}}}{{\left( {v - {v_s}} \right)}}\)
- C graph \(A\) with slope \( = \,\frac{{{f_0}}}{{\left( {v - {v_s}} \right)}}\)
- D graph \(B\) with slope \( = \,\frac{{{f_0}}}{{\left( {v + {v_s}} \right)}}\)
Answer & Solution
Correct Answer
(C) graph \(A\) with slope \( = \,\frac{{{f_0}}}{{\left( {v - {v_s}} \right)}}\)
Step-by-step Solution
Detailed explanation
According to Doppler's effect, Apparent, frequency \(f=\left(\frac{V+V_{0}}{V-V_{S}}\right) f_{0}\) Now, \(f=\left(\frac{f_{0}}{V-V_{S}}\right) V_{0}+\frac{V f_{0}}{V-V_{s}}\) So, slope \(=\frac{f_{0}}{V-V_{S}}\) Hence, option \((c)\) is the correct answer.
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