A man is running between two identical sources producing sound of same frequency as shown in figure. If beat frequency obtained by him is 2 then what will be the new beat frequency obtained by him if he doubles his speed

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Solution

Beat $= f_{1} - f_{2}$

$f_{1} = f_{0} \frac{(v_{s} + v)}{v_{s}}$ and $f_{2} = f_{0} \frac{(v_{s} - v)}{v_{s}}$

$f_{1} - f_{2} = 2 = \frac{f_{0}}{v_{s}} [v_{s} + v - v_{s} + v] = \frac{f_{0}}{v_{s}} \times 2 v \dots \dots (1)$

Again $f_{1} - f_{2} = x = \frac{f_{0}}{v_{s}} [v_{s} + 2 v - v_{s} + 2 v] = \frac{f_{0}}{v_{s}} \times 4 v \dots \dots (2)$

By (1) $\div$ (2) $\Rightarrow x = 4$

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A man is running between two identical sources producing sound of same frequency as shown in figure. If beat frequency obtained by him is 2 then what will be the new beat frequency obtained by him if he doubles his speed

Beat =f1−f2 f1=f0(vs+v)vs and f2=f0(vs−v)vs f1−f2=2=f0vs[vs+v−vs+v]=f0vs×2v……(1) Again f1−f2=x=f0vs[vs+2v−vs+2v]=f0vs×4v……(2) By (1) ÷ (2) ⇒x=4