Discussion Forum

1)

When the observer moves towards the stationary source with velocity, v₁ the apparent frequency of emitted note is f₁. When the observer moves away from the velocity v₁, the apparent frequency is f₂ . if v is the velocity of sound in air and 

$\frac{f_{1}}{f_{2}}=2, then \frac{v}{v_{1}}=$  ?

Answer:

Option B

Explanation:

 According to question ,

 $\frac{f_{1}}{f_{2}}$=2   (given)

[ $f_{1}$ =  apparent frequency when velocity $v_{1}$ is towards the observer.

  $f_{2}$  = apparent frequency when velocity $v_{1}$ is away from the observer]

Now, the apparent frequency of sound  when the observer move towards the source is given by

   $f_{1}=\left(\frac{V}{V-V_{1}}\right)f_{0}$ ............(i)

 [ Symbols have their usual meanings]

 Similarly when observer moves away from the source, apparent frequency is given by

  $f_{2}=\left(\frac{V}{V+V_{1}}\right)f_{0}$  .........(ii)

 From Eqs.(i) and (ii), we get

$\frac{f_{1}}{f_{2}}=\frac{\left(\frac{V}{V-V_{1}}\right)f_{0}}{\left(\frac{V}{V+V_{1}}\right)f_{0}}=\frac{V+V_{1}}{V-V_{1}}$

 $\Rightarrow$      $=\frac{V+V_{1}}{V-V_{1}}=2\Rightarrow2V-2V_{1}=V+V_{1}$

$\Rightarrow$      $V=3V_{1}\Rightarrow \frac{V}{V_{1}}=3$