When a train approaches a listener , the apparent frequency of the whistle is 100 Hz , while the frequency appears to be 50 Hz when the train recedes . Calculate the frequency when the listener is in the train .

Explanation

If the listener is in the train , he will listen to the actual frequency of the whistle (n) .
If V is the velocity of sound , ` u_(s)` is the velocity of the source and ` u_(o) ` is the velocity of the listener ,
the apparent frequency , ` n. (V + u)/(V - u_(s)) xx n `
In the given problem , in both the cases the listener is at rest
So ` u_(o) = 0 `
In the first case , the motion of the train is towards the listener .
so , ` u_(s) ` is positive .
Again in hte second case , the train recedes from the listener .
So , ` u_(s) ` is negative . Therefore , in the two cases we have
` 100 = (V + 0)/(v - u_(s)) xx n `
or ` (V) /(V - u_(s)) xx n = 100` ...(1)
and ` 50 = (V + 0)/(V - (-u_(s))0 xxn `
or ` (V)/(V + u_(s)) xx n = 50 ` ...(2)
From equation (1) ,
` (n) /(100) = (V - u_(s)) /(V) = 1- (u_(s))/(V) or , (u_(s))/(V) = 1 - (n) /(100) =(100- n)/(100)`
From equation (2) , `(n)/(50) = (V + u_(s))/(V) = 1 + (u_(s))/(V) or , (u_(s))/(V) = (n) / (50) - 1 = (n - 50)/(50) `
` therefore (100 - n)/(100) = (n -50)/(50) or , 100 - n = 2n - 100`
3n = 200 or , n = 66(2)/(3) `Hz .