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 .

When a train approaches a listener , the apparent frequency of the whi

1.	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 .
Answer» Solution :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 , inboth the cases the listener is at rest So ` u_(o) = 0 ` In the firstcase , 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 .