The third overtone of an open organ pipe of length `l_(0)` has the same frequency as the third overtone of a closed pipe of length `l_(c)`. The ratio `l_(0)//l_(c)` is equal to

Options

A. 2
B. `(3)/(2)`
C. `(5)/(3)`
D. `(8)/(7)`

Answer

D

Explanation

Let `upsilon_(o)andupsilon_(c)` be the fundamental frequencies of the open and closed pipes respectively.
`therefore upsilon_(o)=(v)/(2L_(o))andupsilon_(c)=(v)/(4L_(c))`
Frequency of third overtone of the closed pipe = `7upsilon_(c)`.
Frequencies of third overtone of the open pipe = `4upsilon_(o)`.
According to question
`4upsilon_(o)=7upsilon_(c) therefore (4v)/(2L_(o))=(7v)/(4L_(c))or(L_(o))/(L_(c))=(16)/(14)=(8)/(7)`