An open organ pipe kept in air at 27^ C resonates at a fundamental frequency f . It is then dipped vertically in water such that 1 4 of its length is submerged. To make the pipe resonate at the same fundamental frequency f , the air inside the pipe is heated. What should be the new temperature of the air?

Step-by-step solution

Let the original length of the open pipe be L and the initial speed of sound at T₁ = 27^ C = 300 K be v₁ . The fundamental frequency of the open pipe is: f = v₁ 2L When 1 4 of the pipe is submerged, it becomes a closed pipe. The length of the remaining air column is: L' = L - L 4 = 3L 4 Let the new speed of sound at the higher temperature T₂ be v₂ . The new fundamental frequency of this closed pipe is: f' = v₂ 4L' = v₂ 4 ( 3L 4 ) = v₂ 3L We are given that the new frequency is equal to the original frequency ( f' =