The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm , the length of the closed pipe will be :

Step-by-step solution

λ 4 = L 1 2 λ 2 = L 2 As , v = f λ ⇒ f 2 = 2 v 2 L 2 ⇒ v = f 1 4 L 1 ⇒ f 2 = v L 2 ⇒ f 1 = v 4 L 1 As per question, f 1 = f 2 v 4 L 1 = v L 2 ⇒ L 2 = 4 L 1 ⇒ 60 = 4 × L 1 ⇒ L 1 = 15 cm