
4λ=L12(2λ)=L2
As,v=fλ⇒f2=2L22v
⇒v=f1(4L1)⇒f2=L2v
⇒f1=4L1v
As per question, f1=f2
4L1v=L2v
⇒L2=4L1
⇒60=4×L1
⇒L1=15cm
JEE Main 2024 — Physics Waves & Oscillations
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 60cm, the length of the closed pipe will be :
Held on 31 Jan 2024 · Verified 6 Jul 2026.
60cm
45cm
30cm
15cm
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