AP EAMCET · PHYSICS · Waves and Sound
A open pipe of length \(l\) is vibrating in \(3 \mathrm{rd}\) overtone with maximum amplitude \(A\). The amplitude at a distance of \(\frac{l}{16}\) from any open end is
- A A
- B 0
- C \(\frac{A}{\sqrt{2}}\)
- D \(\frac{\sqrt{3} A}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{A}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
For open organ pipe, wavelength of \(n^{\text {th }}\) overtone \[ \begin{aligned} & \lambda_n=\frac{2 l}{n+1} \quad(n=3) \\ & \lambda=\frac{2 l}{4}=\frac{l}{2} \end{aligned} \] As pipe is open, so antinode (maximum amplitude) will form at open ends. At \(l=0\), amplitude \(=A\)…
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