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KCET · Physics · Waves and Sound

First overtone frequency of a closed pipe of length ' \( l_{1}{ }^{\prime} \) is equal to \( 2^{\text {nd }} \) hormonic frequency of an open pipe of length ' \( l_{2} \) '. The ratio \( \frac{l_{1}}{l_{2}} \) is

  1. A \( \frac{3}{4} \)
  2. B \( \frac{4}{3} \)
  3. C \( \frac{3}{2} \)
  4. D \( \frac{2}{3} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( \frac{3}{4} \)

Step-by-step Solution

Detailed explanation

Given, first overtone frequency of closed pipe of length \( l_{1}=2^{\text {nd }} \) harmonic frequency of on open pipe of length \( l_{2} \)
Now, first overtone frequency of close pipe \( \frac{3 u}{4 l_{1}} \)
\( 2^{\text {nd }} \) harmonic frequency of open pipe \( =\frac{2 u}{2 l_{2}} \)
\( \therefore \frac{3 u}{4 l_{1}}=\frac{2 u}{2 l_{1}} \)
On rearranging, we get
\( \frac{l_{1}}{l_{2}}=\frac{3 u}{4} \times \frac{1}{u}=\frac{3}{4} \)
Therefore, ratio is \( \frac{l_{1}}{l_{2}}=\frac{3}{4} \).