Question

\(n_{1}\) is the fundamental frequency of the pipe closed at one end and \(n_{2}\) is the fundamental frequency of the pipe open at both ends. If both are joined end to end, the fundamental frequency of closed pipe so formed is

• A. \(\frac{n_{1}n_{2}}{n_{2}+2n_{1}}\)
• B. \(\frac{n_{1}n_{2}}{2n_{2}+n_{1}}\)
• C. \(\frac{n_{1}+2n_{2}}{n_{1}n_{2}}\)
• D. \(\frac{2n_{1}+n_{2}}{n_{1}n_{2}}\)

Hint:

Write each fundamental in terms of its length; the joined pipe is closed with length equal to the sum.

Answer 1

The Correct Option is A

Step 1: Reciprocal Form
Fundamental frequency is inversely proportional to the effective length, so $\dfrac{1}{n}=\dfrac{4(L_{1}+L_{2})}{v}$.

Step 2: Substitute
$\dfrac{4L_{1}}{v}=\dfrac{1}{n_{1}}$ and $\dfrac{4L_{2}}{v}=\dfrac{2}{n_{2}}$, so $\dfrac{1}{n}=\dfrac{1}{n_{1}}+\dfrac{2}{n_{2}}$.

Step 3: Result
$n=\dfrac{n_{1}n_{2}}{n_{2}+2n_{1}}$, option (A).

Final Answer:
Adding the reciprocal frequencies gives the same result. \[ \boxed{\text{(A) }\dfrac{n_{1}n_{2}}{n_{2}+2n_{1}}} \]