Q. $n_{1}\ $Is the frequency of the pipe closed at one end and $n_{2}$ is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed
A) $\frac{n_{1}n_{2}}{n_{2} + 2n_{1}}$
B) $\frac{n_{1}n_{2}}{2n_{2} + n_{1}}$
C) $\frac{n_{1} + 2n_{2}}{n_{2}n_{1}}$
D) $\frac{2n_{1} + n_{2}}{n_{2}n_{1}}$
Explanation
(a) Frequency of closed pipe
$n_{1} = \frac{v}{4\iota_{1}} \Longrightarrow \iota_{1} = \frac{v}{4n_{1}}$
Frequency of open pipe,
$n_{2} = \frac{v}{2\iota_{1}} \Longrightarrow \iota_{2} = \frac{v}{2n_{2}}$
When both pipes are joined then length of closed pipe
$\iota = \iota_{1} + \iota_{2}$
$\frac{v}{4n} = \frac{v}{4n_{1}} + \frac{v}{2n_{2}}$
Or
$\frac{1}{2n} = \frac{1}{2n_{1}} + \frac{1}{2n_{2}}$
Or
$\frac{1}{2n} = \frac{n_{2} + 2n_{1}}{2n_{1}n_{2}}$
Or
$n = \frac{n_{1}n_{2}}{n_{2} + 2n_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.