The fundamental frequency of a closed pipe is $400 \mathrm{~Hz}$. If $1 / 3^{\text {rd }}$ pipe is filled…
The fundamental frequency of a closed pipe is $400 \mathrm{~Hz}$. If $1 / 3^{\text {rd }}$ pipe is filled with
water, the frequency of $2^{\text {nd }}$ harmonic of the pipe will be (Neglect end correction)
$600 \mathrm{~Hz}$
$1800 \mathrm{~Hz}$
$1200 \mathrm{~Hz}$
$300 \mathrm{~Hz}$
Solution
Fundamental frequency of closed pipe,
$\begin{array}{l}
n=\frac{V}{4 L}=400 \\
V=400 \times 4 L
\end{array}$
If $\frac{1}{3}$ rd of pipe is filled with water, then remaining length of air column is $L-\frac{L}{3}=\frac{2 L}{3}$. Now, fundamental frequency $=\frac{v}{4\left(\frac{2 L}{3}\right)}=\frac{3 v}{8 L}$
The second harmonic of the pipe $=3 \times$ fundamental frequency
$\begin{array}{l}
=3 \times\left(\frac{3 \mathrm{v}}{8 \mathrm{~L}}\right) \\
=3 \times \frac{3 \times 400 \times 4 \mathrm{~L}}{8 \mathrm{~L}} \\
=1800 \mathrm{~Hz}
\end{array}$