Question

An organ pipe $P_1$ is closed at one end and vibrating in its first overtone and another pipe $P_2$ opened at both ends vibrating in its third overtone are in resonance with a given tuning fork. Then the ratio of $P_1$ and $P_2$ is :

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{8}{3}$
• D. $\frac{3}{8}$

Answer 1

Answer: (D)

In an organ pipe closed at one end there will be an antinode at the open end and a node at closed end. The distance between an antinode and nearest node is $\frac{\lambda }{4}$. If $i_1$ be the
length of pipe, and $\lambda$ the wavelength then,
$l_1=\frac{\lambda }{4},$ or ${\lambda }_1=4l_1$

If $n$ be the frequency of note emitted and $v$ the velocity of sound in air, then
$n_1=\frac{v}{{\lambda }_1}=\frac{v}{4l_1}$
Also, closed pipe produces only odd harmonics, hence first overtone of closed pipe is
$=3\left(\frac{v}{4l_1}\right)$ ... (i)
For open pipe antinodes are formed at both ends, frequency is given by
$n_2=\frac{v}{2l_2}$
Also open pipe produces both even and odd harmonics,

hence we have freqency of third overtone of open pipe is
$=4\left(\frac{v}{2l_2}\right)$ ... (ii)
At resonance both the frequencies are equal
$\therefore \frac{3v}{4l_1}=\frac{4v}{2l_2}$
$\Rightarrow \frac{l_1}{l_2}=\frac{3}{8}$