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}=4 l_{1}$

If $n$ be the frequency of note emitted and $v$ the velocity of sound in air, then
$n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{4 l_{1}}$
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
$n_{2}=\frac{v}{2 l_{2}}$
Also open pipe produces both even and odd harmonics,

hence we have freqency of third overtone of open pipe is
$=4\left(\frac{v}{2 l_{2}}\right)$ ... (ii)
At resonance both the frequencies are equal
$ \therefore \frac{3 v}{4 l_{1}} =\frac{4 v}{2 l_{2}} $
$\Rightarrow \frac{l_{1}}{l_{2}} =\frac{3}{8}$