Question

In case of closed organ pipe which harmonic the $P^{th}$ overtone will be

• A. 2p + 1
• B. 2p - 1
• C. p + 1
• D. p - 1

Answer 1

The correct option is A 2p + 1

In the fundamental mode of vibration, the air column vibrates with an antinode A at the open end and a node N at the closed end as shown in figure (a). Since the distance between a node and an antinode is $\frac{\lambda }{4}$, the length of the tube l in this case will be equal to $\frac{\lambda }{4}$.
$l=\frac{\lambda }{4}$
$\lambda =4l$

The velocity of wave is given by:
$v={\nu }_0\lambda$
${\nu }_0=\frac{v}{\lambda }=\frac{v}{4l}$

This is (${\nu }_0$) the frequency of the fundamental note is called the 'first harmonic'. (For zero overtone).

Figure (b) shows the first overtone in a closed pipe. Two nodes and two antinodes are formed. The wavelength and the frequency of the sound corresponding to this mode of vibration will be different from those corresponding to the fundamental mode. the length of the tube l in this case will be equal to $\frac{3\lambda }{4}$.
$l=\frac{3\lambda }{4}$
$\lambda =\frac{4l}{3}$

The velocity of wave is given by:
$v={\nu }_1\lambda$
${\nu }_1=\frac{v}{\lambda }=\frac{3v}{4l}$

This is (${\nu }_1$) the frequency of the 'second harmonic'. (For first overtone).
Further, the frequency ${\nu }_2=\frac{5v}{4l}$ is of the 'third harmonic'. (For second overtone).
Hence, for $p^{th}$ overtone the frequency will be only an odd harmonic(integral multiple) of fundamental frequency.
${\nu }_p=(2p+1)\frac{v}{4l}=(2p+1){\nu }_0$

In case of closed organ pipe the $p^{th}$ overtone will be $(2p+1)$ harmonic.