14. waves and sound — Physics STD 11 Science — Question

$\lambda=4 l$

The velocity of wave is given by$:$

$v=\nu_{0} \lambda$

$\nu_{0}=\frac{v}{\lambda}=\frac{v}{4 l}$

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 lin this case will be equal to $\frac{3 \lambda}{4}$.

$l=\frac{3 \lambda}{4}$

$\lambda=\frac{4 l}{3}$

The velocity of wave is given by$:$

$v=\nu_{1} \lambda$

$\nu_{1}=\frac{v}{\lambda}=\frac{3 v}{4 l}$

This is $\left(\nu_{1}\right)$ the frequency of the'second harmonic: (For first overtone).

Further, the frequency $\nu_{2}=\frac{5 v}{4 l}$ is of the third harmonic. (For second overtone).

Hence, for $p^{t h}$ overtone the frequency will be only an odd harmonic(integral multiple) of fundamental frequency.

$\nu_{p}=(2 p+1) \frac{v}{4 l}=(2 p+1) \nu_{0}$

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