Question

e) i) Show that only odd harmonics can be generated in a closed-end organ pipe.
ii) For gravity waves, the phase velocity is given by
vp = Cl1/ 2
Show that group velocity for these waves is half of their phase velocity.

Answer 1

Answer on Question #61833-Physics – Mechanics | Relativity

i) Show that only odd harmonics can be generated in a closed-end organ pipe.

Solution

Consider a close pipe of length $L$, which encloses certain specific amount of air called as air column. This air column is set into vibrations by holding a vibrating tuning fork near its mouth. These longitudinal waves of frequency $u$ wavelength $\lambda$ travel with a velocity $v$ through the pipe and gets reflected at the other end, because the other end acts like a boundary and this reflected wave travelling in opposite direction superpose with incident wave forming a stationary wave such that a node is formed at closed end and antinodes at the open end, which is characterized by a set of natural frequencies called as normal modes of oscillation.

The stationary waves formed is given by

$$y = 2 a \sin k x \cos \omega t,$$

where $2a \sin kx$ gives its amplitude, therefore the positions of nodes are given by

$$\sin k x = 0$$

And hence $kx = n\pi, n = 0,1,2,3 \ldots$. Since $k = \frac{2\pi}{\lambda}$ we get

$$x = \frac{n\lambda}{2}, n = 0,1,2,3 \ldots$$

In the same way, the position of antinodes are given by

$$\sin k x = 1$$

And hence $kx = \left(n + \frac{1}{2}\right)\pi, n = 0,1,2,3 \ldots$. Since $k = \frac{2\pi}{\lambda}$ we get

$$x = \left(n + \frac{1}{2}\right) \frac{\lambda}{2}, n = 0,1,2,3 \ldots$$

Taking the closed end of pipe to be $x = 0$, the condition for node is satisfied and the other end of pipe to be $x = L$ where antinode is formed, requires that the length $L$ to be related to wavelength $\lambda$ given by

$$L = \left(n + \frac{1}{2}\right) \frac{\lambda}{2}, n = 0,1,2,3 \ldots$$

Thus, the possible wavelengths of stationary wave formed in different modes of oscillation are given by

$$\lambda = \frac{2L}{\left(n + \frac{1}{2}\right)}, n = 0,1,2,3 \ldots$$

The corresponding frequencies are

$$u = \frac{v \left(n + \frac{1}{2}\right)}{2L}, n = 0,1,2,3 \ldots$$

Thus, for the fundamental mode of oscillation ($1^{\text{st}}$ harmonic) $n = 0$, and the corresponding frequency is given by

$$u _ {1} = \frac {v}{4 L}.$$

For the second mode of oscillation ( $3^{\text{rd}}$ harmonic) $n = 1$ , and the corresponding frequency is given by

$$u _ {2} = \frac {3 v}{4 L} = 3 u _ {1}.$$

For the third mode of oscillation ( $5^{\text{th}}$ harmonic) $n = 2$ , and the corresponding frequency is given by

$$u _ {3} = \frac {5 v}{4 L} = 5 u _ {1}.$$

Therefore, $u_{1}:u_{2}:u_{3} = 1:3:5$ .

Thus, a closed pipe produces only odd harmonics i.e., the ration of frequencies of overtone to fundamental frequency are odd natural numbers.

ii) For gravity waves, the phase velocity is given by $\mathrm{vp} = \mathrm{C}\lambda 1 / 2$ . Show that group velocity for these waves is half of their phase velocity

Solution

Starting from the dispersion $\omega (k)$ , the phase velocity is

$$v _ {p} = \frac {\omega}{k} = C \lambda^ {\frac {1}{2}}$$

and the group velocity is

$$v _ {g} = \frac {\partial \omega}{\partial k}.$$ $$\lambda = \frac {2 \pi}{k}.$$

Therefore,

$$v _ {p} = \frac {C \sqrt {2 \pi}}{k ^ {\frac {1}{2}}}.$$

Thus,

$$\omega = k v _ {p} = k \frac {C \sqrt {2 \pi}}{k ^ {\frac {1}{2}}} = C \sqrt {2 \pi k}$$

The group velocity is

$$v _ {g} = \frac {\partial \omega}{\partial k} = \frac {\partial}{\partial k} \left(C \sqrt {2 \pi k}\right) = \frac {1}{2} C \sqrt {\frac {2 \pi}{k}}.$$

We can write it as

$$v _ {g} = \frac {1}{2} C \lambda^ {\frac {1}{2}} = \frac {1}{2} v _ {p}.$$

We can see that group velocity for these waves is half of their phase velocity.

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