Stationary Waves in Air Column

Stationary waves in air column

Open pipe: If both ends of a pie are open and a system of air is directed against an edge, standing longitudinal wave can be set up in the tube. The open ends are a dilacement antinodes and pressure nodes

a) For fundamental mode of vibrations,

$L=\frac{{\lambda }_1}{2}\therefore {\lambda }_1=2L$

$v={\lambda }_1{f}_1\therefore v=2L{f}_{1}or{f}_1=\frac{v}{2L}$ …(xxii)

b) For the second harmonic or first overtone,

$L={\lambda }_{2}or{\lambda }_2=L$

$v={\lambda }_2{f}_2\therefore v=L{f}_{2}or{f}_2=\frac{v}{L}=\frac{2v}{2L}=2f$ …(xxiii)

c) For the third harmonic or second overtone,

$L=3\times \frac{{\lambda }_3}{2}\therefore {\lambda }_3=\frac{2}{3}L$

$v={\lambda }_3{f}_3\therefore v=\frac{2}{3}L{f}_{3}or{f}_2=\frac{2v}{2L}$ …(xxiv)

From (xxii), (xxiii) and (xxiv) we get, f₁ : f₂ : f₃ : ……………. = 1 : 2 : 3 : ………….

i.e. for a cylindrical tube, open at both ends, the harmonics excitable in the tude are all integral multiples of its fundamental.

$\therefore$ In the general case, $\lambda =\frac{2L}{n}wheren=1,2........$

Frequency = $\frac{v}{\lambda }=\frac{nv}{2L}wheren=1,2........$

Closed pipe: If one end of a pipe is closed the reflected wave is 180⁰ out of phase with the incoming wave. Thus the dilacement of the small volumes elements at the closed end must always be zero. Hence the closed end must be a dilacement node and pressure antinode

a) This represents the fundamental mode of vibration,

$L=\frac{{\lambda }_1}{4}\therefore {\lambda }_1=4L$,

If f₁ is the fundamental frequency, then the velocity of sound waves is given as,

$v={\lambda }_1{f}_1\therefore v=4L{f}_{1}or{f}_1=\frac{v}{4L}$ ….(xxv)

b) This is the third harmonic or first overtone.

$L=3\times \frac{{\lambda }_2}{4}\therefore {\lambda }_2=\frac{4}{3}L$

$v={\lambda }_2{f}_2\therefore v=\frac{4}{3}L{f}_{2}or{f}_2=2\frac{v}{4L}=3f$ ….(xxvi)

c) This is the fifth harmonic or seconds overtone.

$L=5\times \frac{{\lambda }_2}{4}\therefore {\lambda }_3=\frac{4}{5}L$

$v={\lambda }_3{f}_3\therefore v=\frac{4}{5}L{f}_{3}or{f}_3=\frac{5v}{4L}=5f$ ….(xxvii)

From (xxv), (xxvi) and (xxvii) we get, f₁ : f₂ : f₃ : …………. = 1 : 3 : 5 : ………..

In general, $\lambda =\frac{4l}{\left(2n+1\right)}$ where n = 0, 1, 2 ……..

Velocity of sound = v

Frequency = $\frac{\left(2n+1\right)v}{4L}$ where n = 0, 1, 2 ………

Illustration 1: A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be dilacement node. It is found that when the frequency of the diaphragm is 2000 Hz, a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, the stationary wave pattern disappears but another stationary wave pattern reappears at a frequency of 1600 Hz. Calculate

(i) the speed of sound in air,

(ii) the distance between adjacent nodes at a frequency of 1600 Hz,

(iii) the next lower frequencies at which stationary wave patterns will be obtained.

Solution: Since the node-to-node distance is $\lambda$/2

$\lambda$/2 = 0.08 or = 0.16 m

(i) c = n $\lambda$

$\therefore$ c = 2000 $\times$ 0.16 = 320 ms^–1

(ii) 320 = 1600 $\times$ $\lambda$ or $\lambda$ = 0.2 m

$\therefore$ distance between nodes = 0.2/2 = 0.1 m = 10 m

(iii) Since there are nodes at the ends, the distance between the closed end and the membrane must be exact integrals of $\lambda$/2.

$\therefore$ 0.4 = 2$\lambda$/2 = n^' x 0.2/2 $\Rightarrow$ $\frac{n}{n^{{}^{\prime }}}=\frac{5}{4}$

. When n = 5, n^' = 4

I = n $\times$ 0.16/2 = 0.4 m = 40cm.

(iv) For the next lower frequency n = 3, 2, 1

$\therefore$ 0.4 = 3$\lambda$/2 or $\lambda$= 0.8/3

Since c = n$\lambda$, $n=\frac{320}{0.8/3}=1200$ Hz

Again 0.4 = 1. $\lambda$/2 or $\lambda$ = 0.4 m

$\therefore$ n = 320 / 0.4 = 800 Hz

Again 0.4 = 1. $\lambda$/2 or $\lambda$ = 0.8 m

$\therefore$ n = 320 / 0.8 = 400 Hz.