Sound Waves Concept Page - 9

Example

Understand and use the relation between fundamental frequency and higher overtones in organ pipes

Example: Speed of sound in air is 320 m/s. A pipe closed at one ends has a length of 1 m and there is another pipe open at both ends having a length of 1.6 m. Neglecting end corrections, Find the frequency of sound for which both the air columns in the pipes can resonate.
Solution:

The closed organ pipe can vibrate at frequency of f=vλ$f={\displaystyle \frac{v}{\lambda }}$λ14=1${\displaystyle \frac{{\lambda }_1}{4}}=1$ ⇒f1=3204=80 Hz$\Rightarrow f_1={\displaystyle \frac{320}{4}}=80Hz$
3λ24=1${\displaystyle \frac{3{\lambda }_2}{4}}=1$ ⇒f2=3×3204=240 Hz$\Rightarrow f_2={\displaystyle \frac{3\times 320}{4}}=240Hz$
5λ34=1${\displaystyle \frac{5{\lambda }_3}{4}}=1$ ⇒f3=5×3204=400 Hz$\Rightarrow f_3={\displaystyle \frac{5\times 320}{4}}=400Hz$
The open organ pipe can vibrate at frequency of f=vλ$f={\displaystyle \frac{v}{\lambda }}$
λ12=1.6${\displaystyle \frac{{\lambda }_1}{2}}=1.6$ ⇒f1=3203.2=100 Hz$\Rightarrow f_1={\displaystyle \frac{320}{3.2}}=100Hz$
λ2=1.6${\lambda }_2=1.6$ ⇒f2=3201.6=200 Hz$\Rightarrow f_2={\displaystyle \frac{320}{1.6}}=200Hz$
3λ32=1.6${\displaystyle \frac{3{\lambda }_3}{2}}=1.6$ ⇒f3=3×3201.6×2=300 Hz$\Rightarrow f_3={\displaystyle \frac{3\times 320}{1.6\times 2}}=300Hz$
2λ2=1.6$2{\lambda }_2=1.6$ ⇒f4=3200.8=400 Hz$\Rightarrow f_4={\displaystyle \frac{320}{0.8}}=400Hz$
Hence both pipes can vibrate at 400 Hz$400Hz$ frequency

Diagram

Draw diagrams of standing waves in organ pipes for pressure and displacement for different configurations

Result

Understand the boundary conditions for a standing wave in a tube open at both ends

Open pipe (flute). In the above diagram, the red curve has only half a cycle of a sine wave. So the longest sine wave that fits into the open pipe is twice as long as the pipe. A flute is about 0.6 m long, so it can produce a wavelength that is about twice as long, which is about 2L = 1.2 m. The longest wave is its lowest note, so let's calculate. Sound travels at about c = 340 m/s. This gives a frequency (speed divided by wavelength) of c/2L = 280 Hz. Given the crude approximations we are making, this is close to the frequency of middle C, the lowest note on a flute.

We can also fit in waves that equal the length of the flute (half the fundamental wavelength so twice the frequency of the fundamental), 2/3 the length of the flute (one third the fundamental wavelength so three times the frequency of the fundamental), 1/2 the length of the flute (one quarter the wavelength so four times the frequency of the fundamental)

Example

Problem on possible wavelengths on an open organ pipe in resonance

An open pipe is in resonance in its fundamental mode. Air, hydrogen, ethane are filled in succession in this pipe. The velocity of sound is different in these three media on account of which, we know, V = f λ$\lambda$, where V = speed of sound; f = frequency and λ$\lambda$ = wavelength
The resonance is a standing wave phenomenon in the air column and occurs when the column length is λ$\lambda$/2, λ$\lambda$, 3λ$\lambda$/2 ... ( in case of open pipe)
So in resonance condition amplitude and wavelength cannot change.
Only frequency changes for different kind of gases.

Example

Find the nodes of a given sinusoidal standing wave in a pipe open at both ends

Example: A standing wave in second overtone is maintained in a open organ pipe of length l$l$. The distance between consecutive displacement node and pressure node is l/x$l/x$. Find x$x$.

Solution:
Frequency of second overtone (i.e n=3)$(i.en=3)$ in an open organ pipe,   ν3=3v2l${\nu }_3={\displaystyle \frac{3v}{2l}}$
Now  as        v=ν3λ$v={\nu }_3\lambda$
ν3=3ν3λ2l⟹λ=2l3${\nu }_3={\displaystyle \frac{3{\nu }_3\lambda }{2l}}⟹\lambda ={\displaystyle \frac{2l}{3}}$
The distance between the consecutive displacement node and pressure node is
equal to λ4${\displaystyle \frac{\lambda }{4}}$  where λ$\lambda$ is the wavelength
of the wave.Thus lx=λ4${\displaystyle \frac{l}{x}}={\displaystyle \frac{\lambda }{4}}$
lx=2l34⟹x=6${\displaystyle \frac{l}{x}}={\displaystyle \frac{{\displaystyle \frac{2l}{3}}}{4}}⟹x=6$

Example

Find the antinodes of a given sinusoidal standing wave in a pipe open at both ends

Example: In open pipes, find the positions of antinodes are obtained.

Solution:
In open pipes, position of  antlnodes   is  nλ2${\displaystyle \frac{n\lambda }{2}}$
Putting vales of n as 0, 1, 2 we get position as 0, λ/2$\lambda /2$, λ$\lambda$

Definition

First harmonic in a pipe open on both sides

The oscillation with lowest frequency mode with lowest frequency is called the fundamental mode or the first harmonic. The fundamental (first harmonic) for an open end pipe needs to be an antinode at both ends, since the air can move at both ends.
Two open pipes contain different gases. If they vibrate with fundamental frequencies, the ratio of velocities of sound in the gases is 2:1, then the ratio of lengths of tube is given=v1v2=21$={\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{2}{1}}$
Fundamental frequency in open νo=v2L${\nu }_o={\displaystyle \frac{v}{2L}}$
So, νo1=v12L1${\nu }_{o_1}={\displaystyle \frac{v_1}{2L_1}}$
νo2=v22L2${\nu }_{o_2}={\displaystyle \frac{v_2}{2L_2}}$
By problem
νo1=νo2${\nu }_{o_1}={\nu }_{o_2}$
or, v12L1=v22L2${\displaystyle \frac{v_1}{2L_1}}={\displaystyle \frac{v_2}{2L_2}}$
v1v2=L1L2${\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{L_1}{L_2}}$
L1L2=21(∵v1v2=21)

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