Sound Waves Concept Page - 10

Example

Find higher harmonics and overtones for standing wave in a pipe open at both ends

Example: Air is blown at the mouth of a tube of length 25cm and diameter equal to 2 cm open at both ends. If velocity of sound in air is 330 ms−1${}^{-1}$ the sound emitted will have all the frequencies in the group (neglect end correction)

Solution:
As it is open at both ends, antinodes are formed at both ends
⇒$\Rightarrow$ length l=n(λ2)⇒λ=2ln$l=n(\frac{\lambda }{2})\Rightarrow \lambda ={\displaystyle \frac{2l}{n}}$
Velocity v=fλ$v=f\lambda$
⇒f=vλ=(v2l)n.$\Rightarrow f={\displaystyle \frac{v}{\lambda }}=({\displaystyle \frac{v}{2l}})n.$
=(3302×25×10−2)n$=({\displaystyle \frac{330}{2\times 25\times {10}^{-2}}})n$
=n(660)$=n(660)$ where n = 1,2,3,4..

Example

Use the relation between fundamental frequency and higher overtones in organ pipes open at both ends

Example: A cylindrical metal tube has a length of 50 cm$50cm$ and is open at both ends. The frequencies between  1 kHz$1kHz$  to 2 kHz$2kHz$ at which the air column in the tube resonates are ($($the temperature on that day is 20oC(v0=330 m/s)${20}^{o}C(v_0=330m/s)$

Solution:
For an organ pipe open at both ends
fn=nv2L...................(1)$f_n={\displaystyle \frac{nv}{2L}}...................(1)$
But v$v$ is dependent  on temp.
v=v0(1+t273)−−−−−−−−√=330(1+20273)−−−−−−−−−√=342$v=v_0\sqrt{(1+{\displaystyle \frac{t}{273}})}=330\sqrt{(1+{\displaystyle \frac{20}{273}})}=342$
Substituting v$v$ in (1),$(1),$ we get,
fn=342n2L=342n2×50×10−2$f_n={\displaystyle \frac{342n}{2L}}={\displaystyle \frac{342n}{2\times 50\times {10}^{-2}}}$
The values of frequencies allowed between  1kHz$1kHz$ and and 2 kHz$2kHz$ are for n=3,4,5$n=3,4,5$
The values are 1026,1368,1710 Hz$1026,1368,1710Hz$

Definition

Understand the boundary conditions for a standing wave in a tube closed at one end

The closed end of a pipe acts as a displacement node because the air molecules at the very end cannot displace into the closed end. Thus it is pressure antinode as it has to exert a pressure not to displace air at the closed pipe end.At an open pipe end there must be a pressure node such that pressure and displacement are π/2$\pi /2$ out of phase, so that the open end is also a displacement antinode.

Definition

End Correction

End correction is a short distance applied or added to the actual length of a resonance pipe, in order to calculate the precise resonance frequency of the pipe.
A simple notion is that the fundamental resonance of a pipe occurs when the resonator length is half or a quarter of the sound wavelength. It is however well recognized that the practical frequency comes out lower than this, you have to apply an end correction, the pipe appears to be acoustically somewhat longer than its physical length.
When end correction is incorporated a precise value of resonance frequency of the pipe can be estimated.

Example

Find and use the end correction for a pipe closed at one end

Example: An open pipe 30$30$ cm long and a closed pipe 23$23$ cm long, both of the same diameter, are each sounding its first overtone and these are in unison. Find the end correction of these pipes.

Solution:
2νo=2v2L1$2{\nu }_o={\displaystyle \frac{2v}{2L_1}}$   ( In Ist$I^{st}$ overtone open pipe )
3νc=3v4L2$3{\nu }_c={\displaystyle \frac{3v}{4L_2}}$  ( In Ist$I^{st}$ overtone  closed pipe )
By problem, 
2νo=3νc$2{\nu }_o=3{\nu }_c$
⇒2v2L1=3v4L2$\Rightarrow {\displaystyle \frac{2v}{2L_1}}={\displaystyle \frac{3v}{4L_2}}$
L1L2=43${\displaystyle \frac{L_1}{L_2}}={\displaystyle \frac{4}{3}}$
Here L1=30+0.6×2×r$L_1=30+0.6\times 2\times r$ and L2=23+0.6×r$L_2=23+0.6\times r$
( ∵$\because$ both have same diameter, radius will be same )
⇒30+0.6×2×r23+0.6×r=43⇒r=21.2$\Rightarrow {\displaystyle \frac{30+0.6\times 2\times r}{23+0.6\times r}}={\displaystyle \frac{4}{3}}\Rightarrow r={\displaystyle \frac{2}{1.2}}$
End correction is 0.6r=0.6×21.2=1cm$0.6r=0.6\times {\displaystyle \frac{2}{1.2}}=1cm$.

Example

Problem on relation between frequencies for a pipe closed at both ends

Example: A closed pipe has a certain frequency. Now its length is halved. Considering the end correction, find its new frequency.

Solution:
n1=v4(l+x)$n_1={\displaystyle \frac{v}{4(l+x)}}$,( n1$n_1$ is fundamental frequency of the closed organ pipe)
n2=v4(l2+x)=2v4(l+2x),$n_2={\displaystyle \frac{v}{4({\displaystyle \frac{l}{2}}+x)}}={\displaystyle \frac{2v}{4(l+2x)}},$ where x$x$ is end correction
Clearly n2$n_2$ is less than double of n1$n_1$.

Example

Solve problems on standing waves in a tube

Example: You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length l$l$ and diameter l10${\displaystyle \frac{l}{10}}$. You have stretched a string of mass per unit length μ$\mu$ across the open end of the tube. The other end of the tube is closed. To produce the musical effect you are looking for, you want the frequency of the third harmonic standing wave on the string to be same as fundamental frequency for sound waves in the air column in tube. Velocity of sound in the air column is v$v$.

Solution:
f=v4l=32lTμ−−√$f={\displaystyle \frac{v}{4l}={\displaystyle \frac{3}{2l}\sqrt{{\displaystyle \frac{T}{\mu }}}}}$

Or T=v2μ36$T={\displaystyle \frac{v^2\mu }{36}}$

Example

Problems on flutes

Example: A flute produces a musical sound travelling at a speed of 320 m/s. The frequency of the note is 256 Hz. Find the period of the note?

Solution:
Let, Speed of sound is v=320ms−1$v=320m{s}^{-1}$
The frequency of the note is ν=256Hz$\nu =256Hz$The frequency of musical note is given by ν=1T$\nu ={\displaystyle \frac{1}{T}}$, where T is period of propagation of musical note.
Hence,
T=1ν$T={\displaystyle \frac{1}{\nu }}$
T=1256s$T={\displaystyle \frac{1}{256}}s$

Definition

Experimental demonstration of resonance in air column

The apparatus for the experiment consists of a long cylindrical plastic tube attached to a water reservoir. The length of the water column may be changed by raising or lowering the water level while the tuning fork is held over the open end of the tube. Resonance is indicated by the sudden increase in the intensity of the sound when the column is adjusted to the proper length. The resonance is a standing wave phenomenon in the air column and occurs when the column length is:
λ/4,3λ/4,5λ/4$\lambda /4,3\lambda /4,5\lambda /4$

Result

Descirbe the setup of resonance column method in finding speed of sound in air

Following is the procedure to find speed of sound in air using Resonance tube.

1. First fill the tube nearly full of water and then strike one of the tuning forks with the rubber mallet supplied and hold it above the water column.
 
Precautions to be taken: do not touch the tube with the tuning fork - the rapidly moving fork can break the plastic.

2. By using the moveable water reservoir, lower the water surface slowly, listening for amplification of the tone. When a resonance is found, a pronounced reinforcement of the sound will be heard. Now move the water surface up and down multiple times to locate the point of maximum sound intensity and mark that point using a rubber band on the outside of the tube.

3. Lower the water further to find the next resonant length. One has to Continue in this manner as far as the length of the tube will permit. Obtain the lengths L/4, 3L/4, etc. in meters from your measurements. You will need to check to see if your column lengths follow the progression 1, 3, 5, 7, -- since you may have missed a resonance or counted one of the fainter spurious resonances which sometimes occur. Calculate the wavelength and velocity of sound.

4. Now repeat the procedure for the other tuning forks supplied in the manner described. The velocity in miles per hour may be found by multiplying the velocity in m/sec by the factor 2.24. Please record the room temperature for reference since the velocity of sound increases with increasing air temperature.

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