How Are Sound Waves and Frequencies Affected in Musical Instruments and Rooms?

[physics_06er]

Hi could someone please tell me if I have answered these correctly?

1)-Explain how traveling waves can be formed by traveling waves, and hence how the freq's. of a standing waves in an open tube are related to the speed at which sound travels along the tube?

travelling waves formed when to waves traveling in opposite directions with same freq. and amplitude cancel each other out. frequencies of standing waves in open tubes are related to speed by their wavelength. As one whole wavelength is 4 times the length of the tube-we must change the length of tube for different speeds to occur...is something like ok?

2)-in string instruments, unlike most wind instruments, resonances of the body of the instrument play no part in determining pitch. what role do these resonances play in a string instrument, and how is it desirable for them in be distrubuted in freq.?

they amplify sound right?...hmm what else

3)sound level produced by flut is 53dB
a)what pressure amp. is this...can i justuse formula Lp=20log(P/Po) where Lp=53 and Po=2x10^-5

Last one-this is from textbook solution but am unsure why thye do a certain step..

3)Calculate reverberation time for a room 20m x 15m and 8m high..I get all the steps except they when they calculate the area of wall they go..
A=(2x15x8+2x20x8)...why?

Thanks very much
physics_06er

 

[OlderDan]

Hi could someone please tell me if I have answered these correctly?

1)-Explain how <travelling> standing waves can be formed by traveling waves, and hence how the freq's. of a standing waves in an open tube are related to the speed at which sound travels along the tube?

<travelling> standing waves formed when to waves traveling in opposite directions with same freq. and amplitude cancel each other out. frequencies of standing waves in open tubes are related to speed by their wavelength. As one whole wavelength is 4 times the length of the tube-we must change the length of tube for different speeds to occur...is something like ok?

2)-in string instruments, unlike most wind instruments, resonances of the body of the instrument play no part in determining pitch. what role do these resonances play in a string instrument, and how is it desirable for them in be distrubuted in freq.?

they amplify sound right?...hmm what else

3)sound level produced by flut is 53dB
a)what pressure amp. is this...can i justuse formula Lp=20log(P/Po) where Lp=53 and Po=2x10^-5

Last one-this is from textbook solution but am unsure why thye do a certain step..

3)Calculate reverberation time for a room 20m x 15m and 8m high..I get all the steps except they when they calculate the area of wall they go..
A=(2x15x8+2x20x8)...why?

Thanks very much
physics_06er

For #1 you must mean standing waves. You realize that interference is involved, but you make it sound like the two tracelling waves anihilate one another. They do not. A standing wave results because at any given position, the phase relationship between the two waves is constant. At some points the two waves are always exactly out of phase, while at other points the two waves are always exactly in phase. This creates the nodes and anti-nodes. At points in between there is a gradual change in the phase and a gradual change in the amplitude of the motion or sound pressure at a fixed point.

The length and diameter of a tube have no significant effect on the speed of sound. Speed is the one constant in the situation. Changing the length of the tube changes the resonant frequency of the tube because the open end of the tube is very close to an antinode while the closed end is a node. The tube length controls the wavelength of the resonant wave, and the wavelength determines the frequency.

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/clocol.html#c1

For #2, I suggest you look here

http://hyperphysics.phy-astr.gsu.edu/hbase/music/viores.html#c3

This is one example of the resonance of a stringed instrument. It should help you elaborate on your conclusion.

#3 looks good if you have the right number for the reference pressure

#4 There are 4 walls of the same height (8), and two widths (15 and 20), typical of a rectangular room. A is the total surface area of the 4 walls. There are only two different size walls, so they multiply the area of each size by 2 and add.

 

[physics_06er]

and if a room had a circular floor area eg: 1285m^2
(approx radius 20m) and ceiling height 15m??--would would the calculation there be??

Thanks in advance
physics_06er

 

[OlderDan]

and if a room had a circular floor area eg: 1285m^2
(approx radius 20m) and ceiling height 15m??--would would the calculation there be??

Thanks in advance
physics_06er

If you are looking for the surface area of the walls, think about unwrapping the cylinder wall and laying it flat. What would it look like? What would the dimensions be?

 

[physics_06er]

Hi

can u please see if i have the calculations right-

room has circular floor area of 1285m^2 (radius approx. 20m) height=15m the floor is wooden, walls/ceiling mainly plastered and but quarter of wall covered by hearvy drapery
i have ti estimate reverberation time at 2000Hz
absorption coefficients at 2000Hz
plaster on lath=0.04
wood floor=0.06
draper, heavyweight 0.70

First I have to calculate absorption of each surface using A=Sa (S=area a=coeffecient)-then-T60=0.161*V/A

so calculating volume of room first=1285x15=19275m^3 (areaxheight right?)
then A(Floor)=1285x0.06 (as alrady given floor area)
A(ceiling)=1285x0.04 (ceiling also circular-hence area given)
A(walls)=hmm this is the calculation I am unsrue about-after I work out this part all I do is add up A's and use the reverberation forumla aye?

but can you please explain the A(wall) does it have something to do with 'pi'

 

[OlderDan]

Hi

can u please see if i have the calculations right-

room has circular floor area of 1285m^2 (radius approx. 20m) height=15m the floor is wooden, walls/ceiling mainly plastered and but quarter of wall covered by hearvy drapery
i have ti estimate reverberation time at 2000Hz
absorption coefficients at 2000Hz
plaster on lath=0.04
wood floor=0.06
draper, heavyweight 0.70

First I have to calculate absorption of each surface using A=Sa (S=area a=coeffecient)-then-T60=0.161*V/A

so calculating volume of room first=1285x15=19275m^3 (areaxheight right?)
then A(Floor)=1285x0.06 (as alrady given floor area)
A(ceiling)=1285x0.04 (ceiling also circular-hence area given)
A(walls)=hmm this is the calculation I am unsrue about-after I work out this part all I do is add up A's and use the reverberation forumla aye?

but can you please explain the A(wall) does it have something to do with 'pi'

The area of a circle is S=pi*r^2 (using S for area as you have done). Using the given area of 1285m^2 you can calculate the radius to find r = 20.22m, so the given information of the radius being about 20m is consistent. The circumference of a circle is c=pi*d or c=2*pi*r. The area of the walls is the circumference times the height of the wall, S=c*h=2*pi*r*h . If it is not clear to you why the area is the circumference times the height, imagine papering the wall with strips running horizontally around the room. Each strip would have to go all the way around the room, or one circumference. If you had a strip 15m wide and as long as the circumference of the room, it would perfectly cover the wall. If you laid that strip out flat, it would be a rectangle with dimension c by h, having area S=c*h