How do I apply basic physics principles to solve problems in sound?

[GabrielStigmatic]

Hey guys, my name is Gabriel. I no experience whatsoever with physics except this class I've enrolled in called the Basic Physics of Sound. It's suppose to be completely algebra based, I'm just horrible at math in general. I need some help with a couple of problems. Also, if you could tell me the formula you use and how you recognize to use that formula I would appreciate it. Also, here are the formulas that my physics teacher gave us, I'm not sure if he gave us all of the formulas that are necessary to do the work or not.

http://i6.photobucket.com/albums/y231/BrielStigmatic/equations.jpg"

http://i6.photobucket.com/albums/y231/BrielStigmatic/equations2.jpg"
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P.1 A guitar string 60 cm long has a mass of 3.2 x 10-3 kg and vibrates at the fundamental frequency of 390 Hz. The wavelength λ = ______ m and the string tension T = _______ N.

Work for P.1

I thought this was the wavelength formula 2L=2(.60m)=1.2m which when I input that it says that part is right. and then (1,2m)(390)= 468, now... my teacher gave us the same problem on our worksheet but the numbers are different on the online homework. The original number that changed was 300 Hz... On that particular problem he got 691 Newtons for the second part of the answer. and I can't figure out how the hell he got that. That's what I need help with on this one.

Now I'm not really sure how to find the tension. I'm aware that a Newton= 1kgm/s^2.
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P.2 A mass of 2.0 kg is raised to a height of 5.0 m above ground. The work required = ______ J. The mass is then dropped. The velocity of the mass as it hits the ground = ____ m/s and it takes ________ s.

Work for P.2
On this particular problem he told us to use the formula 1.2m to the square root of bot^2=mgh
the square root of bot^2=2mgh/m
the square root of bot^2=2gh
the sqaure root of bot=the square root of 2gh
now I got 9.8 J for the first part, 9.9m/s for the second part and 1.01 sec for the 3rd part it keeps saying it's wrong and I can't remember what I did to get those answers, I keep trying it over and over and it's just not working.
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P.3 The frequency limits of human hearing are about 20 Hz to 20 kHz. The periods of these oscillations are _______ ms and __________ µs, respectively. The corresponding wave lengths of the sound waves in air at 20°C are ________ m and ______ m, respectively.

Work for P.3

I know on this one that the low end is 20 Hz and the high end is 20 kHz, that's common sense. I got the second part which is 50 µs and the the fourth part which is 0.17 m. For some reason whatever I plug in doesn't work. I know that the speed of sound is 343.3 m/s in the air at 20 degrees C, so what do I need to do to find the other two?
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Thanks again guys, any help before 11 oclock would be greatly appreciated. Sorry I didn't put the work before I didn't get to read all of the sticky threads.

 

[GabrielStigmatic]

Hey Guys, I edited the work in there so that you could possibly help. I had forgotten to read the sticky threads about posting your work. Sorry about that.

 

[Astronuc]

Try this for vibrations of strings - http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html

 

[GabrielStigmatic]

So for this one, A guitar string 60 cm long vibrates at the fundamental frequency of 218 Hz. The frequency of the third harmonic = 654 Hz and the wavelength of these oscillations = ___________ m you convert it to meters? I keep getting the wrong answer. I'm confused.

 

[GabrielStigmatic]

Oh wait, you multiply length times 2 and then convert it to meters. And that should be the wavelength of oscillations. I was just getting confused with wavelength and oscillations I think.

 

[GabrielStigmatic]

Please guys I need help with at least this, so on the problem P.1 A guitar string 60 cm long has a mass of 3.2 x 10-3 kg and vibrates at the fundamental frequency of 390 Hz. The wavelength λ = ______ m and the string tension T = _______ N. Ok on the website that Astronauc gave me it shows the formula v= the square root of T/(m/L). So I squared both sides to get rid of the square root. now it's v^2=T/(m/L). I plugged in the given and this is what I have so far v^2=T/(.0032)(.60) which goes to
v^2= T/.0053 So I multiplied both sides by .0053 and got (.0053)v^2=T Now... I'm stuck, is there anything I can do from here?

 

[GabrielStigmatic]

Nevermind, to find velocity you take frequency times wavelength... Thanks for the help (rolls eyes).