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Simple wavelength problem

by Carnivean
Tags: wavelength
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Carnivean
#1
Jul11-11, 04:39 PM
P: 12
Not sure how to figure this one out, or where to start:

The diagram represents a snapshot of a standing transverse wave on a flexible guitar string taken when the displacement is at a maximum. (Note: the vertical scale is very exaggerated, the actual sideways displacement of the string is negligibly TINY compared to the length of the string!) The string is 1.30 m long with tension 12 N. The total mass of the string is 0.010 kg. Find the wavelength of this standing wave in meters (but do not enter units)

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Pi-Bond
#2
Jul11-11, 05:28 PM
P: 304
The diagram is to help figure the mode of the vibrations, since you know that the string is fixed at both ends. Using this you can obtain the frequency of vibration. Using the tension, length and mass you can get the velocity. After this it is a simple matter to obtain the wavelength.
Carnivean
#3
Jul11-11, 05:43 PM
P: 12
Okay so how exactly would I figure out what mode i am in? Is it how many anti-nodes I see, which is 3 i'm guessing, therefore putting it in the 3rd mode? Which would then mean that Frequency would be equal to 3 x v\2L?

I'm still not clear on this at all...

Okay I took a shot at it... I used the square root of tension/(mass/length) to get 39.49 for the velocity. I used the 3rd mode or harmonic and used 3 x V/2L for the frequency and got 45.56. I divided 39.49 by 45.56 to get 0.8667 m for the wavelength. Anywhere close to right?

Pi-Bond
#4
Jul11-11, 06:03 PM
P: 304
Simple wavelength problem

The nth mode of a string fixed on both ends has n antinodes. Your diagram has 6 antinodes (points of maximum displacement) so it is the sixth mode of vibration. You can use the formula for frequency to relate this to the length.
Carnivean
#5
Jul11-11, 06:31 PM
P: 12
So 6 antinodes.. okay so that would be 6 x V/2L to get the frequency? Is that the equation you would use?

Velocity = 39.49
Frequency = 6 x 39.49/2(1.3) which would equal 91.130.

So... 39.49/91.130 = .433333 for the wavelength? I need more specificity in your answer plz..
Pi-Bond
#6
Jul11-11, 06:33 PM
P: 304
Actually I just realized you could just find the wavelength by using nL/2, with n=6 and L the given length. I don't understand why there is so much information (unless I'm thinking wrong)...do you know the answer?
Carnivean
#7
Jul11-11, 06:49 PM
P: 12
I dont know the answer yet, its online homework and the professor is rather old and terrible at teaching...
Pi-Bond
#8
Jul11-11, 06:57 PM
P: 304
I definitely think there is too much information here. In your answer you mixed the positioning of 6 and 2; 6 goes in the denominator and 2 in the numerator. The answer should be 3.90 m. Anyhow, the more simpler method is simply using nL/2, the velocity, tension, density etc. are not required, most probably your professor put them there to confuse you.

I suggest you look over your textbook/notes to verify..
Redbelly98
#9
Jul11-11, 08:19 PM
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P: 12,064
Carnivean, (hopefully) easy question for you: in this figure, how many wavelengths fit along the length of the string?



The answer to that question should lead you to figuring out what the actual wavelength is.

EDIT: Actually, you have already figured out the correct answer:
Quote Quote by Carnivean View Post
So 6 antinodes.. okay so that would be 6 x V/2L to get the frequency? Is that the equation you would use?

Velocity = 39.49
Frequency = 6 x 39.49/2(1.3) which would equal 91.130.

So... 39.49/91.130 = .433333 for the wavelength? I need more specificity in your answer plz..
Looks good

Quote Quote by Pi-Bond View Post
I definitely think there is too much information here. In your answer you mixed the positioning of 6 and 2; 6 goes in the denominator and 2 in the numerator.
No it doesn't, not for calculating the frequency.
The answer should be 3.90 m
No, from the figure in Post #1 the wavelength is clearly shorter than the 1.3 m length of the string.
Pi-Bond
#10
Jul12-11, 06:31 AM
P: 304
Ah ok - I got confused myself, sorry about that!

EDIT: The formula I meant was 2L/n, which gives the correct wavelength (0.433.. m) - as I thought the other data is not really needed.


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