Wavelength Problem Homework: 2L or 4L for 0.5m String?

[PhizKid]

Homework Statement

A uniform string of length 0.5 m. is fixed at one end and free at the other end. Find the wavelength of the fundamental mode of vibration.

Homework Equations

λ = (2L) / mode

The Attempt at a Solution

λ = (2(0.5 m.)) / 1
λ = 1 m.

But the solution says it's 2 m. because λ = 4*L. Why is it 4L when the formula is 2L?

 

[Delphi51]

The wave is like figure 4 in this link: http://www.studyphysics.ca/newnotes/20/unit03_mechanicalwaves/chp141516_waves/lesson51.htm

The string is fixed at the left end, free to vibrate at the other end. The fundamental frequency shown has 1/4 wavelength on the Length, so λ/4 = L. You almost have to draw the diagram to answer questions like this - too much to remember otherwise.

 

[PhizKid]

The wave is like figure 4 in this link: http://www.studyphysics.ca/newnotes/20/unit03_mechanicalwaves/chp141516_waves/lesson51.htm

The string is fixed at the left end, free to vibrate at the other end. The fundamental frequency shown has 1/4 wavelength on the Length, so λ/4 = L. You almost have to draw the diagram to answer questions like this - too much to remember otherwise.

I don't understand why the fundamental is 1/4 wavelength in my problem. Maybe it might help if I understood where the formula λ = (2L) / mode came from, because it's the only one involving wavelength and length that I know.

 

[Delphi51]

I have not seen λ = (2L) / mode but it suggests a string fixed at both ends:
Then mode = 1 would be one wavelength spread over two lengths or half a wavelength on the string length L. At mode = 2, a complete wavelength on the length L. If you draw pictures of these, you'll see that both ends of the string are fixed in each case. See http://www.phys.unsw.edu.au/jw/strings.html.

 

[Nickie]

I would like to clarify that the formula for calculating the wavelength of a fundamental mode of vibration in a uniform string is indeed λ = (2L)/mode. This means that for a string of length 0.5 m, the wavelength should be 0.5 m for a mode of 1.

However, the solution stating a wavelength of 2 m is also correct, but it is for a different mode of vibration. In this case, the mode is equal to 2, which means that the formula becomes λ = (2L)/2 = L. Therefore, for a mode of 2, the wavelength would be equal to the length of the string, which is 0.5 m.

In conclusion, both answers are correct depending on the mode of vibration being considered. It is important to pay attention to the given information and the formula being used in order to arrive at the correct solution.