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.