# physics - standing wave stuff

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just wanted to make sure i am doing this correctly....

(1)

$\lambda_n=\frac{2L}{n}, \ n=1,2,3,...$

(2)

$v=\sqrt{\frac{F}{\mu}}$

$\mu=\frac{M}{L} \ \ \ \ (mass \ per \ unit \ length)$

$f=\frac{v}{\lambda}=\frac{n\sqrt(\frac{F}{\mu})}{2L}=\frac{n\sqrt(F)}{2L\sqrt(\mu)}$

ok yep thats it (hopefully the latex stuff works, i cannot see it on my computer at the moment, so just tell me if it shows up incorrectly (or not at all) )

Cheers

Sarah

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Just one really tiny correction :

$f_n=\frac{v}{\lambda _n}=\frac{n\sqrt(\frac{F}{\mu})}{2L}=\frac{n\sqrt(F)}{2L\sqrt(\mu)}$

I added a subscript to the frequency to indicate which mode it is for. The rest is all good.

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cool! thanks DQW!

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ive never understood the symbols, i understand physics in plain english
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A small detail

shouldn't the question be a wire between two points at distance L instead of a wire of L length?

Because how can there be an amplitude if the whole wire is used up as n/2 times wave length.

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Perhaps you (kedas) are neglecting the fact that the wire is capable of stretching elastically to accomodate the amplitude ?

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Perhaps you (kedas) are neglecting the fact that the wire is capable of stretching elastically to accomodate the amplitude ?

yes elasticity is needed.

I only want to say that the L in the formula is the distance between A and B.

it's only a small detail like I said.

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