physics - standing wave stuff

[Sarahisme]

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

 

ok my answers are:

 

(1)

[math]

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

[/math]

 

(2)

 

[math]

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

[/math]

 

[math]

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

[/math]

 

[math]

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

[/math]

 

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

[DQW]

Just one really tiny correction :

[math]

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

[/math]

 

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

[Sarahisme]

cool! thanks DQW!

[reyam200]

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

[Kedas]

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.

[DQW]

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

[Kedas]

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.