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OpenStudy (somy):

Quick question, just want to clear out something

OpenStudy (somy):

is it because for a stationary wave to form you need minimum of 2 waves?

OpenStudy (somy):

@KissMyAxe @ParthKohli @thushananth01

OpenStudy (anonymous):

okay same here use mathematics u know that \[\frac{\lambda}{2} \times 1 (loop) = x(wire length)\] \[\lambda = 2x\] Hence \[v = f \lambda\] \[\frac{v}{2x} = f\]-----1 this is the first equation Now you need two loops so form an equation...why dont u try :)

OpenStudy (anonymous):

clarify your doubts

OpenStudy (anonymous):

@Somy did u try?

OpenStudy (somy):

yup thnx i originally did speed=frequency* wavelength and i just thought that i'll need 2 waves for stationary wave to form :D

OpenStudy (somy):

yes i did)

OpenStudy (somy):

i need 4 loops thats why im getting 2 waves

OpenStudy (anonymous):

oh okay :)

OpenStudy (anonymous):

Did u get the answer

OpenStudy (somy):

yes i did) @KissMyAxe

OpenStudy (anonymous):

Well i did it like this and it worked: 1 loop frequency is equal to f therefore for two loops it will be 2f :) ratio method

OpenStudy (anonymous):

Hey guys :) I just want to know when they say the word 'loop' I don't really know how to imagine the wave to look like, any ideas anyone?

OpenStudy (anonymous):

A loop has two nodes and antinodes between the node

OpenStudy (anonymous):

@KissMyAxe like the first 1st image in this picture? http://ipodphysics.com/resources/21603.gif

OpenStudy (somy):

|dw:1402081689327:dw|

OpenStudy (somy):

lmao

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ummm the yes is directed towards? lol

OpenStudy (anonymous):

To the burger which Somy drew

OpenStudy (somy):

ahahahah i see you are hungry @KissMyAxe

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

|dw:1402081983372:dw| what you see is a wave of wavelenth lambda, and therre are 2 loops

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