Physics 11

Waves travelling along a string have a wavelength of 2.4m. When the waves reach the fixed end of the string, they are reflected to produce a standing wave pattern. How far from the end are the first 2 antinodes?

I know that each wavelength has 2 antinodes so each antinode has a length of 1.2m. I feel like I need either the amount of antinodes total or the total length of the string to get that value...

Thanks a bunch :)

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  1. the first antinode is 1/4 wavelength from the fixed end.
    the second one is half a wavelength further
    so at L/4 and 3L/4
    = .6 and 1.8

    the fixed end is a node x = 0
    the ANTInode is 1/4 L away, x = L/4
    the next node is at L/2 , x = L/2
    next ANTInode is at 3L/4 x=(3/4)L
    the next node is at x = L
    DRAW IT

    math
    y = sin2pi (x/L - t/T) + sin2pi(x/L+t/T)

    that is a wave going right plus a wave going left and not that at x = 0 the sum is always zero, that is a NODE
    now do the trig
    sina+sin b = 2sin.5(a+b)cos.5(a-b)
    so here
    y= 2 sin 2pi x/L cos 2pi t/T
    now look at 2sin 2pi x/L
    when x = 0 we have 2*0 = 0
    when x = L/4 we have 2sinpi/2 = 2
    our first antinode
    when x = L/2 we have 2 sinpi = 0
    wen x = 3L/4 we have 2 sin3pi/2 =-2
    our second antinode
    when x = L we have 2 sin 2pi = 0

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  2. Oh I get it!! Thank you very much :)

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  3. LOL i read this maybe... 6 times, then like you say - I DREW IT - and then it made sense < 3

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