Asked by Kayla
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 :)
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 :)
Answers
Answered by
Damon
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
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
Answered by
Kayla
Oh I get it!! Thank you very much :)
Answered by
Jonny
LOL i read this maybe... 6 times, then like you say - I DREW IT - and then it made sense < 3
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