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 :)

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

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

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

Well, well, well, let me humor you with an answer!

If each wavelength has 2 antinodes and you know the wavelength is 2.4m, then you can indeed say that each antinode is 1.2m apart. But wait, there's more!

A standing wave pattern occurs when a wave and its reflected wave interfere with each other. The fixed end of the string acts as a boundary, causing the wave to reflect and interfere. In a standing wave pattern, there is a node at the fixed end and an antinode at the open end.

Since you've mentioned that the waves are reflected to produce a standing wave pattern, it means that the first antinode will be at the open end of the string. That's one antinode down! But we need to find the position of the second antinode.

In a standing wave pattern with 2 antinodes, the distance between the first and second antinode is half a wavelength. So, half of 2.4m is... drumroll, please... 1.2m!

Therefore, the first antinode is at the open end of the string, while the second antinode is 1.2m away from the fixed end. I hope that tickles your funny bone and answers your question!

To determine the distance from the end to the first 2 antinodes, you need to consider the fact that a standing wave pattern is formed when the incident wave and the reflected wave interfere constructively.

In a standing wave pattern, there are always nodes and antinodes. Nodes are the points where the displacement of the string is zero, while antinodes are the points where the displacement of the string is maximum.

For a wave traveling along a string and reflecting at a fixed end, the wavelength of the wave is equal to twice the distance between consecutive antinodes. So in this case, the wavelength of the wave is 2.4m, which means that the distance between consecutive antinodes is half of the wavelength, or 1.2m.

Since you mentioned that each wavelength has 2 antinodes, this means that there is one antinode at each end of the wavelength. Therefore, the distance from the fixed end to the first antinode is equal to half the distance between consecutive antinodes, which is 0.6m. The distance from the fixed end to the second antinode is then 1.2m.

Therefore, the first antinode is located 0.6m from the end, and the second antinode is located 1.2m from the end of the string.