- #1

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How can I measure the length of the actual curve of a sine wave. The actual curve not the wavelength.

Raavin :)

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- Thread starter Raavin
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- #1

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How can I measure the length of the actual curve of a sine wave. The actual curve not the wavelength.

Raavin :)

- #2

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for a function f(x),

arc length S =[inte] [squ](1+ [f'(x)]

and then of course you just need to integrate and evalute it for your boundries.

- #3

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y=sin(x)

dl^2=dx^2+dy^2 (Pythagorean theorem applied at the infinitesimal scale where dl is the length of the infinitesimal segment of the curve)

dl=√(dx^2+dy^2)

dl=√(1+dy^2/dx^2)dx

l=∫0…2π √(1+(dy/dx)^2)dx

l=∫0…2π √(1+cos(x)^2)dx

l=6.28 (approximatly)

Here is a link that may help

http://mathworld.wolfram.com/ArcLength.html

- #4

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Originally posted by Suicidal

y=sin(x)

dl^2=dx^2+dy^2 (Pythagorean theorem applied at the infinitesimal scale where dl is the length of the infinitesimal segment of the curve)

dl=√(dx^2+dy^2)

dl=√(1+dy^2/dx^2)dx

l=∫0…2π √(1+(dy/dx)^2)dx

l=∫0…2π √(1+cos(x)^2)dx

l=6.28 (approximatly)

Here is a link that may help

http://mathworld.wolfram.com/ArcLength.html

no, 6.28 is 2 pi, the circumference of a circle of radius 1

the sin curve, from 0 to pi has a length of approximently 3.8201977890278

- #5

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I assumed he wanted the length of a single wave which is produce over the interval [0,2π]Originally posted by Raavin

How can I measure the length of the actual curve of a sine wave.

- #6

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suicidal, i think you should check your numbers, ORW is correct.

- #7

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Raavin :)

- #8

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Originally posted by On Radioactive Waves

for a function f(x),

arc length S =[inte] [squ](1+ [f'(x)]^{2})dx

and then of course you just need to integrate and evalute it for your boundries.

Could you explain this to me like I'm an idiot???

Explaining what I said above, I want to be able to create a sine wave with one 'period??', calculate the arc length, then create another wave with two periods with the same boundry and the same arc length. So the amplitude will be less on the second one, then do the same with 3 periods, 4 periods etc. Then I want to see if the relationship of the amplitudes corresponds to any other physics relationships.

Get what I mean???

Raavin :)

- #9

Hurkyl

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If you scale a figure by a ratio r, all lengths in the figure are multiplied by r.

So suppose our initial sine wave has one period in [0, 2π], amplitude 1, and an unknown arclength L.

If we scale the figure by one half, it will have one period in [0, π], amplitude 1/2, and an unknown arclength L/2.

So, if we put two copies of the diminished figure together, we get two periods in [0, 2π], amplitude 1/2, and unknown arclength L.

Similarly if we scaled by 1/3; you'd fit 3 periods into [0, 2π] with an amplitude of 1/3 and arclength L.

(Note: the theorem for scaling only works if you scale the same amount in all directions; if you, say, stretch in the x-direction but leave the y-direction unchanged, there isn't an easy formula for lengths)

- #10

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i used a calculator.

is there a way to evaluate ∫√(1+cos(x)^2)dx by hand

- #11

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Originally posted by Suicidal

i used a calculator.

is there a way to evaluate ∫√(1+cos(x)^2)dx by hand

this is an elliptic integral, so, no, not really.

- #12

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But if you keep the same total length, then the area under the curve reduces by the same proportions as the amplitude and the wavelength? eg. halve the wavelength and amplitude and you halve the area under the curve because you are dividing by four and multiplying by 2.

Is that right. If it is, what happens when you reduce the length by less than a full wavelength or reduce the wavelength and amplitude by a non integer amount. Does this stuff the scaling???

Raavin

- #13

Hurkyl

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So yes, if you scale by 1/2, all areas are multipled by 1/4.

And since you take two copies of the scaled figure, the resulting area would be half as much as the original.

You can scale the figure by any real number, but as you noted you might not be able to fit an integral number of scaled periods into your original period... and there's no reason to think that the arclength or area under the sine wave should be proportional to the (fractional) number of copies because the replications aren't complete.

- #14

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Ok, so we scale so that 2.5 or even 2.25 waves fit, it would seem logical that we get the same ratio of area scaling because we have the extra full half wave or part wave with a symmetrical division. The problem arises when we don't have a full half because for example the area of the 1st 1/4 of a node is not 1/4 of the area of the full node (is node the right term??). So during the scaling period in between the discreet 1/4 steps there is incongruence in area and length scaling.

The reason I'm interested in this is in trying to find a simple connection between this effect and discreetness in quantum physics.

This is probably pushing it, but if one was to plot the variation in the area to length ratio in this case, would it fit a standard function????

Raavin [?]

The reason I'm interested in this is in trying to find a simple connection between this effect and discreetness in quantum physics.

This is probably pushing it, but if one was to plot the variation in the area to length ratio in this case, would it fit a standard function????

Raavin [?]

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