wavelength problem . need an explanation .

The stationary wave pattern shown in figure 6 (please open the attachment to see the figure) is set up on a rope of length 3.0m. How would you calculate the wavelength of these waves?i'm all confused . I mean what formula would you choose to calculate the wavelengths .
Thanks .

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Reply 1

uberteknik

Original post by Alen.m

L = \frac{N\lambda}{2}

Where

L

= distance between end attachment points

\lambda

= wavelength

N

= number of Anti-nodes set up on the rope.

(edited 9 years ago)

Reply 2

lerjj

Original post by uberteknik

L = \frac{N\lambda}{2}

Where

L

= distance between end attachment points

\lambda

= wavelength

N

= number of Nodes set up on the rope (excluding the end attachment points).

Are you sure that's right? Given two nodes that computes a wavelength of 3m for this scenario- which doesn't seem right. Shouldn't it be number of antinodes?

Reply 3

uberteknik

Original post by lerjj

Are you sure that's right? Given two nodes that computes a wavelength of 3m for this scenario- which doesn't seem right. Shouldn't it be number of antinodes?

Ha ha, well spotted.

It should be the harmonic number which indeed corresponds to the anti-nodes! Corrected accordingly.

Reply 4

Alen.m

in terms of phase different guys, how would you calculate phase difference between the particle vibrating at O and particle vibrating at B( there's odd number of nudes between them so the answer should be 180 but the text book says 225). any help would be much appreciated .

Reply 5

uberteknik

Original post by Alen.m

It's a standing wave. The particles vibrate up and down in a vertical direction only. There can only ever be 2 answers, 0^o or 180^o.

Points on opposite sides of a node always vibrate in anti-phase. Two points between the same node or between alternate nodes (every other node) will always vibrate in phase.

(edited 9 years ago)

Reply 6

Alen.m

didn't actually get that :frown:

how would you get 225 as the answer?

Reply 7

Stonebridge

Original post by Alen.m

didn't actually get that :frown:

how would you get 225 as the answer?

The book is wrong.
The answer is 180, not 225.

All the points in any one loop are in phase in a stationary wave. In adjacent loops they are 180 degs out of phase.

Reply 8

Alen.m

so should i consider the answer 180 as there's odd number of nodes between OB?

Reply 9

uberteknik

Original post by Alen.m

so should i consider the answer 180 as there's odd number of nodes between OB?

Yes.

The OC and AB pairs are in phase (0^o): Both points are between the same nodes or there are an even number of nodes between the points.

OA, OB, AC and BC pairs are out of phase (pi/2 or 180^o): The points are immediately either side of a single node or there are an odd number of nodes between the points.

(edited 9 years ago)

millions thanks

But in the textbook there are 2 progressive waves which make up the standing wave pattern it’s specifically asking for the progressive wave which is not the dotted line but the filled one, if it was asking for the standing wave phase difference, a much larger wave would be shown in the diagram with different points of interestPlease correct me if I’m wrong

Reply 12

Stonebridge

Original post by BigMuma

The diagram is the standard way of representing a stationary wave pattern. Though looking back (this question is 6 years old now) we don't have the whole question, just this diagram and the poster's text. A stationary wave pattern like this is produced when you have a progressive wave interfering with it's own reflection. So in effect you have 2 progressive waves with the same frequency travelling in opposite directions on the string. The diagram is supposed to show the pattern of nodes and antinodes as described in the various replies above. One segment (between 2 nodes) of the stationary wave is half a wavelength of the progressive wave that has produced this. All points on the rope within that section of the stationary wave are moving in phase.