Displacement of String Waves at Point of Cross

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In the discussion on the displacement of string waves at the point of crossing, it is established that when a string is rigidly fixed, the net displacement at the crossing point is zero due to the formation of a node, as the reflected waves are inverted and cancel each other out. Conversely, if the string is free to move, the net displacement at the crossing point is the sum of the two displacements, resulting in a total displacement of 2A, or 0.300 m, since the reflected waves maintain the same sign. The participants confirm that the addition of the same-sign displacements leads to this outcome. The conversation highlights the importance of understanding wave behavior in different boundary conditions. Overall, the key takeaway is the contrasting effects of fixed versus free boundaries on wave displacement.
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Homework Statement


A series of pulses, each with an amplitude of .150 m are sent down a string that is attached to a post at one end. The pulses are reflected back and travel back along the string with no loss of amplitude. What is the net displacement of an element at a point on the string where the 2 pulses cross a. if the string is rigidly attached, b. if the end at which reflection occurs is free to slide up and down.


Homework Equations


I know that in a rigidly fixed string, reflection inverts the waves, and if the string is free to move up and down, the reflected pulse is not inverted.


The Attempt at a Solution


I am thinking that in a rigid system, the net displacement where the two pulses cross would be zero, because that point would be a node. If the system is free to move, then I think that the net displacement at the point where the waves meet would be plus or minus A. Is my brain working right?
 
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aliciagu said:

The Attempt at a Solution


I am thinking that in a rigid system, the net displacement where the two pulses cross would be zero, because that point would be a node.

Yes, the incoming and outgoing pulses are negatives of each other. Since waves add linearly, their sum is zero.

If the system is free to move, then I think that the net displacement at the point where the waves meet would be plus or minus A. Is my brain working right?

Hold up. Now the displacements of the incoming and outgoing waves have the same sign. How do numbers of the same sign add up?
 
So in the second one, the displacement would be 2A, because the sum of the 2 diplacements is (in this case) .300?
 
Now you've got it.
 

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