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I am a student of class 11 and I have a doubt about reflection of waves on a string tied to a massless ring which can slide on a frictionless rod. When the wave reaches its end, the ring overshoots its amplitude because there is a lack of restoring force on it, as is written in the book Concepts of Physics. And in some places, it is written that this is due to superposition of the incident and reflected waves. But I am not able to understand what causes reflection of the wave in physical terms - like, which force creates the reflected wave of the same amplitude?

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  • $\begingroup$ Hi happy sharma, I've edited your question to trim it down to ask only one thing, since we prefer to have one question per post. You're welcome to make another post to ask the other part of your question. $\endgroup$ – David Z May 14 at 8:47
  • $\begingroup$ That question has completely wrong answers, and was asked at university level it seems. Here, it would be nice to have a correct answer, and at high school level too. $\endgroup$ – Kostas May 17 at 20:22
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The normal explanation is that if the ring is massless then by Newton's law the slightest force acting on it along the rod will create infinite acceleration. What that really means is that the the string will always be normal (at 90') to the rod. In terms of the incoming and outgoing waves we usually write: $$ \frac{d}{dx} [ \sin(x-vt) + c \sin(x+vt+\phi) ] =0 , ~~at~~ x=0 ~~and ~~for ~~all ~~t$$ Using trigonometry try to find c and $\phi$ such that this is true. You should be able to see that this sum of incoming and outgoing waves is indeed at 90' to the rod.

But to visualize the forces is hard. What really happens is that the furthermost piece of the string (say the last 1mm attached to the ring) has a small but finite mass. The string comes to it at almost 90', so the component of the force acting on it along the rod is also small. Newton says $a=F/m$, so the acceleration is a ratio of two small numbers. Sure, $a$ could be finite but within your knowledge of calculus in high school you will not be able to see why that is.

Doubly hard is to visualize what happens and where the force comes from in a transient bump - when you send a single disturbance, not necessarily a sine wave along the string and it reflects back. The incredible result is that the shape of the disturbance doesnt change after reflection, whereas if the ring was fixed the shape comes back the same but flips to the opposite side. Watch carefully and compare the video for a free endpoint and the fixed endpoint:

https://www.youtube.com/watch?v=1PsGZq5sLrw

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The ring is only there so the end of the spring can move sideways only. Another way to do it is with a very light string, which is how I saw it in year 11 many years ago.

As a rising wave passes most bits in the spring they are pulled up by the bit behind and down by the bit in front. The end bit has nothing in front so it moves further than an ordinary bit. This means the bit next to it is now not being pulled down as much, so it starts to go higher as well. Then the third bit is allowed to go higher than normal and so on.

The result is the same as a wave moving back along the spring on the same side as the incoming wave. When the top of the wave arrives the end bits will be much higher than usual, so as the wave leaves they will be pulled down faster as well until they reach their neutral positions.

A little more precisely -

Suppose the wave in the spring is able to travel from your right towards your left, with the spring itself moving up or down. It is the left end of the spring which is held by the ring on the rod.

Imagine the spring is divided along its length into many small equal segments. I will call the one at the left end, attached to the ring, segment 1. Each of these segments is acted on by two forces, one from each of the two adjacent segments except at the ends where one force comes from a neighbouring segment and one comes from the spring's supports.

Each of these forces will have components both along the line of the spring (left-right) and at right angles to the line (up/down). For simplicity I will ignore the left/right components (along the line of the spring). This is a good approximation as long as the amplitude of the wave is small. The effect of the ring at the left end is to set the up/down component of the force to be zero, while still providing the force along the length.

Suppose the spring is undisturbed except for an upward pulse traveling towards the left hand end (the end with the ring). We first look at what happens to a segment (say segment n) near the middle of the spring when the leading slope of the wave arrives.

When the wave arrives segment n experiences an upward force from the segment to its right (segment n+1). As it moves upwards it is still pulled upward by the segment to its right but it is also pulled down by the segment to its left (segment n-1), which has not yet risen as far. Segment n is now pulling segment n-1 upwards, just as segment n itself was pulled upwards a short time before.

When the wave reaches the end segment (1) that segment is pulled upwards by the segment to its right, but it is not pulled down by any segment to its left. Having no downward force on it, segment 1 rises higher than segment n did in the middle of the spring. Because segment 1 is higher, segment 2 immediately to its right is not pulled down by as much, so it is also higher than normal. This has a similar effect on segment 3 to its right, which in turn affects segment 4 to its right, and so on.

It can be shown mathematically that this is the same result as superposing (adding) the wave arriving at the left end, and a reflected wave returning on the same side of the spring. However I do not intend to do the actual Maths here.

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  • $\begingroup$ spring -> string? And, "third bit is allowed to go higher than normal" what does this even mean? $\endgroup$ – Kostas May 17 at 20:20
  • $\begingroup$ @Kostas, If we divide the spring into many sections (bits in less formal English) the section nearest the end would be the first bit, then second, then third. There is no intention to produce a precise description, but only to help people understand the strange behaviour of the spring as the wave is reflected. $\endgroup$ – Peter May 18 at 0:08
  • $\begingroup$ "The end bit has nothing in front so it moves further than an ordinary bit" - how, e.g. in this case, do i distinguish between an imprecise description and incorrect understanding? surely this is not factually true? First end bit goes higher, then the third bit ... what is smb to make of this gibberish?@Peter $\endgroup$ – Kostas May 19 at 9:56

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