Suppose ends of a string are attached to two different points on a ceiling. If a ring is hung freely, why does that ring move to the mid point of the string usually? Is it because the tension along two parts of the string with respect to the ring are same? If that is so, why are those equal?

Again if the ring is released from one end of the string, before reaching the midpoint the two parts of the string with respect to ring are unequal in length. Are tension along those two parts equal too?

If the string has mass then why does the mid portion remain as the trough of a wave?(will be really helpful if this question is answered too,can be ignored as well)


1 Answer 1


The tension will be equal no matter where the ring is, if it is frictionless.

The ring is pulled downwards by gravity, and moves to the lowest point it can reach. It will not take a huge amount of trig to determine that this is the midway point.

To do it without maths, attach string to the edge of a piece of paper, put a pencil against it, pull the string tight, and draw the set of points you can reach. You will get the edge of an ellipse, and it will be clear where the rest point is.

  • $\begingroup$ Why is tension equal evem if the string is bent? $\endgroup$
    – MSKB
    Sep 26, 2021 at 16:48
  • $\begingroup$ The string pulls on the string. If there were more tension on one side it would pull the other side through the ring until it reached equilibrium. $\endgroup$ Sep 28, 2021 at 8:02

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