In order to send a pulse and to propagate it, the string must be under tension.$^\text{1}$

Why is the tension necessary? Why should the string be stretched/taut for the transmission of the pulse?

$^\text{1}$ Principles of Physics by Walker,Resnick,Halliday.

  • $\begingroup$ You put "In order to send a pulse and to propagate it, the string must be under tension" in a box. Why did you do that? Is this a quotation from a book? $\endgroup$ – DanielSank Feb 5 '15 at 8:46
  • $\begingroup$ Hint: vibrations are caused by elasticity. $\endgroup$ – Phoenix87 Feb 5 '15 at 9:09
  • $\begingroup$ The tension in the rope provides a restoring force - in order for waves to propagate, each element of rope must experience a force from its neighbors bringing it back towards equilibrium when it is displaced, and a force away from equilibrium when its neighbors are displaced; that is the process by which a wave propagates. $\endgroup$ – Brionius Feb 5 '15 at 9:44
  • $\begingroup$ @Brionius: Sir, I urge you to please answer this. $\endgroup$ – user36790 Feb 5 '15 at 11:14
  • $\begingroup$ @Brionius: When the disturbed part provides tension to another undisturbed element, the net force is upwards but soon after restoring force starts to act to the previously undisturbed element thus decelerating it to zero when it reaches the highest amplitude. ,right? $\endgroup$ – user36790 Feb 8 '15 at 11:38

A wave on a string is like a harmonic oscillator (think mass on a spring). The oscillation happens because of the interplay between the restoring force, which tries to bring the mass to equilibrium, and the inertial quality of the mass, which tends to overshoot equilibrium. If you take away either, you don't get an oscillation.

If you work out the problem, you will find that the velocity of the pulse is proportional to the square root of the tension. If not under tension, there is no restoring force to counteract the inertial one. All you get is a moving string, not a pulse.

  • $\begingroup$ ". . .there is no restoring force to counteract the inertial one." Can you please elaborate this? I'll be grateful to you. $\endgroup$ – user36790 Feb 5 '15 at 11:42

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