3
$\begingroup$

In deriving wave equation or power transmission of wave transmitted by a string, it is usually stated (with some assumptions) that the transverse force on a point of the string is proportional to the slope at that point. An example is given in p.20 of this notes.

If the slope is zero the transverse force is also zero. It can also be seen in the way that if some portion of the string is horizontal the tensions on both side are also horizontal and thus cancel out, therefore no transverse force.

However, in the case that the wave is sinusoidal, the points at the amplitude of the wave should have greatest acceleration and should experience the greatest force because every point is performing SHM. There seems like a contradiction here. Why?

$\endgroup$
  • $\begingroup$ "points at the amplitude of the wave"... what does that mean? $\endgroup$ – DanielSank Jul 1 '15 at 20:30
  • $\begingroup$ @DanielSank It is the portion of string at the maximum position of the string in a sinusoidal wave. $\endgroup$ – Kelvin S Jul 2 '15 at 10:33
3
$\begingroup$

In those notes, it states that the left part of the string pulls the dot with a force proportional to the slope. However, the right side pulls the dot in the other direction with a force proportional to the slope, so that if the slope is constant, there is no net force on the dot. The only way to get a net force is for the slopes to be different on the left and right sides, that is there must be a non-zero second derivative.

With this in mind, the crest of the sinusoidal wave is the point with the greatest curvature and so it's the point with the greatest acceleration, as you expected.

$\endgroup$
  • $\begingroup$ You mentioned that crest of the sinusoidal wave is the point with the greatest curvature. However if you take derivative of sinusoidal wave the slope at the amplitude should be zero. $\endgroup$ – Kelvin S Jul 2 '15 at 10:35
  • 1
    $\begingroup$ I got the answer now. We need to consider the forces on both side of the small portion of the string. The net transverse force is proportional to the change of the slope, i.e., the second derivative of the string shape, which describes the difference of the forces on each side of a small string element. $\endgroup$ – Kelvin S Jul 2 '15 at 12:24
  • $\begingroup$ @DanielSank should I add something to my answer? $\endgroup$ – ragnar Jul 2 '15 at 14:53
  • $\begingroup$ @ragnar Nope, looks great. $\endgroup$ – DanielSank Jul 2 '15 at 15:06
  • $\begingroup$ Summary: Tension is proportional to curvature. $\endgroup$ – ja72 Mar 23 '16 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.