Solve the Tension Paradox - Find the Flaw in Reasoning

[Zorodius]

I realize that there is a flaw in my reasoning. I don't know what it is. Please point it out

A string is oscillating in a transverse sinusoidal wave. Consider a small element of that string.

When the element of the string has displacement zero, it has the greatest possible kinetic energy and elastic potential energy, and is stretched by the greatest amount possible.

When the element of the string has displacement equal to the amplitude of the wave, it has zero kinetic energy, zero elastic potential energy, and is stretched by the smallest amount possible.

Where is the tension greatest?

The elastic potential energy of the element is greatest when the displacement is zero. Therefore, the tension must be greatest there also.

But wait: F = ma, and the acceleration at displacement zero is also zero, while the acceleration at maximum displacement is the greatest possible amount. The only force the string is subject to is due to tension. Therefore, since the acceleration is greatest when the displacement is equal to the amplitude, the tension must be greatest there also.

Both of these sound perfectly well reasoned to me, and yet are completely mutually exclusive. One or both of them must therefore have a logical error, but what is it?

 

[Moose352]

Can you explain to me how the elastic PE is greatest when the displacement is zero?

 

[Zorodius]

Can you explain to me how the elastic PE is greatest when the displacement is zero?

When the string element is at its maximum displacement, its length has its normal undisturbed value, so its elastic potential energy is zero. However, when the element is rushing through its zero displacement position, it is stretched to its maximum extent, and its elastic potential energy then is a maximum.

If that sounds unconvincing, I can't really disagree, that's just paraphrased out of my textbook.

 

[HallsofIvy]

When the string element is at its maximum displacement, its length has its normal undisturbed value, so its elastic potential energy is zero. However, when the element is rushing through its zero displacement position, it is stretched to its maximum extent, and its elastic potential energy then is a maximum.

If that sounds unconvincing, I can't really disagree, that's just paraphrased out of my textbook.

No, that's exactly backwards. The "normal undisturbed" length is the straight line distance between its endpoints. Minimum potential energy occurs when the elastic is lying along that straight line, maximum stretch and maximum potential energy occur when the element is at its maximum displacement.

 

[Zorodius]

No, that's exactly backwards. The "normal undisturbed" length is the straight line distance between its endpoints. Minimum potential energy occurs when the elastic is lying along that straight line, maximum stretch and maximum potential energy occur when the element is at its maximum displacement.

Thanks a lot for the reply, Halls! I thought that sounded very weird. I'm surprised that this book would make such an emphatically stated point of their error.

Just to double-check, is it true that the following description of energy transport along a string wave - copied straight from my textbook - is completely wrong?

The oscillating string element thus has both its maximum kinetic energy and its maximum elastic potential energy at y= 0. In the snapshot of Fig. 17-11 (I recreated this using paint) the regions of the string at maximum displacement have no energy, and the regions at zero displacement have maximum energy. As the wave travels along the string, forces due to the tension in the string continuously do work to transfer energy from regions with energy to regions with no energy.

Figure 17-11 has this text beneath it:
A snapshot of a traveling wave on a string at time t = 0. String element a is at displacement $y = y_m$, and string element b is at displacement y = 0. The kinetic energy of the string element at each position depends on the transverse velocity of the element. The potential energy depends on the amount by which the string element is stretched as the wave passes through it.

 

[ArmoSkater87]

I must say that's a job well done if you used pain to draw that sine wave. :D

 

[Zorodius]

I must say that's a job well done if you used pain to draw that sine wave. :D

Technically, I wrote a program in OpenGL to graph the sine wave, then used Paint to draw most of the diagram, and GIMP to compress it. But that seemed like more information than was necessary

So, is the book right or wrong?

 

[Petrushka]

Technically, I wrote a program in OpenGL to graph the sine wave, then used Paint to draw most of the diagram, and GIMP to compress it. But that seemed like more information than was necessary

So, is the book right or wrong?

If the book says "The oscillating string element thus has both its maximum kinetic energy and its maximum elastic potential energy at y= 0." where y is the displacement then it is precisely wrong.

At maximum displacement, the string will have maximum potential energy (elastic potential energy) and zero kinetic energy, as at maximum displacement the oscillation comes to instantaneous rest.

At zero displacement, where velocity is maximum, the oscillating string will have maximum kinetic energy and zero elastic potential energy, as the string would essentially be at its natural length and would not be taut, hence not store any potential energy.

 

[Zorodius]

I notice that, from the diagram, it seems like they're trying to show that a string element would be stretched the most at zero displacement because, if you cut out all the string contained in either of two columns of equal width - one around the point where the string crosses y=0 and the other around the point where the string reaches its maximum displacement - you get more string from the first than from the second. (at that particular instant, I mean. In real life, cutting it free of the rest of the string would, of course, lower the tension on the string element to zero )