Vibrating strings and potential energy

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SUMMARY

The discussion centers on the behavior of vibrating strings, specifically addressing why the midpoint of a string fixed at both ends has zero potential energy in its fundamental mode of vibration. The energy of vibration is stored in the elastic stretching of the string, with maximum potential energy occurring at points of maximum displacement (antinodes), where the string is stretched the most. At the midpoint, where the string is horizontal, there is no stretching energy, and thus no potential energy. The kinetic energy at the mean position is converted to potential energy as the string oscillates.

PREREQUISITES
  • Understanding of basic physics concepts, particularly oscillation and energy conservation.
  • Familiarity with the wave equation and its derivation.
  • Knowledge of tension forces in strings and their effects on motion.
  • Concept of potential energy in elastic materials.
NEXT STEPS
  • Study the derivation of the wave equation for vibrating strings.
  • Explore the relationship between kinetic energy and potential energy in oscillatory systems.
  • Learn about the effects of tension and displacement on the behavior of strings.
  • Investigate the differences between nodes and antinodes in wave motion.
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Physics students, educators, and anyone interested in the mechanics of vibrating strings and wave phenomena.

zorro
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I read that the middle point of a string fixed between two rigid supports set in its fundamental mode of vibration has no potential energy. I would like to know the reason behind it. At the max displacement from the mean position, it must have max. potential energy.
 
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I know it seems paradoxical, but this is how the maths of the situation works out.

I will try to explain in physical terms.

The length of the vibrating string is obviously longer than the length of the still string.
The energy of vibration is stored in the elastic stretching of the string.
If you think of the string as a series of connected short elements, each element will have a different slope from the horizontal as they align with their local tangent to the stretched shape.
The stretching forces are perpendicular to the string, and zero horizontally.
So a horizontal element will be unstretched and have zero stretching energy.
The element at maximum deviation from the horizontal (maximum slope) is stretched the most and occurs where the string crosses the horizontal axis. This has the greatest stretching energy.

The stretching energy is potential energy.
So an element with zero stetching energy has zero potential energy.

Elements at maximum transverse displacement (nodes) are horizontal and therefore posess zero potential energy.

go well
 
Studiot said:
The stretching forces are perpendicular to the string, and zero horizontally.

Perpendicular to every small string element?

Studiot said:
So a horizontal element will be unstretched and have zero stretching energy.

It (middle point) has some kinetic energy at the mean position which causes it to displace. If no stretching, where does this energy go?

Studiot said:
Elements at maximum transverse displacement (nodes) are horizontal and therefore posess zero potential energy.

Antinodes?
 
Perpendicular to every small string element?

Perpendicular to the still string.

Antinodes?

My mistake, sorry, yes antinodes.

 
Studiot said:
Perpendicular to the still string.

I am confused here. By stretching forces do you mean the restoring force once the string is stretched or the initial force which we apply to stretch the string?
 
Let the string be stretched horizontally ( x axis) and plucked vertically (y axis)

If we ignore gravity then the only force acting after we have plucked the string and let go is the tension in the string.

The tension is constant in magnitude throughout the string, but since it is parallel to the tangent to the string, its direction changes from point to point. It is this change of direction that introduces actual changes within a string element dx.

If \varphi is the angle made by the element with the x-axis at the beginning of a small element dx and (\varphi+\delta\varphi) at the end.

Bearing in mind that the tensions in the small element point in opposite directions at each end, the components are

Vertically

Tsin(\varphi+\varphi\delta) - Tsin(\varphi)

It is this difference, which is a rotation, that provides the vertical displacement force.

Note that when \varphi is zero the element is horizontal and sin(\varphi) is zero so there no displacement force

Horizontally


The stretching force is

Tcos(\varphi+\varphi\delta) - Tcos(\varphi)

Note that when the element is horizontal T points horizontally at both ends so there is no net horizontal foce acting.


Here is a derivation of the wave equation using this information.

attachment.php?attachmentid=30310&d=1291335688.jpg
 
That made it clear about forces.
One question- When you pluck the string and let it perform oscillations (ignore external forces), the middle point of the string has maximum kinetic energy at the mean position. When it reaches the extreme position, you proved that its potential energy is 0. Where does the initial kinetic energy go then?

If you analyse other points, there is a stored potential energy due to stretching which compensates for the loss of kinetic energy.
 
Good question and I'm sorry I didn't answer first time.

The string is continuous so is the stretching. The PE is stored throughout the string in an uneven manner, just as the KE is distributed throughout the elements in an uneven manner.

If fact as the antinode element passes through y=zero the string is unstretched so all the PE has been returned as KE.
Similarly as the antinode element rises it slows, the whole string stretches, and the KE is converted to PE in the whole string. Eventually at the antinode the element halts and reverses, Thus the KE passes through zero and the PE is a max.
 
Thanks. Its clear now :)
 
  • #10
Vibrating strings are actually more tricky than many other forms of oscillation.

:wink:

go well
 

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