Stationary waves energy problem

[pisatajay2009]

hi, i had this doubt regarding stationary waves
consider a string held between two points and vibrating in fundamental mode, so that there will be nodes at the fixed points.
then consider the datum of the potential energy at the mean position.

now when all particles have gone up by their respective amplitudes,the PE(gravitational) of the wave as a whole increases,KE is zero and there is some elastic potential energy say "k" joules.

now when the particles go to the other extreme position their PE(gravitational) is negative,
KE is zero and by symmetry, the stretching energy is the same ie "k" joules.

now my doubt is where does the difference in the energy appear?
assume experiment is conducted in vacuum.

 

[Stonebridge]

hi, i had this doubt regarding stationary waves
consider a string held between two points and vibrating in fundamental mode, so that there will be nodes at the fixed points.
then consider the datum of the potential energy at the mean position.

now when all particles have gone up by their respective amplitudes,the PE(gravitational) of the wave as a whole increases,KE is zero and there is some elastic potential energy say "k" joules.

now when the particles go to the other extreme position their PE(gravitational) is negative,
KE is zero and by symmetry, the stretching energy is the same ie "k" joules.

now my doubt is where does the difference in the energy appear?
assume experiment is conducted in vacuum.

The flaw is in the question. The stretching of the string at the top is less than at the bottom.
At the bottom the string is pulled downwards by gravity and its own weight and upwards by the restoring force due to the elasticity of the material.
At the top it is pulled downwards by all three forces. The string will stretch just a little more at the bottom than it does at the top.
The difference in the potential energy in the two cases is balanced by the difference in the stretching energy.
In other words, because of the greater stretching force at the bottom, the motion/amplitude of the string is not quite symmetrical. This gives rise to the apparent anomaly.

 

[pisatajay2009]

i got your answer.thanks a lot!

i never accounted for gravity also pulling down the string even further.
but then shouldn't the amplitudes be different ?
I mean we use the formula y=2Asin(wt)cos(kx), where the amplitude on either sides is the same.
is this an approximation of some sort?

 

[takeTwo]

i got your answer.thanks a lot!

i never accounted for gravity also pulling down the string even further.
but then shouldn't the amplitudes be different ?
I mean we use the formula y=2Asin(wt)cos(kx), where the amplitude on either sides is the same.
is this an approximation of some sort?

Why do this with gravity? Let the vibration direction x have nodes along x. I used to play the guitar. Seems to me that the length of the string must change when it is vibrating. Thus the energy is a combination of kinetic and potential. When you "pluck" the string, all the potential energy is in the elastic lengthening of the string. This is converted into kinetic energy (when the string is released) and after some transient, the mode or tone reaches steady state (one k dominates).

More generally, I'd think that delta_x(x,t) = 2A(t,k_j) sin[w(k)*t] * cos(k_j x) may well describe the motion for each mode "k_j" (j=0, 1,2,3,...). The k_j's would be the "over-tones" plus the fundamental, I think. As t -> infinity, all the k_j's damp out except one. k_0 is pi/length-of-string (the fundamental mode). In a modal analysis (FFT?), each k_j would have a unique temporal history.

The approximation in this case is that delta(x, t=0) doesn't break the string ; ).

Many guitar solos rely on A(t) being audibly interesting (the "attack" and the location of the pluck along x).

If it's really a gravitational problem with a "continuous-wave (CW)", external source of vibration of the end-nodes (you didn't say how this vibration is sustained, you imply A(t) is constant), I would agree with Stonebridge. What approximation? I'd say your equation does not account for gravity. It is a general simple harmonic oscillator with no external forces (no gravity!), no damping, no transients. It is the linear, "homogeneous" response from a "wave equation". This means that y is close to zero (small-signal approximation). If y is ~ zero it is symmetric, as your equation shows.

 

[Stonebridge]

i got your answer.thanks a lot!

i never accounted for gravity also pulling down the string even further.
but then shouldn't the amplitudes be different ?
I mean we use the formula y=2Asin(wt)cos(kx), where the amplitude on either sides is the same.
is this an approximation of some sort?

The effect of gravity on the motion of the string is disappearingly negligible.
As a physics problem, the gravitational potential energy considerations at the top and bottom of its vertical motion are unimportant. (The test would be: does the string sag in the middle when at rest?)
As a problem for mathematicians to get their teeth into, it would be one where the string is not performing perfect simple harmonic motion, because the restoring force is not the same at the top and bottom. The amplitude would be different at the top and bottom - slightly less at the top - and so the usual equations would not be valid.
However, as I said before, this effect is negligible. It does, however, explain why in your original question, there appears to be a problem with potential energy considerations.
The answer is, as I stated, that the elastic potential energy would be slightly more at the bottom and slightly less at the top. This compensates for the fact that there would be a small amount of gravitational pe at the top. The total energy, as always, does not vary. (Assuming no losses due to friction)

 

[pisatajay2009]

thank you for all the detailed replies!