Can Algebra Solve String Wave Problems?

  • #1
Saitama
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Homework Statement


I am posting three problems together as I believe they are quite short.

1.Show that the particle speed can never equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by ##2\pi##.

2.Two wave pulses identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At the instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of two pulses?

3.Show that for a wave traveling on a string
$$\frac{y_{max}}{v_{max}}=\frac{v_{max}}{a_{max}}$$​
where the symbols have usual meanings. Can we use componendo and dividendo taught in algebra to write
$$\frac{y_{max}+v_{max}}{y_{max}-v_{max}}=\frac{v_{max}+a_{max}}{v_{max}-a_{max}}$$​

(By usual notation, y is the displacement, v is transverse velocity and a is the acceleration of any particle on the string)

Homework Equations





The Attempt at a Solution



Attempt for 1.
Let the string wave be represented by the equation ##y=A\sin(\omega t-kx)##. The transverse velocity of particle at position at any instant of time t is given by ##∂y/∂t=A\omega \cos(\omega t-kx)##. The speed is hence given by:

$$\sqrt{A^2\omega^2\cos^2(\omega t-kx)+\frac{\omega^2\lambda^2}{4\pi^2}}$$

where the second term is the wave speed given by ##\omega/k=\omega\lambda/(2\pi)## and ##\lambda## is the wavelength. Since ##A \leqslant \lambda/(2\pi)##, we have

$$\sqrt{A^2\omega^2\cos^2(\omega t-kx)+\frac{\omega^2\lambda^2}{4\pi^2}}\leqslant \frac{\omega \lambda}{2\pi}\sqrt{1+\cos^2(\omega t-kx)}\leqslant \frac{\sqrt{2}\omega \lambda}{2\pi}$$
The above does not agree with the problem. :confused:

Attempt for 2.
I honestly have no idea about this one. :(

Attempt for 3.
It is very easy to find the relations presented in the problem statement by assuming the wave equation to be ##y=A\sin(\omega t-kx)##. I am not sure why the author asks the second part of the problem. I don't see why it would be wrong to do some simple algebra on the relation obtained.

Any help is appreciated. Thanks!
 
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  • #2
For #2

I think that all the kinetic energy of the particles will be converted into potential energy. The particles of the string will move a bit away from each other.

For #3- I am not sure but-

One thing to know is that the dimension of Vmax in LHS is [L]. Also the equation isn't valid (for obvious reasons) when Ymax=Vmax (similar thing for RHS)
 
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  • #3
Can you describe the motion of a particle of the string?
 
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  • #4
Hi consciousness and TSny! :)

consciousness said:
For #3- I am not sure but-

One thing to know is that the dimension of Vmax in LHS is [L]. Also the equation isn't valid (for obvious reasons) when Ymax=Vmax (similar thing for RHS)

Yes, the dimensions don't match and I think this must be the reason, thank you! :)

TSny said:
Can you describe the motion of a particle of the string?

Umm...I am not sure but do you ask the following?

$$y(x,t)=A\sin(\omega t-kx)$$
$$x(t)=\frac{\omega}{k}t$$
 
  • #5
Set aside the equations for the moment. Can you describe in words the motion of a point of the string? In particular, can you describe (at least roughly) the vertical component of its motion? the horizontal component of its motion?
 
  • #6
TSny said:
In particular, can you describe (at least roughly) the vertical component of its motion?
The vertical component of motion is simple harmonic with the mean position on the x-axis.

the horizontal component of its motion?

The horizontal component of motion is simply translational motion with a constant velocity in horizontal direction.
 
  • #9
That's a fundamental property of wave propagation in matter: the only thing that propagates is the wave itself; the matter oscillates only.
 
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  • #10
voko said:
That's a fundamental property of wave propagation in matter: the only thing that propagates is the wave itself; the matter oscillates only.

Thanks voko and TSny, I think I get the first problem. :)

What about problem 2? :confused:
 
  • #11
Think about a small element on the string. When it oscillates, when is its potential energy maximal? Its kinetic energy?
 
  • #12
voko said:
Think about a small element on the string. When it oscillates, when is its potential energy maximal? Its kinetic energy?

The potential energy is maximum when it is at the maximum amplitude and kinetic energy is maximum when it is at its mean position.
 
  • #13
So when the string is straight, where is the energy of the pulses?
 
  • #14
voko said:
So when the string is straight, where is the energy of the pulses?

It is converted to kinetic energy as the particles are on their mean position, right?
 
  • #15
Can it be anywhere else?
 
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  • #16
voko said:
Can it be anywhere else?

Nope. :D

Thanks a lot voko! :smile:
 
  • #17
voko said:
Think about a small element on the string. When it oscillates, when is its potential energy maximal? Its kinetic energy?

When a wave propagates on a string the velocity of the particles is maximum when they are in their mean position. But when two waves having the same amplitude destructively interfere, the string will become completely straight (principle of superposition is valid) so the kinetic energy becomes zero! A similar thing can't happen here?

Edit- After thinking some more I think that the velocity of the particles is indeed highest when they superimpose but this problem is mathematically solved IMO.
 
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  • #18
consciousness said:
this problem is mathematically solved IMO.

What does that mean?
 
  • #19
voko said:
What does that mean?

When two traveling waves interfere, it is possible that the string becomes straight and remains that way. All the particles are at rest. The same is not true for this situation but I got confused somehow.
 
  • #20
consciousness said:
When two traveling waves interfere, it is possible that the string becomes straight and remains that way.

No, that is not possible. Both waves carry energy. It cannot just disappear.
 
  • #21
voko said:
No, that is not possible. Both waves carry energy. It cannot just disappear.

Consider 2 sine waves with same frequency and amplitude oscillating in opposite phase-

Y1=Asin(kx-ωt)
Y2=Asin(kx-ωt+∏)

Y=Y1+Y2=0 at all points and times.

The catch is I think, that the total work being done on the string is zero. The positive work on one side is equal to the negative work on the other side. So the string never carries any energy
 
  • #22
Those are not two counter-propagating waves as in the original problem. Those are two co-propagating waves in anti-phase, cancelling each other identically everywhere, so there is no energy transfer.

I should also warn against using pure sinusoidal waves in analyzing energy transfer. They exists everywhere in space, and so carry infinite energy. This is very different from real waves, which carry finite energy, and so are localized in space, or fall off rapidly with distance, or both.
 
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