Can Algebra Solve String Wave Problems?

In summary: Sorry for the confusion. :confused:In summary, the conversation discusses three problems regarding particle speed and wave speed in a sine wave, the energy of two wave pulses on a string, and the use of algebra in wave equations. The first problem is solved using equations and it is concluded that the particle speed can never equal the wave speed if the amplitude is less than the wavelength divided by ##2\pi##. The second problem involves a discussion on the energy of the pulses when they reach the middle of the string and it is determined that the kinetic energy is maximum when the string is straight. The third problem is solved using equations and it is stated that it is possible for the string to become straight and remain that way when two traveling waves interfere
  • #1
Saitama
4,243
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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?
 
Last edited:
  • #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.
 

1. What are string wave problems?

String wave problems refer to the study and analysis of waves that propagate through a string or rope. These types of problems often involve finding the properties of the wave, such as its speed, frequency, and wavelength, as well as understanding how the wave behaves when encountering different obstacles or boundary conditions.

2. What is the wave equation and how is it used to solve string wave problems?

The wave equation is a mathematical equation that describes how waves propagate through a medium. It is used to solve string wave problems by relating the wave's properties, such as its speed, frequency, and wavelength, to the properties of the string, such as its tension, density, and length. By solving this equation, we can determine the behavior of the wave and predict its motion.

3. What are the boundary conditions for solving string wave problems?

The boundary conditions for solving string wave problems refer to the conditions at the ends of the string, which can affect the behavior of the wave. These conditions can include fixed or free ends, as well as the presence of obstacles or discontinuities. These conditions are important to consider as they can affect the amplitude, frequency, and wavelength of the wave.

4. How do we determine the speed of a wave in a string?

The speed of a wave in a string is determined by the tension and density of the string, as well as the frequency and wavelength of the wave. This can be calculated using the wave equation, where the speed is equal to the square root of the tension divided by the density of the string. Alternatively, the speed can also be measured experimentally by measuring the time it takes for the wave to travel a known distance.

5. What are some real-life applications of solving string wave problems?

There are many real-life applications of solving string wave problems, including understanding the behavior of musical instruments, such as guitars and violins, as well as the transmission of sound through strings in telecommunication systems. Additionally, string wave problems are important in fields such as engineering, acoustics, and seismology, where understanding wave behavior is crucial for designing structures, predicting earthquakes, and detecting underground oil and gas reserves.

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