Waves on a String -- Minimizing reflections from the far end (ring on a bar)

In summary: This will ensure that the reflected wave has the same amplitude as the incident wave.In summary, the string should be stretched to a tension T such that the energy flux in the reflected wave is minimized. The energy reflection coefficient at this tension is ω m/b <<1.
  • #1
Loubrainz
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Homework Statement


A string of linear density ρ terminates in a small ring of mass m. A horizontal bar is passed through the ring, perpendicular to the direction of the string at rest, and the string is stretched to a tension T. Friction between the ring and the bar generates a resistive force –bvy, where vy is the transverse velocity of the ring and b is a constant. A harmonic wave of angular frequency ω is directed along the string towards the ring. To what value of T should the string be stretched to minimise the energy flux in the reflected wave, and what is the energy reflection coefficient at this tension? (Assume that ω m/b <<1).

Homework Equations


F = ma (Newton's Second Law)
y(x,t)i = Aei(ωt-kx) (mathematical description of incident wave)
y(x,t)r = Bei(ωt+kx) (mathematical description of reflected wave)
gif.gif
(wave equation)
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(wave speed in terms of tension and linear density)
gif.gif
(Power passing through a point on a string with some amplitude 'a')
Energy Reflection Coefficient = Reflected Energy/Incident Energy

The Attempt at a Solution


My initial thoughts are, firstly, that there is a geometrical boundary condition: the displacement of the incident and reflected wave at the ring must be continuous:
yi(x=0) = yr(x=0) (letting the position of the ring be at x=0);
therefore, A=B. (?)

Secondly, the net transverse force acting on the ring must be the combined transverse components of the tension in the string from both the incident and reflected wave balanced by the friction between the ring and rod:
gif.gif


I tried substituting in the mathematical description of the incident and reflected wave, but it doesn't seem to rearrange into anything immediately useful that can be plugged into the power equation.

What do you think?
 

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  • #2
Loubrainz said:
My initial thoughts are, firstly, that there is a geometrical boundary condition: the displacement of the incident and reflected wave at the ring must be continuous:
yi(x=0) = yr(x=0) (letting the position of the ring be at x=0);
therefore, A=B. (?)
This is not correct. The amplitudes can only be the same if the wave is completely reflected. You need to get the boundary condition from the force balance on the ring instead.

I suggest that you instead write the full wave as ##y(x,t) = A e^{i(\omega t - kx)} + B e^{i(\omega t + kx)}##, where ##A## is the amplitude of the incoming and ##B## the amplitude of the reflected wave. Also note that you can express the resistive force in terms of the wave function ##y(x,t)## and its derivatives at ##x = 0##.

Also note that you can normalise the incoming wave amplitude to one.
 
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Likes berkeman

1. How do waves on a string minimize reflections from the far end?

Waves on a string can minimize reflections from the far end by adjusting the tension and mass of the string. The tension should be high and the mass should be low to minimize reflections.

2. What is the relationship between tension and reflections in waves on a string?

The higher the tension in a string, the fewer reflections will occur at the far end. This is because higher tension results in a faster wave speed and a larger impedance mismatch, reducing the amount of reflected energy.

3. Can changing the mass of the string also reduce reflections?

Yes, changing the mass of the string can also help minimize reflections. A lower mass results in a higher wave speed, which reduces the impedance mismatch and therefore reduces reflections at the far end.

4. Are there any other factors that can affect reflections in waves on a string?

Other factors that can affect reflections include the shape and material of the string. A thinner, more flexible string will have less reflections than a thicker, stiffer string. Additionally, using a material with a higher density can also help reduce reflections.

5. Is it possible to completely eliminate reflections in waves on a string?

No, it is not possible to completely eliminate reflections in waves on a string. However, by adjusting the tension and mass of the string, reflections can be minimized to a very low level.

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