# Ring Impedance & String Connection: Is It Valid?

• gasgas

#### gasgas

Homework Statement
Suppose there are two strings with equal tension but different impedances attached to a ring on a vertical rod. If we shake the first string, it will make the ring slide and transmit the wave to the other string. As the ring slides on the rod, there is friction acting on it which is equal to F=b*v where b is coefficient of friction and v the speed of the ring. What is the value of b if there is no wave reflection?
Relevant Equations
F=Z*V
R=(Z1-Z2)/(Z1+Z2)
If we consider the coefficient b as the rings impedance, we can consider the effective impedance on the right to be b+Z2 where Z2 is the impedance of the second string. Then because there is no reflection it follows that Z1=b+Z2 or b=Z1-Z2.
Is this a valid solution? My professor went through a more complicated derivation which we concluded was wrong so I tried this. Is it correct? Can we suppose that the ring and string are connected in series and can we equate b with impedance? They match dimensionally so i don't see why not. Any help appreciated :)

My doubt would be whether there may be a constant multiplier needed to convert b to a comparable impedance. I see no way other than to analyse the motion fully.

berkeman
I agree with @haruspex that this is a complicated problem, and I'd be inclined to analyze it fully from a motion and wave equation standpoint, instead of trying to simplify it to the impedance/reflection equation. Perhaps there is a simpler way to do it, but I'm not seeing it intuitively.

Also, it doesn't look like it's specified in the problem statement, but it seems clear to me that this can only work if the impedance of the 2nd string is lower than the first. The friction at the ring would seem to add impedance, so if the 2nd string had higher impedance than the first, I don't think there is any way to avoid reflections at the ring junction.

I also don't understand this part of the problem statement, about the frictional force depending on the velocity. That's not normally how frictional forces work, but maybe it helps with the math somehow...
As the ring slides on the rod, there is friction acting on it which is equal to F=b*v where b is coefficient of friction and v the speed of the ring.

maybe it helps with the math somehow.
It will. With b negative, it arranges that the frictional force always opposes the motion.

But speed dependent?

But speed dependent?
That's the price of making the math more manageable.
Bear in mind that standard damped oscillation analysis assumes F∝v.

berkeman