Electrical Equivalent

The simplest equivalent would be a free electron near a free proton in space, far from any other influences.

In terms of an electric circuit, I suppose a Crooke's Tube device would be the starting point for a circuit that models this behavior.

Electrons don't "fall freely".

Let us assume the falling body is a single electron by analogy.

R is resistance - nothing to change here. The air is the resistive medium.

L is inductance - Gravity acceleration would "induce" a change in position of the "electron".

C is capacitance, and for this Height would be the equivalent in the analogy. As the height decreases so to does the capacitance or the height that one may fal.

In an inductive circuit, the voltage leads the current, and in your problem the voltage is already at peak. In a capacitive circuit current leads voltage, so it is safe to assume we are on the down side of an already charged cap.

The free-fall scenario will then equate to 1/4 cycle in an capacitive discharge of one electron through an air resistor of X ohms for a time and velecity determined by the total voltage in height and gravity acceleration.

The circuit will be discharged when the electron hits the ground.

Should I consider air resistance when I am dropping a mass of 10g from a height of say 20cm..? And also should I take R, L, and C in series or parallel..??

Resistance is air resistance.

I think R is in series, and L and C are in parallel. But also remember that if you are going to equate this to a circuit, that this scenario is only 1/4 of a cycle. You see only the fall from peak potential to zero potential.

The movement of the mass is also like separating the plates of the capacitor. It also looks like a discharge through a conductor and resistor in series.

The problem as presented is that the analogy is not one to one. Electrons do not accelerate, they are essentially at full velocity the moment they are released. so, your falling and accelerating mass is more like an increase in current over time as the equivalent to the accelerating mass. So when the mass is sitting still it is the same as 0 Amps. When it reaches it's peak velocity, it is the same as peak amps. Same for the voltage. At the starting height the voltage is at maximum. At the bottom of the fall, voltage is at minimum. What you see in the falling mass is the phase angle between the voltage and the current. That angle is 90deg and is why you are only modeling 1/4 wave or 1/4 cycle.

My university memory is getting very rusty, but here is what I recall:

Resistor-Inductor-Capacitor (RLC) Electrical Circuit

Vl = L di/dt (Inductor voltage equals inductance multiplied by the time-rate-of-change of the current)
Vr = i R (Resistor voltage equals current multiplied by the resistance)
Vc = 1/C integral{i dt} (Capacitor voltage equals 1/C multiplied by the time-integral of the current)

L di/dt + i R + 1/C integral(i dt) = 0 (Loop equation)

i" + R/L i' + 1/LC i = 0 (2nd order differential equation)

Spring-Mass-Damper Mechanical System

Fs = -k x (Spring force equals spring constant k multiplied by displacement)
Fd = -d v = -d x' (Damping force equals damping factor d multiplied by velocity)

m a = -Fs - Fd (Sum of forces equation)
m x" = -k x - d x'

x" + d/m x' + k/m x = 0 (2nd order differential equation)


Going by the system differential equations:

V ~ F (Voltage is equivalent to force)
L ~ m (Inductance is equivalent to mass)
C ~ 1/k (Capacitance is equivalent to inverse spring constant 1/k)
i ~ x (Current is equivalent to displacement)
R ~ d (Electrical resistance is equivalent to damping factor d)

I did a quick scan of the following 2 wiki pages and believe my recollection above is correct. Please check the wiki links and other sources for verification. Hope this helps :-)

http://en.wikipedia.org/wiki/RLC_circuit
http://en.wikipedia.org/wiki/Damping

Dear Akash,

I guess you mean a "freely falling" body in the earth's gravitational field? It could also be a magnetic body and simultaneously affected by the earth's magnetic field.

Sticking to the gravitational effect alone, assuming the body has 0 initial velocity, and is near the earth's surface and subject to standard gravity (gravity varies with position on earth)...

1. mechanically v = gt, where v is velocity, g is acceleration due to gravity and t is time. The approximate standard value of g is 9.81 metres per second velocity increase per second.
2. Electrically, for a capacitor, Q = CV, where Q = charge, C = capacitance, V = voltage (units Coulombs, Farads and Volts). Also for a constant charging current I, Q = It, where I is amps and t is seconds. So also, substituting It for Q in first formula, It = CV.
3. Note similarity between v = gt and CV = It. Dividing both sides of CV = It by C, one gets V = (I/C)t - compare v = gt.
4. So one can make an equivalent circuit in which the voltage on a capacitor charged at constant current is analogous to velocity of a body under steady acceleration.

Since it is a "free" falling body, one can argue there is no friction, which is analogous to electrical resistance [note that a resistor in parallel with the capacitor would decrease charging current in proportion to 1/V and could be compared to friction proportional to velocity]. Also, there is no need for inductance in this case, so R and L would be zero in an RLC analogy.

In a more complicated case of a mass hanging on a spring under gravity, an electrical analogy would set capacitive stored energy with potential mechanical energy variation with height, resistance with friction and inductive stored energy with kinetic energy of the moving mass. The solution would be a damped oscillation.

I have no more time now, I hope this helps.

67model