Coupling mechanical to electrical energy in the domain of electric guitars

I'm working on a project for university where I want to find the coupling between the mechanical energy of an oscillating string and the electric energy in the pickup due to the current induction.

My idea was that I apply a certain input mechanical energy of the string, measure the output of the guitar for the electric energy, and set both equal. Of course, we need to consider that there are losses since the coupling of the mechanical to electric energy is not perfect.

So basically

$$F x \eta = \frac{1}{2} L I^2$$ where $$F$$ is the force applied to displace the string, $$x$$ is the displacement of the string, $$\eta$$ is the efficiency, $$L$$ is the inductance of the pickup, and $$I$$ is the current induced in the pickup.

I tied a spring scale to the string with a thread so that the string was displaced. I then cut through the thread. The trigger of the scope would then go off and measure the output signal of my guitar.

I then took the RMS value of the output voltage and the impedance of my pickup (at the corresponding frequency) to calculate $$I_\text{rms}$$ and use it in the formula above.

1. In the feedback that I got, I was told that this is wrong since the electrical energy I used only applies for coils without losses charged with a DC current (I would have though that using RMS accounted for that). However, I have a system with AC current, so I have watt-less power or oscillating energies, i.e. energy is oscillating between the magnetic field and source back and forth. This can be described with the (blind) current but does not have any information about the transmitted power, which equates to an active current (Ohmic).

2. Moreover, the way my measurements were done are not correct. The suggestion is to tie the supply current of the coil to the resulting amplitude of the string.

A friend of mine made a suggestion about the electric energy, which in hindsight makes a lot of sense. I used the textbook formula which applies for certain conditions (as described in feedback), but I probably need to derive the equation myself.

With the help of µPad I went about it like this:

\begin{align} i(t) &= A \sin(\omega t) \\ u(t) &= L \frac{d\,i(t)}{dt} \, . \end{align} Since the electric power is known as $$P=ui$$ I could get the energy through integration with $$W(t) = \int P(t) \,d t =-\frac{A^{2} L \cos(2 \omega t)}{4} \, .$$

Do you think that this is correct?

I'm not sure what that really means. I can't actuate my pickups with current because I might risk damaging them. That's why I want to calculate the coupling in the first place. Also I don't see the difference to what I did. The coupling should be more or less the same in both directions. Or wouldn't it?

@V.F.: Thank you for this in depth reply. It is not exactly the answer I was looking for, but that's my fault. I realize I left out my intention in the original question. However your reply is very valuable to me because of the underlying concept you explained.

If it's okay I'd like to summarize it to make sure whether I understood the principle correctly, or not:

For my problem there are at least two types of energy losses that are of interest to me.

1) The mechanical losses Which indeed I have measured and calculated in a similar fashion as you described, because I need to know the envelope of the decaying function of the string due to the friction and clamping forces that hold the string in place.

2) The losses due to the coupling of the strings with the transducer The kinetic energy of the string is converted to electric energy, resulting in a current. The energy oscillates back and forth between the string and the pickup. This energy conversion has its losses.

The losses of point 2) might be the coupling I'm looking for. Is it legit to estimate the electric energy in the pickup due to the energy I apply with a spring scale and aforementioned loss factor?

My main goal is to calculate the current I'd need to apply to the pickup to generate a force large enough to compensate for the decay of the guitar string and keep it oscillating at a certain amplitude.

This I could do with calculating the mechanical energy I want to produce and then use the loss factor to get the electrical energy I'd need, or am I missing something again?

• "My postulation was that I apply a certain input energy (mechanical) of the string and measure the output of the guitar for the electric energy and set both equal" That doesn't make sense. If all the mechanical energy went into the pickup, because of conservation of energy, the string would immediately stop vibrating - but it continues to vibrate for a long time. I'm struggling to give you any advice beyond "all this looks completely wrong" though. You seem very confused about the basic ideas of energy and power. and what the pickup transducers actually do in an electric guitar. – alephzero Oct 6 '18 at 19:00
• I would get a big kick out of working with you on this project and have some ideas which may be useful. can we connect outside of this forum? – niels nielsen Oct 6 '18 at 19:04
• I've converted a response to one answer into an edit on this question. DanMan, you might like to re-edit so that the question reads more straightforwardly with your additional remarks. – rob Oct 7 '18 at 19:46

Here is a possible way to approach this problem.

A vibrating string changes the magnetic flux going through a pickup coil, which induces some emf into the coil.

If the coil is not shorted or loaded by some impedance to complete the circuit, there won't be any current flowing though it and, therefore, there won't be any power transfer from the string to the coil.

If you observe the voltage on the output on the unloaded coil, you'll see that it will be exponentially decaying. This reflects the exponential decay of the string damped vibrations, which happens because, every cycle, the string is losing a certain fraction of its energy due to mechanical losses. The higher the percentage of the losses, the faster the decay. But, since the fraction of energy loss over each cycle is very small, it takes lots of cycles to exhaust the initial string energy.

Let's say that the string loses a certain fraction $$x$$ of its energy every cycle. Then, after n cycles, its energy will be $$E_n=E_0(1-x)^n$$. Since the energy is proportional to the square of the amplitude, the amplitude, after $$n$$ cycles, will be $$A_n=A_0 \sqrt {(1-x)^n}$$. So, if we know the initial amplitude, $$A_0$$, and the amplitude after $$n$$ cycles $$A_n$$, we can calculate the fraction of energy loss per cycle as $$x=1-{\big (\frac {A_n}{A_0}}\big)^{\frac 2 n}$$.

If we assume that the amplitude of the voltage (emf) induced into the coil is proportional to the amplitude of the string vibrations, by measuring the amplitude of the voltage on the coil in a couple of different points (say, at the beginning of vibrations and $$100$$ cycles later) and applying the above formula, we can roughly estimate the energy loss of the string per cycle.

If you complete the circuit by loading the coil with a resistor or with an amplifier, some current will flow in the circuit and some power will be consumed. Naturally, this power will come from the vibrating string and, therefore, it will cause some additional energy losses in the string, which should accelerate the decay of the string vibrations.

If you observe the voltage on the output of the loaded coil, you may notice that the voltage will be lower and that it will decay faster than it was with an open circuit. By measuring the voltage in a couple of different points, as before, and applying the same formula, you'll find that the fraction of energy loss per cycle has increased.

Now, since you know the total fractional energy loss per cycle and the energy loss associated with mechanical losses, you can estimate the energy loss due to the pickup coil, which will amount to the energy transfer rate from the string to the coil, which, I believe, is what you are looking for.