# Deriving relationship between LVDT and mass spring damper

I have this system below.

It is the schematic of a linear accelerometer moving horizontally, where $m$ is the total mass of the slide, $b$ denotes the viscous damping, and $k$ represents the spring constant. The relative position between the moving mass and the case is measured by a linear variable differential transformer (LVDT).

How do I derive the dynamic equation describing the relationship between $V$, the voltage output of the LVDT, and $x_1$ , the external position, and show that $V$ indeed can be used to measure the external acceleration? I'm supposed to also state any assumptions made and any possible problems and remedies.

I know that the mass spring damper system has the equation $mx_1'' + bx_1' + kx = 0$, and the moving core that is pulled by the mass spring damper system will induce a voltage in the LVDT. If I was given a transfer function for the LVDT, $G$, then I have $V = Gx_1$. But I don't know how to link the 2 concepts together to derive the relationship.

As for the assumptions, I would say the damping force is constant, i.e. the case is smooth. I'm wondering how the mass hitting the stopper will affect the system. Should it be made of rubber to absorb the impact and to minimize the shock wave?

• Isn't the beauty of LVDT that the voltage is a linear scaled version of x1 within the operating range? In other words, V = G*x1 and G is a constant gain? My guess is the stopper prevents the mass from moving outside the linear range of the LVDT, you don't want to mess with it. macrosensors.com/lvdt_tutorial.html – willpower2727 Nov 25 '15 at 17:38
• I think I'm supposed to derive the G from the info given. But I'm not sure how to get it from the force from the spring system. – Rayne Nov 26 '15 at 10:14
• @ Rayne what if G (the transfer function of the LVDT) was a constant gain? What would be sufficient to say the LVDT is appropriate for measuring x1? – willpower2727 Nov 30 '15 at 20:31