An analog VU meter is just a moving-coil galvanometer. The figure below shows a schematic representation:

Schematic representation of a moving-coil instrument.
The needle moves on a pivot and is kept in place using a spiral-shaped torsion spring.
A rectangular coil is fixed to the needle. The coil moves in an air gap between the fixed permanent magnet and a
cylindrical iron core.
This configuration results in a strong, close to radial magnetic field in the air gap.

When current flows through the coil, the force on one half of the coil is given by Lorentz's Force Law:
$${\overrightarrow{F}}_{L}=Ni\overrightarrow{l}\times \overrightarrow{B},$$
where $N$ is the number of windings, $i$ is the current, $\overrightarrow{l}$ is a vector that points along the vertical
part of the wires of the coil, in the direction of the current and with the same length as the height of the coil,
and $\overrightarrow{B}$ is the magnetic field vector.

In the figure above, $\overrightarrow{B}$ is shown using black arrows, the vector $Ni\overrightarrow{l}$ is shown in red. The one on
the left points out of the page, and the one on the right points into the page. The resulting force vectors are
shown in green. You can verify their direction using the right-hand rule (middle finger is the magnetic
field vector $\overrightarrow{B}$, index finger is the current vector $Ni\overrightarrow{l}$, thumb is the force vector
${\overrightarrow{F}}_{L}$).

The magnitude of the Lorentz force is then
$${F}_{L}=\Vert {\overrightarrow{F}}_{L}\Vert =NilB\mathrm{sin}(\phi ),$$
where $\phi $ is the angle between the magnetic field vector and the direction of the current.
The angle is always $90\xb0$, so this simplifies to
$${F}_{L}=NilB.$$

The horizontal part of the coil (parallel to the page) experiences no net force, since the current at the bottom
and at the top of the coil flows in opposite directions and the magnetic field is symmetric.

The coil is wound around a rectangular piece of metal. When the needle moves, the armature (coil and the
rectangular piece of metal) rotates in the magnetic field. This rotation causes the magnetic flux through the
armature to change, which according to Faraday's Law of Induction induces a current in the metal, resulting
in a magnetic field that opposes the rotation (this is Lenz's Law). These induced currents are called Eddy
currents or Foucault's currents. The same principle is used for
induction brakes in high-speed trains,
for example.

The magnitude of this opposing force is proportional to the rate of change of the magnetic flux in the armature,
$\frac{\partial {\mathrm{\Phi}}_{B}}{\partial t}$. In a first order approximation, this is simply proportional to the angular
velocity of the armature, $\dot{\theta}={\displaystyle \frac{d\theta}{dt}}$, therefore
$${F}_{d}=-\mu {\textstyle \phantom{\rule{thinmathspace}{0ex}}}\dot{\theta},$$
where $\mu $ is a proportionality constant that takes into account the area of the armature, the resistance of
the metal it's made of, the strength of the magnetic field, etc.

Recall the formula for torque:
$$\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F},$$
where $\overrightarrow{\tau}$ is the torque, and $\overrightarrow{r}$ is the radius vector from the center of rotation to the
point of application of the force $\overrightarrow{F}$. Using scalar quantities, the formula becomes:
$$\tau =\Vert \overrightarrow{r}\Vert \Vert \overrightarrow{F}\Vert \mathrm{sin}(\psi ),$$
where $\psi $ is the angle between $\overrightarrow{r}$ and $\overrightarrow{F}$.

Because of the radial magnetic field in the air-gap, the angle $\psi $ is always $90\xb0$, so
the torque on the coil caused by the current is
$${\tau}_{L}=2r{F}_{L}=2rNilB=fi,$$
where $r$ is the radius of the coil. The factor $2$ accounts for the two halves of the coil (there are two
green force vectors that have to be summed together). Since the number of windings, the dimensions of the coil,
and the strength of the magnet are all constant, we introduce a constant $f\triangleq 2rNlB$.

The damping force ${F}_{d}$ caused by induction in the armature causes an opposing torque, proportional to the
velocity of the needle:
$${\tau}_{d}=-r\mu {\textstyle \phantom{\rule{thinmathspace}{0ex}}}\dot{\theta}=-c\dot{\theta}.$$
Again, a constant $c\triangleq r\mu $ is introduced to make future calculations easier.

When the needle moves away from its center position, there is a third torque, caused by the torsion spring. This
torque can be approximated using Hooke's law:
$${\tau}_{H}=-k\theta ,$$
where $k$ is the torsion coefficient (spring constant), and $\theta $ is the angle of the needle.

Newton's Second Law ($F=ma$) has a rotational equivalent that relates the net torque to the angular
acceleration of an object:
$$\tau =I\alpha ,$$
where $\tau $ is the net torque, $I$ is the moment of inertia, and
$\alpha =\ddot{\theta}={\displaystyle \frac{{d}^{2}\theta}{d{t}^{2}}}$ is the
angular acceleration.

Applying this law to the three torques derived above:
$$\begin{array}{rl}{\tau}_{L}+{\tau}_{d}+{\tau}_{H}& =I\alpha \\ fi-c\dot{\theta}-k\theta & =I\ddot{\theta}\\ I\ddot{\theta}+c\dot{\theta}+k\theta & =fi.\end{array}$$

Both the angle $\theta $ and the current $i$ vary over time. This results in the differential equation
$$\begin{array}{}\text{(1)}& I\ddot{\theta}(t)+d\dot{\theta}(t)+k\theta (t)=fi(t).\end{array}$$

Recall that $I$, $c$, $k$ and $f$ are physical constants, $\theta (t)$ is the angle of the needle, and
$i(t)$ is the current applied to the coil. Overdots indicate derivatives with respect to time.