EMI BYU 4 - Physics.lt

Solution to Problem 4

Solution: Oscillations of a Rod on Parallel Rails

1. Calculation of Current (Using Equivalent Battery Method)

Consider the rod displaced by a distance $x$. We can model the circuit using the equivalent EMF ($\varepsilon_{eq}$) and equivalent resistance ($r_{eq}$) of the two parallel rail sections.

Equivalent EMF ($\varepsilon_{eq}$):

$$ \varepsilon_{eq} = \frac{V[(R_0 + 2\rho x) – (R_0 – 2\rho x)]}{2R_0} $$

Simplifying the numerator:

$$ \varepsilon_{eq} = \frac{V(4\rho x)}{2R_0} = \frac{2V\rho x}{R_0} $$

Equivalent Internal Resistance ($r_{eq}$):
Using the formula for parallel resistors $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$:

$$ r_{eq} = \frac{(R_0 + 2\rho x)(R_0 – 2\rho x)}{2R_0} $$

For small displacements where $\rho x \ll R_0$, we can approximate:

$$ r_{eq} \approx \frac{R_0}{2} $$

Net Current ($i$):
Now, calculating the current through the load resistance $R$:

$$ i = \frac{\varepsilon_{eq}}{r_{eq} + R} $$ $$ i = \frac{\frac{2V\rho x}{R_0}}{\frac{R_0}{2} + R} = \frac{4V\rho x}{R_0(R_0 + 2R)} $$

2. Equation of Motion

The magnetic force acting on the rod is $F = i \ell B$. Substituting the expression for current:

$$ F = \left[ \frac{4V\rho \ell B}{R_0(2R + R_0)} \right] x $$

This force acts as a restoring force $F = -kx$, where the effective spring constant $k$ is:

$$ k = \frac{4V\rho \ell B}{R_0(2R + R_0)} $$

Thus, the rod executes Simple Harmonic Motion (SHM).

3. Angular Frequency Calculation

The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}}$:

$$ \omega = \sqrt{\frac{4V\rho \ell B}{m R_0(2R + R_0)}} $$

Using $R_0 = \rho L$ (total resistance of the rails):

$$ \omega = \sqrt{\frac{4V \ell B}{m L (2R + \rho L)}} $$