Problem 42 - Calculus

Question

Resonance in electric circuits leads to the expression $$ \left(\omega L-\frac{1}{\omega C}\right)^{2} $$ where \(\omega\) is the variable and \(L\) and \(C\) are constants. (a) Find \(\omega_{0},\) the value of \(\omega\) making the expression zero. (b) In practice, \(\omega\) fluctuates about \(\omega_{0},\) so we are interested in the behavior of this expression for values of \(\omega\) near \(\omega_{0} .\) Let \(\omega=\omega_{0}+\Delta \omega\) and expand the expression in terms of \(\Delta \omega\) up to the first nonzero term. Give your answer in terms of \(\Delta \omega\) and \(L\) but not \(C\)

Step-by-Step Solution

Answer

\(\omega_0 = \sqrt{\frac{1}{LC}}\) and expansion gives \(L^2 (\Delta \omega)^2\).

Step 1: Set Expression to Zero

We start by setting the expression \(\left(\omega L - \frac{1}{\omega C}\right)^{2}\) to zero. This implies taking the inside part of the square equal to zero: \(\omega L - \frac{1}{\omega C} = 0\). This simplifies to \(\omega L = \frac{1}{\omega C}\).

Step 2: Solve for \(\omega_0\)

Multiply both sides by \(\omega\) to eliminate the fraction: \(\omega^2 L = \frac{1}{C}\). To isolate \(\omega\), divide both sides by \(L\): \(\omega^2 = \frac{1}{LC}\). Finally, solving for \(\omega\) gives \(\omega_0 = \sqrt{\frac{1}{LC}}\).

Step 3: Substitute \(\omega = \omega_0 + \Delta \omega\)

Substitute \(\omega = \omega_0 + \Delta \omega\) into the original expression: \(\left((\omega_0 + \Delta \omega) L - \frac{1}{(\omega_0 + \Delta \omega) C}\right)^{2}\).

Step 4: Expand using Taylor Series

In expressions involving small perturbations, use \(1/(\omega_0 + \Delta \omega) \approx 1/\omega_0 - (\Delta \omega/\omega_0^2)\). Substituting back, replace \((\omega_0 + \Delta \omega) L\) and \(1/(\omega_0 + \Delta \omega) C\).

Step 5: Simplify

The simplified expression becomes \(\left((\omega_0 + \Delta \omega) L - \left(\frac{1}{\omega_0 C} - \frac{\Delta \omega}{\omega_0^2 C}\right)\right)^{2}\). Simplifying this results in an expression of smaller terms including \(\Delta \omega\).

Step 6: Find the First Nonzero Term

The term linear in \(\Delta \omega\) cancels because of our choice of \(\omega_0\). The quadratic term in \(\Delta \omega\) remains, leading to \(L^2 (\Delta \omega)^2\).

Key Concepts

Taylor Series Expansion

Taylor series is a powerful tool in mathematics that helps us approximate functions. It's especially useful when we are looking at how a function behaves around a certain point. For the given exercise, we're interested in the behavior of the expression \((\omega L - \frac{1}{\omega C})^2\) near the resonance frequency \(\omega_0\).

By setting \( \omega = \omega_0 + \Delta \omega \), we introduce a small perturbation \( \Delta \omega \) around \( \omega_0 \). The core idea of using a Taylor series here is to expand the expression to reveal how it changes with respect to \( \Delta \omega \). This is done by approximating the reciprocal component as follows:

\[ \frac{1}{\omega_0 + \Delta \omega} \approx \frac{1}{\omega_0} - \frac{\Delta \omega}{\omega_0^2} \]

This expansion allows us to replace complex terms with simpler expressions involving only \( \omega_0 \) and \( \Delta \omega \). This is useful for focusing on how the circuit behaves when the frequency deviates slightly from its resonant frequency.

Circuit Analysis

In circuit analysis, understanding resonance is crucial because it helps us predict how a circuit will perform at certain frequencies. When we examine the expression \((\omega L - \frac{1}{\omega C})^2\), we're essentially exploring the relationship between the frequency \(\omega\), the inductance \(L\), and the capacitance \(C\) of the circuit.

Resonance occurs when the inductive and capacitive reactances are equal, making the impedance minimal or zero. This creates a situation where the circuit can efficiently transfer energy or respond to oscillations without resistance from reactance.

Finding \(\omega_0 = \sqrt{\frac{1}{LC}}\) identifies the frequency at which resonance happens. This insight is significant because at this point, the overall circuit reactance is zero, and current flow through the circuit is highest, which is often desirable behavior in circuits like tuners and filters.

Nonzero Terms in Expansions

When performing expansions, particularly Taylor series, identifying the "first nonzero term" can be very enlightening. Once we set \( \omega = \omega_0 + \Delta \omega \), and expand the expression \((\omega L - \frac{1}{\omega C})^2\), we are interested to see which terms do not vanish for small \( \Delta \omega \).

In our exercise, we found that the linear term in \( \Delta \omega \) cancels out, which is typical due to selecting \( \omega_0 \) appropriately as the point where resonance happens. The next term, which is quadratic in \( \Delta \omega \), does not cancel and becomes very important.

This term was found to be \( L^2 (\Delta \omega)^2 \). This remaining term tells us how the expression grows or changes when \(\omega\) deviates from \(\omega_0\). Specifically, it implies that small deviations in frequency result in changes to the circuit's behavior, crucial for sensitivity analysis and predictions in practical circuit design.