Tuning an LC Circuit to Span 540 kHz Range

[purduegirl]

Homework Statement

A radio receiver contains an LC circuit whose natural frequency of oscillation can be adjusted, or tuned, to match the frequency of the incoming radio waves. The adjustment is made by means of a variable capacitor. Suppose that the inductance of the circuit is 11.00 μH. What capacitance must the capacitor be adjusted to if the circuit is to span the 540.00 kHz range?

Homework Equations

$$\omega$$ = $$\sqrt{L/C}$$

The Attempt at a Solution

$$\omega$$ = $$\sqrt{L/C}$$
$$\omega$$ = 540.00 kHz
L = 11.00 microH
C = what we're looking for

I solved for C getting C = $$\frac{1}{\omega^2 L}$$
C = $$\frac{1}{(540000Hz)^2 * .000011 H}$$
C = $$\frac{1}{3.20E6}$$
C = 3.1176E-7 F

Where am I going wrong?

 

[purduegirl]

C = 1/3207600

 

[marcusl]

You are mixing up frequency with angular frequency. Remember that

$$\omega=2\pi f$$