Driven resonance v. natural resonance

BunmiFariyike
I am in a physics lab course and we set up the following circuit.

Given that we had a 150 milliHenry capacitor and a .5 microfarad capacitor, we calculated the expected resonance frequency using the formula for the natural frequency of an LC circuit: omega = 1/ sqrt(LC). We then measured the driven resonance frequency through an oscilloscope. Obviously, the two values should differ slightly because the natural resonance isn't the same as the driven one due to damping from the resistor and the internal resistance of the circuit. However, this should make the driven resonance smaller than the natural one, not larger, right? For all of the measured values, our resonance frequency was larger than what was given by the formula and I need to explain why in my report, but I have researched for hours and genuinely don't know. Can anyone help? Thanks.

The 150 mH was for an inductor, correct?

Were the L and C in series or in parallel? Also, what frequency did you expect to see and what did you measure?

BunmiFariyike
Were the L and C in series or in parallel? Also, what frequency did you expect to see and what did you measure?
My apologies. I thought the picture of the circuit would display. L and C were in series. We tested 3 different resonance values: 10, 50, and 500 ohms and I obtained values for angular frequency of 3727, 4242, and 3759, respectively. The expected value was 3651.48. Thank you.

I didn't see any picture with what you submitted. Normally I would calculate the frequenct - f - rather than omega. This will give you 581.2 hertz.

By calculating the frequency, the result is a smaller result so the 3 test values will appear closer. Percentagewise, they will be the same.

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