Question

In the absence of a nearby metal object, the two inductances (LA and LB) in a heterodyne metal detector are the same, and the resonant frequencies of the two oscillator circuits have the same value of 510.0 kHz. When the search coil (inductor B) is brought near a buried metal object, a beat frequency of 6.5 kHz is heard. By what percentage does the buried object reduce the inductance of the search coil?

Answer 1

fo=constant/L

so if fo changes by (510+-6.5)/510, then l must change by the inverse of that.

L=510/516.5 or percent change is -6.5/510

Answer 2

That wasn't the right answer, but thank you anyways!