Part a of the figure shows a heterodyne metal detector being used. As part b of the

Question:

Part a of the figure shows a heterodyne metal detector being used. As part b of the figure illustrates, this device utilizes two capacitor/ inductor oscillator circuits, A and B. Each produces its own resonant frequency, f0A = 1/[2π(LAC)1/2] and f0B = 1/[2π(LBC)1/2]. Any difference between these frequencies is detected through earphones as a beat frequency |f0B - f0A |. In the absence of any nearby metal object, the inductances LA and LB are identical. When inductor B (the search coil) comes near a piece of metal, the inductance LB increases, the corresponding oscillator frequency f0B decreases, and a beat frequency is heard. Suppose that initially each inductor is adjusted so that LB = LA, and each oscillator has a resonant frequency of 855.5 kHz. Assuming that the inductance of search coil B increases by 1.000% due to a nearby piece of metal, determine the beat frequency heard through the earphones.