Two electrical oscillators are used in a heterodyne metal detector to detect bur

Question

Two electrical oscillators are used in a heterodyne metal detector to detect buried metal objects. The detector uses two identical electrical oscillators in the form of LCcircuits having resonant frequencies of 700 kHz. When the signals from the two oscillating circuits are combined, the beat frequency is zero because each has the same resonant frequency. However, when the coil of one circuit encounters a buried metal object, the inductance of this circuit increases by 1.200%, while that of the second is unchanged. Determine the beat frequency that would be detected in this situation.

Answer should be in kHz

Explanation / Answer

Resonant frequency, f = (1/(2*pi))*(1/sqrt(LC))

when L changes by 1.2 percent,

Now, new inductance, L' = 1.012*L

So, f' = (1/(2*pi))*(1/sqrt(1.012*L*C)) = 0.994*f

So, Earlier both of them were identical , so f1 = f , f2 = f

So, beat frequency = f1 - f2 = 0

But, now, f1 = f ,f2 = f' = 0.994*f

So, beat freqeucncy now, B = f1 - f2

So, B= 0.006*f

= 0.006*700

= 4.16 kHz <---------answer

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