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?
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
That wasn't the right answer, but thank you anyways!
To find the percentage by which the buried object reduces the inductance of the search coil, we need to understand the relationship between inductance and resonant frequency in a heterodyne metal detector.
In a heterodyne metal detector, two oscillator circuits are used. One oscillator circuit, connected to an inductor (LA), operates at a fixed resonant frequency of 510.0 kHz. The other oscillator circuit, connected to the search coil (LB), is affected by the presence of a nearby metal object, resulting in a beat frequency of 6.5 kHz.
The beat frequency is the difference between the frequencies of the two oscillator circuits. In this case, we have:
Beat frequency = Resonant frequency of LA - Resonant frequency of LB
6.5 kHz = 510.0 kHz - Resonant frequency of LB
Rearranging the equation to find the resonant frequency of LB:
Resonant frequency of LB = 510.0 kHz - 6.5 kHz
Resonant frequency of LB = 503.5 kHz
Now, we know that the resonant frequency of the search coil (LB) changes due to the presence of the buried metal object. This change in the resonant frequency indicates a change in the inductance of the search coil.
The resonant frequency of an oscillator circuit is given by the formula:
Resonant frequency = 1 / (2π√(LC))
Where L is the inductance and C is the capacitance of the circuit.
Since the two inductances (LA and LB) are initially the same, we can assume that the capacitances are the same as well. Therefore, we can rewrite the equation as:
Resonant frequency = 1 / (2π√L)
Now we can use this equation to find the initial value of inductance (L_initial) for the search coil when there is no metal object nearby:
Resonant frequency_initial = 1 / (2π√L_initial)
Resonant frequency_initial = 503.5 kHz
Solving this equation for L_initial:
L_initial = (1 / (2π√Resonant frequency_initial))^2
L_initial ≈ (1 / (2π√503.5 kHz))^2
L_initial ≈ 1.56 × 10^-7 H
Now, let's consider the inductance of the search coil (LB) when a metal object is present. We'll denote this as L_object.
Using the same equation as before, but with the new resonant frequency:
Resonant frequency_object = 1 / (2π√L_object)
Resonant frequency_object = 510.0 kHz
Solving this equation for L_object:
L_object = (1 / (2π√Resonant frequency_object))^2
L_object ≈ (1 / (2π√510.0 kHz))^2
L_object ≈ 1.52 × 10^-7 H
Now that we have the initial value of inductance (L_initial) and the inductance with the metal object present (L_object), we can calculate the percentage reduction in inductance:
Percentage reduction = ((L_initial - L_object) / L_initial) × 100
Percentage reduction ≈ ((1.56 × 10^-7 H - 1.52 × 10^-7 H) / 1.56 × 10^-7 H) × 100
Percentage reduction ≈ (0.04 × 10^-7 H / 1.56 × 10^-7 H) × 100
Percentage reduction ≈ 25.64%
Therefore, the buried object reduces the inductance of the search coil by approximately 25.64%.