Understanding the relationship between the dimensions of a crystal and its oscillation frequency is fundamental in many electronic applications, especially those involving crystal oscillators.

Crystal Oscillation Principle

The frequency of oscillation of a crystal, particularly a piezoelectric crystal like quartz, is primarily determined by its physical dimensions, especially its thickness. For a simple mode of vibration, the fundamental resonant frequency (\(f\)) of a crystal is inversely proportional to its thickness (\(t\)).

This relationship can be expressed mathematically as:

\(f \propto \frac{1}{t}\)

Or, more precisely, \(f = \frac{K}{t}\), where \(K\) is a constant that depends on the crystal material's properties (like Young's modulus and density) and the specific mode of vibration.

Thickness Reduction and Frequency Impact

The question states that the initial thickness of the crystal is 10 mm and this thickness is reduced by 2%.

Let's denote:

• Initial thickness = \(t_1\)
• Initial frequency = \(f_1\)
• New thickness = \(t_2\)
• New frequency = \(f_2\)

Step-by-Step Calculation:

1. Initial State:

   The initial frequency \(f_1\) is related to the initial thickness \(t_1\) by:

   \(f_1 = \frac{K}{t_1}\) (Equation 1)

2. Calculate New Thickness:

   The thickness is reduced by 2%. So, the reduction amount is \(0.02 \times t_1\).

   The new thickness \(t_2\) will be:

   \(t_2 = t_1 - (0.02 \times t_1)\)

   \(t_2 = t_1 (1 - 0.02)\)

   \(t_2 = 0.98 t_1\)

3. New State Frequency:

   The new frequency \(f_2\) is related to the new thickness \(t_2\) by:

   \(f_2 = \frac{K}{t_2}\) (Equation 2)

4. Relate Frequencies:

   Substitute the expression for \(t_2\) into Equation 2:

   \(f_2 = \frac{K}{0.98 t_1}\)

   We can rewrite this as:

   \(f_2 = \frac{1}{0.98} \times \frac{K}{t_1}\)

   From Equation 1, we know that \( \frac{K}{t_1} = f_1 \). So, substitute \(f_1\) into the equation for \(f_2\):

   \(f_2 = \frac{1}{0.98} f_1\)

5. Calculate Percentage Change:

   Now, let's find the numerical value of \( \frac{1}{0.98} \):

   \( \frac{1}{0.98} \approx 1.0204 \)

   So, \(f_2 \approx 1.0204 f_1\).

   To find the percentage change, we use the formula:

   \(\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%\)

   \(\text{Percentage Change} = \frac{f_2 - f_1}{f_1} \times 100\%\)

   \(\text{Percentage Change} = \frac{1.0204 f_1 - f_1}{f_1} \times 100\%\)

   \(\text{Percentage Change} = \frac{(1.0204 - 1)f_1}{f_1} \times 100\%\)

   \(\text{Percentage Change} = 0.0204 \times 100\%\)

   \(\text{Percentage Change} = 2.04\%\)

Thus, when the thickness is reduced by 2%, the frequency of oscillations increases by approximately 2.04%, which can be rounded to 2%.

Frequency and Thickness Relationship Summary

This direct calculation confirms the inverse relationship: when the thickness of a crystal decreases, its fundamental oscillation frequency increases. Conversely, if the thickness were to increase, the frequency would decrease.

Therefore, a 2% reduction in crystal thickness leads to an approximate 2% increase in its oscillation frequency.