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Crystal oscillator - Wikipedia

A crystal oscillator is an electronic oscillator circuit that **uses** the mechanical
resonance of a vibrating crystal of piezoelectric material to create an electrical
signal ...

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**Physics** Notes on **Units** and **Measurement** for CBSE Class 11

Jun 16, 2017 **...** Class 11 **Physics** notes on **Units** & **Measurement** (Chapter 2 of 11th ... **Error**,
**Relative Error** and **Percentage Error**, **Combination** of **Errors**, ... The result of every
**measurement** by any **measuring** instrument contains some uncertainty. ... **Least**
**count error** belongs to the category of random **errors** but within a ...

For more information, see **Physics** Notes on **Units** and **Measurement** for CBSE Class 11

The relative error in a physical quantity raised to the power k is equal to k times the individual quantity's relative error.

Z = \frac{A^p B^q}{C^r}

\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}

Given that

Time period of oscillations T = 2\pi \sqrt{\frac{L}{g}}

Length L = 20 cm

Least count error of length \Delta L = 1 mm = 0.1 cm

Time period of oscillations t = 90 sec

Least count error of time \Delta t = 1 sec

Step 1: Set up an equation for g

T = 2\pi \sqrt{\frac{L}{g}}

Square on both sides

T^2 = 4\pi^2 \frac{L}{g}

g = 4\pi^2 \frac{L}{T^2}

Relative error \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} ...................(1)

Time taken for n oscillations = t

Time taken for 1 oscillation = ?

Time period one oscillation T = \frac{t}{n}

Error \Delta T = \frac{\Delta t}{n}

We can write, \frac{\Delta T}{T} = \frac{\Delta t}{t}

Substitute \frac{\Delta T}{T} value in equation (1)

Hence, Relative error \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta t}{t}

Step 2: Finding the relative and percentage error

Relative error \frac{\Delta g}{g} = \frac{0.1}{20} + 2 \frac{1}{90}

Relative error \frac{\Delta g}{g} = 0.027

Percentage error \frac{\Delta g}{g} * 100 = 0.027 * 100

Percentage error = 2.7 % \approx3%

Therefore, Accuracy in determination of g = 3%