A student records the period of oscillation of a simple pendulum by measuring the time tak… — JAMB UTME

A student records the period of oscillation of a simple pendulum by measuring the time taken for 20 complete swings as $40.0\text{ s}$ . If the absolute error in this time measurement is $\pm 0.8\text{ s}$ , what is the percentage error in the measured time?

A
$0.2\%$
$2.0\%$ Answer
C
$5.0\%$
D
$8.0\%$

Answer

2.0\%

The correct answer is

2.0\%

. Percentage error is defined as the absolute error divided by the measured value, expressed as a percentage:

\frac{0.8\text{ s}}{40.0\text{ s}} \times 100\% = 2.0\%

Step-by-Step Solution

Identify the given measurement and absolute error.

Measured time

t = 40.0\text{ s}

and absolute uncertainty

\Delta t = 0.8\text{ s}

These quantities are needed to determine the relative error of the time measurement.

Apply the percentage error formula:

\text{Percentage Error} = \left(\frac{\Delta t}{t}\right) \times 100\%

\text{Percentage Error} = \left(\frac{0.8\text{ s}}{40.0\text{ s}}\right) \times 100\% = 0.02 \times 100\% = 2.0\%

Multiplying the fractional error by 100 converts it into a percentage representation.

Key Concept

Percentage Error in Physical Measurements

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