A simple pendulum suspended in a physics laboratory has a length of 0.64\text{ m}. Given t… — JAMB UTME

A simple pendulum suspended in a physics laboratory has a length of $0.64\text{ m}$ . Given that acceleration due to gravity $g = 10\text{ m/s}^2$ and taking $\pi^2 = 10$ , what is the period of oscillation of the pendulum in seconds?

Answer: 1.6 s

Answer

The period of oscillation of the simple pendulum is

1.6\text{ s}

Using the pendulum period relationship

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

, substituting

L = 0.64\text{ m}

g = 10\text{ m/s}^2

, and

\pi = \sqrt{10}

yields

T = 2\sqrt{10}\sqrt{\frac{0.64}{10}} = 2\sqrt{0.64} = 1.6\text{ s}

Step-by-Step Solution

Identify the formula for the period of a simple pendulum.

The period formula is

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

The period depends on the length of the pendulum

L

and acceleration due to gravity

g

Substitute the given values into the formula.

T = 2\pi \sqrt{\frac{0.64\text{ m}}{10\text{ m/s}^2}}

Given parameters are

L = 0.64\text{ m}

and

g = 10\text{ m/s}^2

Simplify the equation using

\pi = \sqrt{10}

T = 2\sqrt{10} \times \sqrt{0.064} = 2 \sqrt{10 \times 0.064} = 2 \sqrt{0.64} = 2 \times 0.8 = 1.6\text{ s}

Using

\pi^2 = 10

simplifies the calculation cleanly without requiring a calculator.

Key Concept

Simple Pendulum Period of Oscillation

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