A simple pendulum of length 2.45 m. Find the period of oscillation (g = 9.8 m/s²) — Physics problems | PhysLabo
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Period of a simple pendulum

Easy 11 grade Oscillations / Mechanical Oscillations
Problem statement

A simple pendulum has a length $l = 2.45$ m. Find its period of oscillation. The acceleration due to gravity is $g = 9.8$ m/s².

Given
  • $l = 2.45$ m
  • $g = 9.8$ m/s²
  • $T = ?$
Hints
Hint 1 / 2
The period depends only on the length: $T = 2\pi\sqrt{l/g}$.
Hint 2 / 2
$2.45 / 9.8 = 0.25$, and the square root of $0.25$ is $0.5$.
Solution

The period of a simple pendulum:

$$T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{2.45}{9.8}} = 2\pi \sqrt{0.25} = 2\pi \cdot 0.5 = \pi \approx 3.14 \text{ s}$$

Answer

$T \approx 3.14$ s

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