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JAMB Physics 2013 Paper

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Question : 4

Total: 43

A simple pendulum of length 0.4m has a period of 2s. What is the period a similar pendulum of length 0.8m at the same place?

8s
4s
2 $\sqrt{2}s$
$\sqrt{2}s$

Solution:

Concept:

The period of a simple pendulum is proportional to the square root of its length.

Explanation:

For a simple pendulum,

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

.

So, at the same location,

T \propto \sqrt{L}

.

Given:

T_1 = 2\,s

,

L_1 = 0.4\,m

, and

L_2 = 0.8\,m

.

Using the relation

\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}

, we get:

T_2 = T_1\sqrt{\frac{L_2}{L_1}} = 2\sqrt{\frac{0.8}{0.4}}

T_2 = 2\sqrt{2}\,s

Answer:

The period is

2\sqrt{2}\,s

, which is option C.

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