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JAMB Physics 2023 Paper

Question : 25

Total: 65

Solution:

Concept:

For a simple pendulum, the square of the period is proportional to its length:

T^2 \propto L

Explanation:

Let the original length be

L

metres.

The new length after shortening by

3\,\mathrm{m}

L - 3

metres.

Given:

T_1 = 5.77\,\mathrm{s}

and

T_2 = 4.60\,\mathrm{s}

Since

T^2 \propto L

, we can write

T^2 = kL

, where

k

is a constant.

So,

\frac{T_1^2}{L} = \frac{T_2^2}{L - 3}

Substitute the values:

\frac{5.77^2}{L} = \frac{4.60^2}{L - 3}

\frac{33.29}{L} = \frac{21.16}{L - 3}

Cross multiply:

33.29(L - 3) = 21.16L

33.29L - 99.87 = 21.16L

12.13L = 99.87

L = \frac{99.87}{12.13} = 8.23\,\mathrm{m}

The new length is

L - 3 = 8.23 - 3 = 5.23\,\mathrm{m}

Answer:

The new length of the pendulum is

5.23\,\mathrm{m}

. Option A.

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