A simple pendulum on Earth (g = 10\text{ m/s}^2) completes 50 full oscillations in 40\text… — JAMB UTME

A simple pendulum on Earth ( $g = 10\text{ m/s}^2$ ) completes $50$ full oscillations in $40\text{ s}$ . The pendulum is then transferred to a lunar station where the acceleration due to gravity is $1.6\text{ m/s}^2$ , and its length is reduced by $64\%$ . What is the time taken, in seconds, for this modified pendulum to complete $30$ oscillations on the lunar station?

Answer: 36 s

Answer

The time taken for the modified pendulum to complete 30 oscillations on the lunar station is 36 s.

The initial period on Earth is

T_1 = \frac{40}{50} = 0.8\text{ s}

. The formula for the period of a simple pendulum is

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

. When length decreases by

64\%

, the remaining length ratio is

\frac{L_2}{L_1} = 0.36

. The ratio of gravity is

\frac{g_1}{g_2} = \frac{10}{1.6} = 6.25

. Taking the ratio gives

\frac{T_2}{T_1} = \sqrt{0.36 \times 6.25} = \sqrt{2.25} = 1.5

. Thus, the new period is

T_2 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

. For

30

oscillations, the total time is

t = 30 \times 1.2\text{ s} = 36\text{ s}

Step-by-Step Solution

Calculate the initial period of oscillation on Earth

T_1 = 0.8\text{ s}

Period

T_1

is total time divided by the number of oscillations:

T_1 = \frac{40\text{ s}}{50} = 0.8\text{ s}

Set up the ratio for period under altered length and gravitational field

\frac{T_2}{T_1} = 1.5

Using

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

, we have

\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1} \cdot \frac{g_1}{g_2}} = \sqrt{(1 - 0.64) \cdot \frac{10}{1.6}} = \sqrt{0.36 \cdot 6.25} = \sqrt{2.25} = 1.5

Determine the new period of oscillation

T_2 = 1.2\text{ s}

T_2 = 1.5 \times T_1 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

Calculate the total time required for 30 oscillations

t_2 = 36\text{ s}

Total time

t_2 = N_2 \times T_2 = 30 \times 1.2\text{ s} = 36\text{ s}

Key Concept

Period of a simple pendulum and its dependence on length and gravitational acceleration

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