A simple pendulum has a period of oscillation of 1.6\text{ s} on the surface of the Earth,… — JAMB UTME

A simple pendulum has a period of oscillation of $1.6\text{ s}$ on the surface of the Earth, where the acceleration due to gravity is $10.0\text{ m/s}^2$ . What is the period of oscillation of this pendulum when placed on a moon where the acceleration due to gravity is $2.5\text{ m/s}^2$ ?

Answer: 3.2 s

Answer

The period of oscillation of the pendulum on the moon is

3.2\text{ s}

The period of a simple pendulum is given by

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

. Because the length

l

is constant, period is inversely proportional to the square root of acceleration due to gravity (

T \propto \frac{1}{\sqrt{g}}

). Reducing the local gravity from

10.0\text{ m/s}^2

2.5\text{ m/s}^2

decreases gravity by a factor of 4, which increases the period by a factor of

\sqrt{4} = 2

. Multiplying the initial period of

1.6\text{ s}

by 2 yields

3.2\text{ s}

Step-by-Step Solution

Relate the period of oscillation of a simple pendulum to gravitational acceleration.

The period formula is

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

, showing that

T

is inversely proportional to

\sqrt{g}

The length of the pendulum

l

remains unchanged.

Formulate a ratio comparing the pendulum's period on the moon to its period on Earth.

\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}}

Dividing the two equations cancels the constant terms

2\pi

and

\sqrt{l}

Substitute the known numerical values and solve for

T_{moon}

T_{moon} = 1.6 \times \sqrt{\frac{10.0}{2.5}} = 1.6 \times 2.0 = 3.2\text{ s}

The ratio of gravities is 4, whose square root is 2, doubling the initial period.

Key Concept

Dependence of Simple Pendulum Period on Gravitational Acceleration

Estimated Time:1m 30s

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