A simple pendulum of length L and bob mass m has a period of oscillation T on the surface … — JAMB UTME

A simple pendulum of length $L$ and bob mass $m$ has a period of oscillation $T$ on the surface of the Earth. The length of the pendulum is increased by $44\%$ , its bob mass is doubled to $2m$ , and the entire apparatus is transported to a planet where the acceleration due to gravity is $36\%$ less than that on Earth. What is the new period of oscillation of the pendulum?

A
$3.0T$
$1.5T$ Answer
C
$1.2T$
D
$0.67T$

Answer

1.5T

The period of a simple pendulum is determined by

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

. It is entirely independent of the mass of the bob. With a length increase of

44\%

(

L' = 1.44L

) and a gravity reduction of

36\%

(

g' = 0.64g

), the new period becomes

\sqrt{\frac{1.44}{0.64}}T = \frac{1.2}{0.8}T = 1.5T

Step-by-Step Solution

State the equation for the period of a simple pendulum

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

The period depends solely on length

L

and acceleration due to gravity

g

, and is independent of mass

m

Determine the new length and new acceleration due to gravity

L' = 1.44L

and

g' = 0.64g

44\%

increase in length yields

1 + 0.44 = 1.44

, while a

36\%

decrease in gravity yields

1 - 0.36 = 0.64

Calculate the factor of change in the period

\frac{T'}{T} = \sqrt{\frac{L'}{L} \times \frac{g}{g'}} = \sqrt{\frac{1.44}{0.64}} = \sqrt{\frac{144}{64}} = \frac{12}{8} = 1.5

Substituting the relative changes into the pendulum period ratio yields the scaling factor.

Express the new period in terms of

T

T' = 1.5T

Multiplying the initial period by the calculated scaling factor gives the final period.

Key Concept

Mass independence and parametric scaling of simple pendulum period in Simple Harmonic Motion

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