A simple pendulum suspended in a terrestrial laboratory has a period of oscillation T when… — JAMB UTME

A simple pendulum suspended in a terrestrial laboratory has a period of oscillation $T$ when fitted with a bob of mass $m$ . If the bob is replaced by another bob of mass $4m$ and the entire setup is moved to a high-altitude station where the acceleration due to gravity is $\frac{g}{4}$ , what is the new period of oscillation of the pendulum?

$2T$ Answer
B
$4T$
C
$8T$
D
$\frac{T}{2}$

Answer

The new period of oscillation is

2T

The period of a simple pendulum is given by

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

. It depends strictly on the length of the string

L

and the local gravitational acceleration

g

, making it independent of the bob's mass

m

. Replacing mass

m

with

4m

does not alter the period. When gravity decreases to

g' = \frac{g}{4}

, the new period becomes

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2 \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

Step-by-Step Solution

Identify the formula for the period of a simple pendulum

The period

T

of a simple pendulum is given by

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

, where

L

is the pendulum length and

g

is the acceleration due to gravity.

This formula defines how physical parameters affect the oscillatory period of a pendulum under simple harmonic motion.

Analyze the effect of changing the bob's mass

Changing the mass of the bob from

m

4m

has no effect on the period.

The equation for the period of a simple pendulum contains no mass term, demonstrating that mass does not influence the period of small-angle oscillations.

Calculate the new period

T'

under the altered gravitational field

g' = \frac{g}{4}

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

Reducing the gravitational acceleration to one-fourth increases the square-root term

\sqrt{\frac{1}{1/4}} = 2

, thereby doubling the period.

Key Concept

Mass Independence and Gravity Dependence of Simple Pendulum Period

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