A simple pendulum of fixed length l carries a bob of mass m and oscillates with a period T… — JAMB UTME

A simple pendulum of fixed length $l$ carries a bob of mass $m$ and oscillates with a period $T$ . If the bob is replaced by another bob of mass $2m$ while maintaining the exact same string length, what is the new period of oscillation?

A
$2T$
$T$ Answer
C
$\sqrt{2}T$
D
$\frac{T}{2}$

Answer

The new period of oscillation remains

T

The period of oscillation for a simple pendulum undergoing small displacement SHM is given by

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

. The mass of the bob

m

is not a parameter in this equation. Therefore, altering the mass while keeping length

l

constant leaves the period unchanged as

T

Step-by-Step Solution

Identify the governing equation for the period of a simple pendulum executing simple harmonic motion.

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

To analyze which physical parameters determine the period of the simple pendulum.

Check for mass dependence in the formula.

The mass variable

m

does not appear in

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

The restoring gravitational force and inertia both scale linearly with mass, causing mass to cancel out entirely.

Determine the period after doubling the mass at constant length.

The new period is equal to

T

Since length

l

and acceleration due to gravity

g

are constant, changing the mass from

m

2m

has no effect on period.

Key Concept

Independence of Simple Pendulum Period from Bob Mass

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