A simple pendulum of length L with a bob of mass m has a period of oscillation of 2.0\text… — JAMB UTME

A simple pendulum of length $L$ with a bob of mass $m$ has a period of oscillation of $2.0\text{ s}$ . If the mass of the bob is increased to $4m$ and the length of the pendulum string is increased to $4L$ , what is the new period of oscillation?

A
$2.0\text{ s}$
$4.0\text{ s}$ Answer
C
$8.0\text{ s}$
D
$16.0\text{ s}$

Answer

The new period of oscillation is

4.0\text{ s}

The period of a simple pendulum undergoing simple harmonic motion is given by

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

. The mass of the bob does not enter the period formula, meaning changes to bob mass have zero effect on the period. When the length of the pendulum is quadrupled (

L' = 4L

), the new period becomes

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

. Since the original period was

2.0\text{ s}

, the new period is

2 \times 2.0\text{ s} = 4.0\text{ s}

Step-by-Step Solution

Identify the formula for the period of a simple pendulum.

The period is given by

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

The period of simple harmonic motion for a simple pendulum depends only on the length of the string

L

and the acceleration due to gravity

g

, and is independent of the mass of the bob

m

Substitute the scaled values into the formula to find the new period

T'

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

Taking the square root of

4L

factors out a multiplier of

\sqrt{4} = 2

, while the change in mass from

m

4m

has no effect on the period.

Calculate the numerical value of the new period.

T' = 2 \times 2.0\text{ s} = 4.0\text{ s}

Multiplying the initial period of

2.0\text{ s}

2

yields

4.0\text{ s}

Key Concept

Mass Independence and Length Relationship of a Simple Pendulum

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