What is the length of a simple pendulum that has a period of oscillation of 2.0\text{ s} a… — JAMB UTME

What is the length of a simple pendulum that has a period of oscillation of $2.0\text{ s}$ at a location where the acceleration due to gravity is $g = \pi^2\text{ m/s}^2$ ?

A
$0.5\text{ m}$
$1.0\text{ m}$ Answer
C
$2.0\text{ m}$
D
$4.0\text{ m}$

Answer

The length of the simple pendulum is

1.0\text{ m}

Using the period equation

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

, squaring both sides yields

T^2 = 4\pi^2 l / g

. Rearranging gives

l = \frac{T^2 g}{4\pi^2}

. Substituting

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

yields

l = \frac{4 \pi^2}{4 \pi^2} = 1.0\text{ m}

Step-by-Step Solution

State the formula for the period of a simple pendulum.

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

This formula relates the oscillation period

T

, pendulum length

l

, and gravitational acceleration

g

Square both sides of the equation to solve for

l

T^2 = 4\pi^2 \left(\frac{l}{g}\right) \implies l = \frac{T^2 \cdot g}{4\pi^2}

Isolating

l

allows direct evaluation using the given numerical values.

Substitute

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

into the expression.

l = \frac{(2.0)^2 \cdot \pi^2}{4\pi^2} = \frac{4\pi^2}{4\pi^2} = 1.0\text{ m}

Simplifying by canceling

\pi^2

and

4

gives the exact length.

Key Concept

Simple Pendulum Period and Length Relationship

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic