Chapter 14, Problem 87A

To determine

The period of a $\text{1}.\text{4}\text{m}$ pendulum.

Answer to Problem 87A

The required answer is $\text{2}\text{.37}\text{s}$ .

Explanation of Solution

Given:

The length of the pendulum is, $l=1.4\text{ m}$ .

The acceleration due to gravity is, $g=9.8{\text{ m/s}}^2$ .

Formula used:

The time period of simple pendulum can be written as,

$T=2\pi {\left(\frac{l}{g}\right)}^{1/2}$..... (1)

Here, $T$ is the time period of the pendulum, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Calculation:

Substituting $1.4\text{ m}$ in $l$ and $9.8{\text{ m/s}}^2$ in $g$ in equation (1),

  $\begin{array}{l}T=2\pi {\left(\frac{1.4\text{m}}{9.8{\text{m/s}}^{\text{2}}}\right)}^{1/2}\\ T=2.37\text{ s}\end{array}$

Conclusion:

Hence, the required answer is $\text{2}\text{.37 s}$ .