Question

Calculate the time period of a simple pendulum of length 1.12 m, when acceleration due to gravity
is 9.8 ms​

Answer 1

The time period of the simple pendulum is 2.12 seconds.

Explanation:

In the question, it's been stated that the length of a pendulum is 1.12 m and we've been asked to determine its time period when the acceleration due to gravity is 9.8 m/s².

Well, a simple pendulum is just a small metal bob suspended by a long thread from a fixed support, such that the bob oscillates periodically.

The time period of a simple pendulum is nothing but the time taken by it to complete a full oscillation. It's formula is given by:

$\qquad \qquad \qquad \quad\boxed{ \sf{T = 2\pi \sqrt{\dfrac{l}{g} } }}$

Using this formula, the time period of the pendulum can be calculated.

• Time period (T) = ?
• Length (l) = 1.12 m
• Acceleration due to gravity (g) = 9.8 m/s²

The value of π can be either taken as 3.14 or 22/7. Here, let's take it as 3.14.

Substituting the values in the formula:

$\twoheadrightarrow \quad\sf{T = 2\pi \sqrt{\dfrac{l}{g} } }$

$\twoheadrightarrow \quad\sf{T = 2 \times 3.14 \times \sqrt{\dfrac{1.12}{9.8} } }$

$\twoheadrightarrow \quad\sf{T =6.28 \times \sqrt{\dfrac{1.12}{9.8} } }$

$\twoheadrightarrow \quad\sf{T =6.28 \times \sqrt{0.114} }$

$\twoheadrightarrow \quad\sf{T =6.28 \times 0.338}$

$\twoheadrightarrow \quad \underline{ \boxed{\bf \red{Time \: period = 2.12 \: s} }}$

∴ The time period of the simple pendulum is 2.12 seconds.

Answer 2

Answer:

The time period of a simple pendulum of length 1.12 m, when acceleration due to gravity $g = 9.8\ m/s^2$  is $2.1241s$

 

Explanation:

• Let T be the time period of a simple pendulum and l be the length of the pendulum. Then time period is given by the expression,

        $T = 2\pi \sqrt{\frac{l}{g} }$       ...(1)

       Where g is the acceleration due to gravity.

• In the question, it is given that,

         $l = 1.12 m$      $g = 9.8\ m/s^2$

        Then substitute these values into equation(1),

         $T = 2\times 3.14\times \sqrt{\frac{1.12}{9.8} }$

         So we get,

• The time period of the simple pendulum, $T =2.1241 s$