Question

calculate the time period of a simple pendulum of length 0.49, when g =9.8 ms–².​

Answer 1

Explanation:

$t = 2\pi \sqrt{l \div g}$

T= 2π√.49/9.8 = 2π√1/20

Answer 2

Answer:

The time period of a simple pendulum of length 0.49, when $g = 9.8\ m/s^2$  will be  $1.4042 s$

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,

$l = 0.49 m$      $g = 9.8\ m/s^2$

Then substitute these values into equation(1),

$T = 2\times 3.14\times \sqrt{\frac{0.49}{9.8} }$

So we get,

The time period of the simple pendulum, $T =1.4042 s$