Question

calculate the time period of a simple pendulum of length 1.45m,when acceleration due gravity is 9.8

Answer 1

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Answer 2

Answer:

The time period of a simple pendulum of length 1.45m, when acceleration due gravity is $9.8m/s^2$ will be $2.4156s$

Explanation:

• The time period of a simple pendulum is the time taken to complete one full oscillation.

• Let T be the time period 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,

         $l = 1.45 m$      $g = 9.8\ m/s^2$

         We know,

         $\pi = 3.14$

        Then substitute these values into equation(1),

         $T = 2\times 3.14\times \sqrt{\frac{1.45}{9.8} }$

             $= 6.28\times0.38465$

             $= 2.4156s$

         So we get,

• The time period of the simple pendulum, $T =2.4156 s$