Question

The period of simple pendulum increases by 10% when its length is increased bt 21 cm. Find the original length and period of pendulum

Answer 1

Explanation:

Let its initial length is L and initial time period of oscillation is T.

Then, T = 2π√(l/g) .......(1)

a/c to question, 

The period of oscillation of simple pendulum increases by 20% when its length is increased by 44cm.

e.g., T + 20% of T = 2π√{(l + 0.44)/g}

6T/5 = 2π√{(l+0.44)/g}.....(2)

From equations (1) and (2), 

5/6 = √{l/(l+0.44)}

25/36 = l/(l + 0.44) 

25l + 25 × 0.44 =

11 = 11l

l = 1m 

Hence, initial length of pendulum is 1m

Now, T = 2π√(l/g)

Put l = 1m , g = 9.8 m/s² and π = 3.14 

Then, you will get, T = 2.007 sec

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Answer 2

The original length of the pendulum is 1 meter,

and, the period of the pendulum is 2 seconds.

Given: The period of a simple pendulum increases by 10% when its length is increased by 21 cm.

To Find: The original length and period of the pendulum.

Solution:

The time period (T) of a simple pendulum is given by,

           T = 2π √(l/g)         [ where, l = length of pendulum ]   ..... (1)

Let original length be ' L1 ' and the time period be ' T1 '.

So, new length ' L2 ' = ( L1 + 0.21 ) m

and new time period ' T2 ' = ( T1 + ( 10 × T1)/100 )

                                             = 1.1 T1

So, from (1), we get;

     ( T1 / T2 ) = √ ( L1 / L2 )

⇒  ( T1 / 1.1 T1 ) = √ ( L1 / L1 + 0.21 )

⇒  ( L1 + 0.21 / L1 ) = ( 1.1 )²

⇒ 1 + 0.21 / L1 = 1.21

⇒ L1 = 0.21 / 0.21

         = 1 m

Now, for finding the period of the pendulum, we use the formula,

 T = 2π √(l/g)  

    = 2π √ ( 1 / 9.8 )

    ≅ 2 s

Hence, the original length of the pendulum is 1 meter,

and, the period of the pendulum is 2 seconds.

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