Question

Compare compare the time period of two simple pendulum of length 4 m and 49m at a place. *​

Answer 1

Answer:

Step-by-step explanation:

        T = 2π√l/g

         means T ∞ √l

            T1/T2 = √4/49 = 2/7

Answer 2

The relation between the time period of the two simple pendulum is given as 7 $T_{1}$ = 2 $T_{2}$

Given :

Length of the first simple pendulum, $L_{1}$ = 4m

Length of the second pendulum, $L_{2}$ = 49m

To find :

Relation between the $time$ period of the two pendulums

Solution :

The $time$ period of a simple pendulum = 2π x √(L/g)

Here, L = length $of$ the pendulum.

g = acceleration due $to$ gravity.

Let  $T_{1}$ be the $time$ period of the first simple pendulum.

$T_{1}$ = 2π x √($L_{1}$/g) = 2π x √(4m/g)

Let $T_{2}$ be the $time$ period of the second simple pendulum.

$T_{2}$ = 2π x √($L_{2}$/g) = 2π x √(49m/g)

Comparing $T_{1}$ and $T_{2}$ by finding their ratio:

$T_{1}$ / $T_{2}$ = [2π x √($L_{1}$/g)] / [2π x √($L_{2}$/g)]

          =  [2π x √(4m/g)] / [2π x √(49m/g)]

          = √(4/49) = 2/7

$T_{1}$ / $T_{2}$  = 2/7

7 $T_{1}$ = 2 $T_{2}$

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