A SIMPLE PENDULUM OF LENGTH 0.4m HAS A PERIOD OF 2s. WHAT IS THE PERIOD OF A SIMILAR PENDULUM OF LENGTH 0.8m AT THE SAME PLACE

The answer to the question is: C. 2√2s

The period of a simple pendulum is given by the formula:

T = 2π√(L/g)

Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Given that the length $L1 = 0.4 m$ and $T_{1} = 2 s$ , we can find $g$ :

T1 = 2π√L1/g

T1 = 2π√0.4/g

2 = 2π√0.4/g

1 = π√0.4/g

1/π = √0.4/g

1/π² = 0.4/g

0.4/g = 1/π²

g = 0.4π² m/s²

Now, for the second pendulum with length $L_{2} = 0.8 m$ , we can find its period $T_{2}$ :

T2 = 2π√L2/g

T2 = 2π√0.8/g

T2 = 2π√0.8/0.4π²

T2 = 2π√2/π²

T2 = 2π√2/√π²

T2 = 2π√2 ÷ π

T2 = 2√2 s.