# If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor of

Jamb 1979

If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor of A. 4 B. π/2 C. 1/4 D. 2π E. 2

EXPLANATION: for simple pendulum;

T = 2π√(L/g)

Where;

T is the Tension

L is the length of the inextensible string

g is the acceleration due to gravity

Since 2π and g are constant, it means we have T to be directly proportional to the square root of L. Then,

T = k√L

Where K is the constant of proportionality.

We can have,

T/√L = k

And since, T/√L is constant, we can now have,

T₁/√L₁ = T₂/√L₂ = T₃/√L₃ = … = Tₙ/√Lₙ

Hope you understand?

Here for this question, we will use

T₁/√L₁ = T₂/√L₂

From the question, it is understandable that our L₂ = 4L₁, so, we will have;

T₁/√L₁ = T₂/√(4L₁)

Cross multiply and divide both sides by √L₁ you have;

(√(4L₁) × T₁) / √L₁ = T₂

Solving this mathematically, you arrive at;

T₂ = 2T₁