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
ANSWER: Option E
EXPLANATION: for simple pendulum;
T = 2π√(L/g)
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
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His Name is Tiamiyu Abdulbazeet Olawale. He is a student at the University of Ilorin studying Electrical and Electronics Engineering. He likes to give the latest updates about Nigeria and international institutions as he aims at helping people to lay their hands on the right information.