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
ANSWER: Option E
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
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