Question

A particle of mass m = 3.04kg is suspended from a fixed point by a light inextensible string of length = 1.11m. You are required to determine the relationship between the period of swing and the length of the pendulum. For small angle approximation, sin ¦È ¡Ö ¦È.

Calculate the period of the swing of this simple pendulum, in seconds. You may use g = 9.81ms-2 and ¦Ð = 3.142.

Answer 1

Newton’s 2 law for rotation:
I•ε = M.
where I =mL^2 is the moment of inertia of material point.
m•L^2•(d2φ/dt2) = -m•g•L•sin φ.
d2φ/dt2 is the derivative of order two,
sin φ ≈φ,
d2φ/dt2 + (g/L) φ = 0.
The solution of this equation is
φ = φ(max) •cos(ω•t+α),
where ω = sqrt(g/L)
T =2π/ ω =2π•sqrt(l/g).
T = 2π•sqrt(1.11/9.81) = 2.114 s.