Posted by **Anonymous** on Friday, April 6, 2012 at 11:09am.

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.

- Physics -
**Elena**, Friday, April 6, 2012 at 2:27pm
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.

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