Applied Maths - Mechanics 60

Define the positive direction as downwards and anti-clockwise. The spring k1 is at the left.
Let the original length of the springs be h1 and h2 respectively.
At equilibrium, let the length of both string be L so that the tensions in the spring just cancel the weights of the objects (M + m)g. In other words k1(L – h1) + k2(L – h2) = (M + m)g … (1)
Also at equilibrium, there is no net torque in the system, k1(L – h1) = k2(L – h2) … (2)
Let the position of the centre of the rod be x' from the ceiling.
Let the position of A be y and the position of B be z, also from the ceiling.
Let the angle of the rod to the horizontal be θ.
Equation of motion for the centre of mass is
(M + m)d2x'/dt2 = – k1(y' – h1) – k2(z' – h2) + (M + m)g
k1(y' – h1) + k2(z' – h2)
= k1(y' – L + L – h1) + k2(z' – L + L – h2)
= k1(y' – L) + k2(z' – L) + k1(L – h1) + k2(L – h2)
= k1(y' – L) + k2(z' – L) + (M + m)g (Refers to 1)
(M + m)d2x'/dt2 = – k1(y' – L) – k2(z' – L)
Geometry equation, x' = (y' + z')/2
Motion about the centre of mass
Id2θ/dt2 = k2(z' – h2)l – k1(y' – h1)l
= k2(z' – L + L – h2)l – k1(y' – L + L – h1)l
= k2(z' – L)l – k1(y' – L)l (Refers to 2)
By shifting the reference position from the ceiling to the equilibrium position, the equation can be rewritten as:
(M + m)d2x/dt2 = – k1y – k2z... (3)
x = (y + z)/2 … (4)
Id2θ/dt2 = k2lz – k1ly … (5)
When m = 0, then I = 0
(5) reduces to k2z = k1y
(3) becomes Md2x/dt2 = – 2k1y … (6)
(4) becomes x = (y + k1y/k2)/2
y = 2k2x/(k1 + k2)
Sub into (6), d2x/dt2 + 4k1k2x/[M(k1 + k2)] = 0 => SHM
2π / T = √{4k1k2/[M(k1 + k2)]}
T = 2π√[(Mk1 + Mk2) / (4k1k2)]
T = π√[(Mk1 + Mk2) / (k1k2)]