Question

Consider a mass m attached to a spring of unstretched length `0. The top of the spring is attached to the ceiling. The mass is stretched down a distance r from it’s unstretched equilibrium length, while at the same time the spring is inclined at an angle θ with respect the the vertical. This makes the spring an elastic pendulum, where not only the angle changes, but also the length! Determine the equations of motion of this pendulum (Hint: there will be two - one for r, and one for θ). Check that you get the correct limit in your θ equation when r → 0, and also the correct limit in the r equation when θ → 0.

Answer 1

F = md²x/dt² = -kx

From here we have a second order differential equation for spring's displacement x with respect to time. This gives a solution of a sinusoidal function:

x(t) = c₁cos(t)+ c₂sin(t) = Acos(t- )

Where = √(k/m) , A = (c₁²+c₂²)^1/2, = tan^-1(c₂/c₁₎

vecocity and acceleration ca be find out as :

V(t) = -Asin(t+)

a(t) = -A²cos(t+)

The time period can be given as:

T= 2π√(m/k)

Now take the simple pendulum problem.Assuming the initial angle less than 1 radian i.e sin() ~.

The small angle approximation yields the harmonic oscillator equation:

d²/dt² +(g/l) = 0

Further approximation must include initial conditions (0) = ₀,

d/dt = 0.

₀ be the maximum angle pendulum rod and vertical.

(t) = ₀ cos(√(g/l)t)

The period of motion will be T = 2π√(l/g)