Question 4.7: Figure 4.16 depicts a simple pendulum with a support that ha...

Assuming small angular oscillations, the displacement of the mass m in the horizontal direction is given by

x = y + lθ = Y_{0} \sin ω_{f}t + lθ
The velocity and acceleration of the mass are then

\dot{x} = \dot{y} + l \dot{θ} = ω_{f}Y_{0} \cos ω_{f}t + l \dot{θ}
\ddot{x} = \ddot{y} + l \ddot{θ} = −ω^{2}_{f} Y_{0}  sin  ω_{f}t + l \ddot{θ}

Let R_x and R_y denote the reaction forces, as shown in the figure. By taking the moments of the applied and inertia forces about O, one can obtain the dynamic equation
−mgl sin θ − c \dot{x}l  cos  θ = m\ddot{x}l cos θ

For small angular oscillations sin θ ≈ θ and cos θ ≈ 1, thus
m\ddot{x}l + c \dot{x}l + mglθ = 0

Using the expressions for x, \dot{x}, and \ddot{x}, one obtains
m(−ω^{2}_{f} Y_{0} sin ω_{f}t + l\ddot{θ} ) l + c(ω_{f}Y_{0} cos ω_{f}t + l \dot{θ}) l + mglθ = 0

which can be written as
ml^{2}\ddot{θ}  = cl^{2} \dot{θ} + mglθ = mω^{2}_{f} Y_{0}l sin ω_{f}t − cω_{f}Y_{0}l cos ω_{f}t

= ω_{f}Y_{0}l[mω_{f} sin ω_{f}t − c cos ω_{f}t ]

= ω_{f}Y_{0}l\sqrt{(mω_{f})^{2} + c^{2}} sin(ω_{f}t − ψ_{b})
This equation can be written in a simple form as

me \ddot{θ} + c_{e}\dot{θ} + k_{e}θ = F_{e} sin(ω_{f}t − ψ_{b})

where m_{e} = ml^{2}, c_{e} = cl^{2}, k_{e} = mgl, F_{e} = ω_{f}Y_{0}l\sqrt{(mω_{f})^{2} + c^{2}}, and  ψ_{b} = tan^{−1}(c/mω_{f}). This equation is in a form similar to Eq. 4.37, therefore its steady state solution can be expressed as

θ_{p} = \frac{F_{e}/k_{e}}{ \sqrt{(1 − r^{2})^{2} + (2rξ)^{2}}} sin(ω_{f}t − ψ − ψ_{b})
where

r = \frac{ω_{f}}{ω}, ω= \sqrt{\frac{k_{e}}{m_{e}}} = \sqrt{\frac{mgl}{ml^{2}}} = \sqrt{\frac{g}{l}},      ξ = \frac{c_{e}}{C_{c}} =\frac{cl^{2}}{C_{c}},

C_{c} = 2m_{e}ω = 2ml^{2} \sqrt{\frac{g}{l}},     ψ = tan^{−1}\left(\frac{2rξ}{1 − r^{2}}\right)