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Chapter 6

Q. 6.15

The equation for the undamped, forced vibration of the pendulum device described in Fig. 6.19, page 150, is given in (6.81e). Solve this equation for the case when both the motion of the hinge support and the angular motion of the pendulum are small. Assume that the pendulum is released from rest at a small angle  \phi_0.

\ddot{\phi}  +  p^2 \sin \phi=\frac{x_O \Omega^2}{\ell} \cos \phi \sin \Omega t \text {, }                            (6.81e)

Screenshot 2022-10-10 154611

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Verified Solution

The differential equation (6.8Ie) describes a complicated nonlinear, undamped, forced vibrational motion of the pendulum. To simplify matters, we consider the case when the angular placement is sufficiently small that terms greater than first order in Φ may be ignored. Then (6.81e) simplifies to

\ddot{\phi}  +  p^2 \phi=\frac{x_O \Omega^2}{\ell} \sin \Omega t                (6.93a)

where  p^2=g / \ell.  This equation has the same form as (6.91); it describes the small, undamped, steady forced vibrational motion of the pendulum. For consistency with the small motion assumption, however, we consider only the case for which the motion of the hinge support O also is small , so that  x_O / \ell \ll 1.  Because the amplitude of the disturbing force in (6.93a) varies with its frequency , for small motions Φ(t), the range of operating frequencies also is limited .

\ddot{x}+p^2 x=Q \sin \Omega t                   (6.91)

The general solution of(6.93a), with  Q \equiv x_O \Omega^2 / \ell,  may be read from (6.92a) :

x(t)=A \cos p t  +  B \sin p t  +  H \sin \Omega t \text {, }                 (6.92a)

\phi(t)=A \cos p t  +  B \sin p t  +  H \sin \Omega t                (6.93b)

in which A and B are constants and the steady-state amplitude, by (6.92b), is

H=\frac{Q}{p^2\left(1  –  \xi^2\right)}=\frac{F_0 / k}{1  –  \xi^2}=\frac{X_S}{1  –  \xi^2}, \quad \xi=\frac{\Omega}{p} \neq 1 .                 (6.92b)

H=\frac{x_O \xi^2}{\ell\left(1  –  \xi^2\right)}, \quad \xi \equiv \frac{\Omega}{p} \neq 1.                       (6.93c)

The assigned initial data determine the constants in (6.93b),

\phi(0)=A=\phi_0, \quad \dot{\phi}(0)=B p  +  H \Omega=0,                 (6.93d)

which then yields the solution for the small angular motion of the pendulum:

\phi(t)=\phi_0 \cos p t  +  \frac{x_O \xi^2}{\ell\left(1  –  \xi^2\right)}(\sin \Omega t  –  \xi \sin p t).                    (6.93e)

It is evident that this small motion solution is meaningful only for sufficiently small values of the driving frequency ratio  \xi;  otherwise, the smallness of  \phi(t)  is violated .