Chapter 4, Problem 4.41P

To determine

The natural frequency of the system by using Rayleigh’s method.

Answer to Problem 4.41P

The natural frequency of the system is $\frac{1}{2\pi }\sqrt{\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}}$.

Explanation of Solution

Write the expression for equality condition of potential energy and kinetic energy for Rayleigh’s criterion if the system is vibrating at natural frequency.

${\left(\text{KE}\right)}_{\max }={\left(\text{PE}\right)}_{\max }$ ...... (I)

Here, the kinetic energy of the system is $\text{KE}$ and the potential energy of the system is $\text{PE}$ Write the law of conservation of energy for simple harmonic motion.

${\text{KE}}_{\text{max}}+{\text{PE}}_{\text{min}}={\text{PE}}_{\text{max}}+{\text{KE}}_{\text{min}}$ ...... (II)

For simple harmonic motion potential energy is at maximum and kinetic energy is zero when the displacement of the system is at maximum.

Substitute $0$ for ${\text{KE}}_{\min }$ in Equation (II).

$\begin{array}{l}{\text{KE}}_{\text{max}}+{\text{PE}}_{\text{min}}={\text{PE}}_{\text{max}}+0\\ {\text{KE}}_{\text{max}}+{\text{PE}}_{\text{min}}={\text{PE}}_{\text{max}}\\ {\text{KE}}_{\text{max}}={\text{PE}}_{\text{max}}-{\text{PE}}_{\text{min}}\end{array}$ ...... (III)

Write the expression of equation of motion for simple harmonic motion.

$x\left(t\right)=A_o\sin \left({\omega }_{n}t+\varphi \right)$ ...... (IV)

Here, the displacement of the pendulum with respect to time is $x\left(t\right)$, the amplitude is $A$, the natural frequency is ${\omega }_n$, the initial phase angle is $\varphi$.

Differentiate equation (IV) with respect to $t$.

$\dot{x}\left(t\right)=A_o{\omega }_n\cos ({\omega }_{n}t+\varphi )$ ...... (V)

Substitute $0$ for $\varphi$ and $x{\left(t\right)}_{\max }$ for $\dot{x}\left(t\right)$ in Equation (V).

$\begin{array}{l}\dot{x}{\left(t\right)}_{\max }=A_o\cos \left({\omega }_{n}t+0\right)\\ \dot{x}{\left(t\right)}_{\max }=A_o{\omega }_n\end{array}$

Write the expression of moment of inertia of reverse pendulum.

$I_o=m{L}_2^2$

Here, the moment of inertia is $I_o$, the mass of the pendulum is $m$ and length of the pendulum is $L_2$.

Write the expression of maximum kinetic energy of the system.

${\text{KE}}_{\max }=\frac{1}{2}{I}_o{\left(\frac{\dot{x}{\left(t\right)}_{\max }}{L_2}\right)}^2$ ...... (VI)

Substitute $A_o{\omega }_n$ for $\dot{x}{\left(t\right)}_{\max }$ in Equation (VII)

${\text{KE}}_{\max }=\frac{1}{2}{I}_o{\left(\frac{A_o{\omega }_n}{L_2}\right)}^2$

Write the expression of potential energy of the system.

$\text{PE}=k{\left(L_1\theta \right)}^2+mg{L}_2(1-\cos \theta )$ ...... (VII)

Here, the angle of deviation is $\theta$.

Since the angle is very small so $\cos \theta \approx 1-\frac{{\theta }^2}{2}$.

Substitute $1-\frac{{\theta }^2}{2}$ for $\cos \theta$ in Equation (VIII)

$\begin{array}{c}\text{PE}=k{\left(L_1\theta \right)}^2+mg{L}_2\left(1-\left(1-\frac{{\theta }^2}{2}\right)\right)\\ =k{\left(L_1\theta \right)}^2+mg{L}_2\frac{{\theta }^2}{2}\\ =\left(k{L}_1^2+\frac{mg{L}_2}{2}\right){\theta }^2\end{array}$ ..... (VIII)

Substitute $\frac{A_o}{L_1}$ for $\theta$ and ${\text{PE}}_{\max }$ for $\text{PE}$ in Equation (VIII).

${\text{PE}}_{\text{max}}=\left(k{L}_1^2+\frac{mg{L}_2}{2}\right){\left(\frac{A_o}{L_1}\right)}^2$

Write the expression for fundamental natural frequency of the system in terms of frequency of the vibration of the cantilever beam.

${\omega }_n=2\pi f_n$ ...... (IX)

Substitute $\left(k{L}_1^2+\frac{mg{L}_2}{2}\right){\left(\frac{A_o}{L_1}\right)}^2$ for ${\text{PE}}_{\text{max}}$ and $\frac{1}{2}{I}_o{\left(\frac{A_o{\omega }_n}{L_2}\right)}^2$ for ${\text{KE}}_{\max }$ in Equation (I).

$\begin{array}{l}\frac{1}{2}m{L}_2^2{\left(\frac{A_o{\omega }_n}{L_2}\right)}^2=\left(k{L}_1^2+\frac{mg{L}_2}{2}\right){\left(\frac{A_o}{L_1}\right)}^2\\ m{L}_2^2{A}_o^2{\left(\frac{{\omega }_n}{L_2}\right)}^2=\left(2k{L}_1^2+mg{L}_2\right)\left(\frac{A_o^2}{L_1^2}\right)\\ m{L}_2^2{\left(\frac{{\omega }_n}{L_2}\right)}^2=\left(2k{L}_1^2+mg{L}_2\right)\left(\frac{1}{L_1^2}\right)\\ {\left({\omega }_n\right)}^2=\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}\end{array}$

${\omega }_n=\sqrt{\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}}$

Substitute $\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}$ for ${\omega }_n$ in Equation (IX).

$\begin{array}{l}\frac{1}{2\pi }\sqrt{\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}}=2\pi f_n\\ f_n=\frac{1}{2\pi }\sqrt{\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}}\end{array}$.

Conclusion:

The natural frequency of the system is $\frac{1}{2\pi }\sqrt{\frac{\left(2k{L}_1^2+mg{L}_2\right)}{m{L}_1^2}}$.