Solved Problem on Harmonic Oscillations
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Determine the equation of motion as a function of time and the period of oscillations for a physical pendulum, it consists of a body of any shape and mass m where the position of the center of mass is known in the small oscillation approximation.
Problem data:
- Mass of the body: m.
On the pendulum acts the following force (Figure 1-A):
- Fg: gravitational force.
The vector r locates the Center of Mass of the body at a distance d relative to the fixed point.
Solution
Applying Newton's Second Law for the rotation motion
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{N}=I\mathbf{\alpha}} \tag{I}
\end{gather}
\]
- N is the torque of the force that acts on the body;
- I is the moment of inertia of the body;
- α is the angular acceleration of the body.
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{N}=\mathbf{r}\times{\mathbf{F}}} \tag{II}
\end{gather}
\]
The only force that acts on the body is the gravitational force Fg e
| r | = d is the distance from the fixed point to the Center of Mass
(C.M.) of the pendulum.Writing the angular acceleration as
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{\alpha}=\frac{d^{2}\mathbf{\theta}}{dt^{2}}} \tag{III}
\end{gather}
\]
substituting the expressions (II) and (III) into expression (I)
\[
\begin{gather}
\mathbf{r}\times{\mathbf{P}}=I\frac{d^{2}\mathbf{\theta}}{dt^{2}} \tag{IV}
\end{gather}
\]
where
\( {\mathbf{F}}_{gT}=-F_{g}\sin \theta\;{\mathbf{e}}_{\theta} \)
\( {\mathbf{F}}_{gN}=F_{g}\cos \theta\;{\mathbf{e}}_{r} \)
\( \mathbf{r}=d\;{\mathbf{e}}_{r} \)
\( \mathbf{\theta}=\theta\;{\mathbf{e}}_{z} \)
Note: Some people find it difficult to understand that the angular displacement vector
points to the direction ez perpendicular to the plane of rotation. When a body
moves from a position r to a position r', we have a displacement s along the
trajectory, the angular displacement θ is contained in the plane, but the angular displacement
vector θ points perpendicular to the plane, this preserves the cross product (Figure 2). The
angular displacement vector indicates that the body is rotated, its magnitude indicates the angular
displacement (scalar) and the direction of the vector indicates the direction of rotation of the body, if
the vector θ is positive the cross product indicates that the body is if moving clockwise, if the
vector is negative the is moving counterclockwise.
Figure 2
This is the same argument used for angular velocity, the velocity vector v is tangent to the trajectory, but the angular velocity vector ω is perpendicular to the trajectory.
This is the same argument used for angular velocity, the velocity vector v is tangent to the trajectory, but the angular velocity vector ω is perpendicular to the trajectory.
\[
\begin{gather}
\mathbf{F}_{g}={\mathbf{F}}_{gT}+{\mathbf{F}}_{gN}\\
\mathbf{F}_{g}=-F_{g}\sin \theta\;{\mathbf{e}}_{\theta}+F_{g}\cos \theta\;{\mathbf{e}}_{r}
\end{gather}
\]
The cross product
\( \mathbf{r}\times {\mathbf{F}_{g}} \)
will be
\[
\begin{gather}
\mathbf{r}\times{\mathbf{F}_{g}}=
\left|
\begin{matrix}
{\mathbf{e}}_{r} & {\mathbf{e}}_{\theta} & {\mathbf{e}}_{z}\\
d & 0 & 0\\
F_{g}\cos \theta &-F_{g}\sin \theta & 0
\end{matrix}
\right|=\\[5pt]
=[0.0-0.(-F_{g}\sin \theta)]{\mathbf{e}}_{r}-[d.0-0.F_{g}\cos\theta]{\mathbf{e}}_{\theta}+[-d F_{g}\sin \theta-0. F_{g}\cos \theta]{\mathbf{e}}_{z}\\[5pt]
\mathbf{r}\times{\mathbf{F}_{g}}=-d F_{g}\sin \theta\;{\mathbf{e}}_{z} \tag{V}
\end{gather}
\]
the gravitational force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg} \tag{VI}
\end{gather}
\]
substituting the value of θ and expressions (V) and (VI) into expression (IV)
\[
\begin{gather}
-dmg\sin \theta\;{\mathbf{e}}_{z}=I\frac{d^{2}\theta}{dt^{2}}\;{\mathbf{e}}_{z}\\[5pt]
\frac{-{dmg}}{I}\sin \theta=\frac{d^{2}\theta}{dt^{2}}
\end{gather}
\]
the equation has only components in the ez direction and setting
\( \omega ^{2}=\frac{mgd}{I} \)
and writing
\( \frac{d^{2}\theta}{dt^{2}}=\ddot{\theta} \)
\[
\begin{gather}
\ddot{\theta}=-\omega ^{2}\sin \theta\\
\ddot{\theta}+\omega ^{2}\sin \theta =0
\end{gather}
\]
as we are working on a small angle approximation oscillations, we can expand the sin θ function in a
Taylor series.
Taylor series expansion of sin θ
\( \displaystyle \frac{f^{0}(0)}{0!}\theta ^{0}=\frac{\sin 0}{1}.1=0 \)
Note: \( f^{0} \) DOES NOT mean the function f to the zero power means the zero derivatives of the function f, that is, the function itself calculated in the point a.
\( \displaystyle \frac{f^{\text{I}}(0)}{1!}\theta ^{1}=\frac{\cos 0}{1}\theta =\theta \)
\( \displaystyle \frac{f^{\text{II}}(0)}{2!}\theta^{2}=\frac{-\sin 0}{2.1}\theta ^{2}=0 \)
\( \displaystyle \frac{f^{\text{III}}(0)}{3!}\theta ^{3}=\frac{-\cos 0}{3.2.1}\theta^{3}=-{\frac{\theta ^{3}}{6}} \)
\( \displaystyle \frac{f^{\text{IV}}(0)}{4!}\theta^{4}=\frac{-(-\sin 0)}{4.3.2.1}\theta ^{4}=0 \)
\( \displaystyle \frac{f^{\text{V}}(0)}{5!}\theta ^{5}=\frac{\cos 0}{5.4.3.2.1}\theta^{5}=\frac{\theta ^{5}}{120} \)
The sine function can be represented by the following series of powers
For an angle of \( 10°=\frac{\pi}{18}=0,1745 \), we have \( \sin \frac{\pi}{18}=0,1736 \), the approach represents an error of 0.5%.
\[ \bbox[#99CCFF,10px]
{f(x)=\sum _{n=0}^{\infty}{\frac{f^{n}(a)}{n!}(x-a)^{n}}}
\]
expanding around the equilibrium point with a = 0, for the first 6 terms of the series
\( \displaystyle \frac{f^{0}(0)}{0!}\theta ^{0}=\frac{\sin 0}{1}.1=0 \)
Note: \( f^{0} \) DOES NOT mean the function f to the zero power means the zero derivatives of the function f, that is, the function itself calculated in the point a.
\( \displaystyle \frac{f^{\text{I}}(0)}{1!}\theta ^{1}=\frac{\cos 0}{1}\theta =\theta \)
\( \displaystyle \frac{f^{\text{II}}(0)}{2!}\theta^{2}=\frac{-\sin 0}{2.1}\theta ^{2}=0 \)
\( \displaystyle \frac{f^{\text{III}}(0)}{3!}\theta ^{3}=\frac{-\cos 0}{3.2.1}\theta^{3}=-{\frac{\theta ^{3}}{6}} \)
\[ \displaystyle \frac{f^{\text{III}}(0)}{3!}\theta ^{3}=\frac{-\cos 0}{3.2.1}\theta^{3}=-{\frac{\theta ^{3}}{6}} \]
\( \displaystyle \frac{f^{\text{IV}}(0)}{4!}\theta^{4}=\frac{-(-\sin 0)}{4.3.2.1}\theta ^{4}=0 \)
\[ \displaystyle \frac{f^{\text{IV}}(0)}{4!}\theta^{4}=\frac{-(-\sin 0)}{4.3.2.1}\theta ^{4}=0 \]
\( \displaystyle \frac{f^{\text{V}}(0)}{5!}\theta ^{5}=\frac{\cos 0}{5.4.3.2.1}\theta^{5}=\frac{\theta ^{5}}{120} \)
\[ \displaystyle \frac{f^{\text{V}}(0)}{5!}\theta ^{5}=\frac{\cos 0}{5.4.3.2.1}\theta^{5}=\frac{\theta ^{5}}{120} \]
The sine function can be represented by the following series of powers
\[
\sin \theta =\theta -\frac{\theta ^{3}}{6}+\frac{\theta^{5}}{120}-...
\]
As we are considering θ a small angle, we can make the approach
\[
\sin \theta \approx \theta
\]
and we neglect higher-order terms.For an angle of \( 10°=\frac{\pi}{18}=0,1745 \), we have \( \sin \frac{\pi}{18}=0,1736 \), the approach represents an error of 0.5%.
\[
\ddot{\theta}+\omega ^{2}\theta =0
\]
Solution of the differential equation \( \displaystyle \ddot{\theta}+\omega^{2}\theta =0 \)
The solution is exponential type, calculating its derivatives
The solution is exponential type, calculating its derivatives
\[
\begin{array}{l}
\theta =\operatorname{e}^{\lambda t} \\
\dot{\theta}=\lambda \operatorname{e}^{\lambda t} \\
\ddot{\theta}=\lambda^{2}\operatorname{e}^{\lambda t}
\end{array}
\]
substituting in the equation
\[
\begin{gather}
\lambda ^{2}\operatorname{e}^{\lambda t}+\omega^{2}\operatorname{e}^{\lambda t}=0\\[5pt]
\lambda^{2}+\omega^{2}=0\\[5pt]
\lambda ^{2}=-\omega^{2}\\[5pt]
\lambda =\pm i\sqrt{\omega^{2}}\\[5pt]
\lambda =\pm i\omega
\end{gather}
\]
the solution is as follows, where C1 and C2 are constant
\[
\theta (t)=C_{1}\operatorname{e}^{i\omega t}+C_{2}\operatorname{e}^{-i\omega t}
\]
using Euler's formula
\( \operatorname{e}^{ix}=\cos x+i\sin x \)
\[
\begin{gather}
\theta (t)=C_{1}\left(\cos \omega t+i\sin \omega t\right)+C_{2}\left(\cos \omega t-i\sin \omega t\right)\\[5pt]
\theta(t)=\left(C_{1}+C_{2}\right)\cos \omega t-i\left(C_{2}-C_{1}\right)\sin \omega t
\end{gather}
\]
defining the following constants
\[
A=C_{1}+C_{2}\quad ,\quad B=i(C_{2}-C_{1})
\]
\[
\theta (t)=A\cos \omega t+B\sin \omega t
\]
setting
\[
\begin{array}{l}
\cos \phi=\dfrac{A}{\sqrt{A^{2}+B^{2}}}\\[5pt]
\sin \phi=\dfrac{B}{\sqrt{A^{2}+B^{2}}}\\[5pt]
\theta_{0}=\sqrt{A^{2}+B^{2}}
\end{array}
\]
substituting in the equation
\[
\begin{gather}
\theta (t)=\left(A\cos \omega t-B\sin \omega t\right)\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{A^{2}+B^{2}}}\\[5pt]
\theta(t)=\sqrt{A^{2}+B^{2}}\left(\frac{A}{\sqrt{A^{2}+B^{2}}}\cos \omega t-\frac{B}{\sqrt{A^{2}+B^{2}}}\sin \omega t\right)\\[5pt]
\theta (t)=\theta_{0}\left(\cos \phi \cos \omega t-\sin \phi \sin \omega t\right)
\end{gather}
\]
The equation of motion will be
\[ \bbox[#FFCCCC,10px]
{\theta (t)=\theta _{0}\cos \left(\omega t+\phi \right)}
\]
The period of oscillations is given by
\[ \bbox[#99CCFF,10px]
{T=\frac{2\pi }{\omega}}
\]
substituting the definition of ω0 made above
\[
T=\frac{2\pi }{\sqrt{\dfrac{mgd}{I}}}
\]
\[ \bbox[#FFCCCC,10px]
{T=2\pi \sqrt{\frac{I}{mgd}}}
\]
Note: The moment of inertia I that appears in this problem, is not the moment of
inertia calculated relative to the center of mass of the body, but the moment of inertia relative to the
point of fixation of the body around which it oscillates.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .