Solved Problem on Harmonic Oscillations
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Determine the equation of motion as a function of time and the period of oscillations to a simple pendulum in the small oscillation approximation.
Problem diagram:
Let's assume that the pendulum consists of a sphere with a mass m swinging on a rope of length
L, inextensible and with negligible mass. Consider the radius of the sphere small, so that it can
be neglected relative to the length of the rope. If the rope mass and the radius of the sphere could not
be neglected we would have a compound pendulum (Figure 1).
The initial angle of the pendulum displacement is θ0, at this point the distance from the mass to the pendulum fixed point, measured vertically, equals y0. When the pendulum oscillates, at any moment, it will have an angle θ of displacement and a distance y relative to the fixed point.
The initial angle of the pendulum displacement is θ0, at this point the distance from the mass to the pendulum fixed point, measured vertically, equals y0. When the pendulum oscillates, at any moment, it will have an angle θ of displacement and a distance y relative to the fixed point.
Solution
The variation of potential energy, ΔU, when it moves from a height h, is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\Delta U=mgh} \tag{I}
\end{gather}
\]
Assuming that the pendulum is initially at rest, the variation of kinetic energy, ΔK,
is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\Delta K=\frac{1}{2}mv^{2}} \tag{II}
\end{gather}
\]
Since energy is conserved, there is no energy dissipation in this system, potential energy is converted to
kinetic energy
\[
\Delta U=\Delta K
\]
equating the expressions (I) and (II)
\[
\begin{gather}
mgh=\frac{1}{2}mv^{2} \tag{III}
\end{gather}
\]
from the expression (III), the tangential speed v is given by
\[
\begin{gather}
v=\sqrt{2gh} \tag{IV}
\end{gather}
\]
The tangential speed is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=\omega r} \tag{V}
\end{gather}
\]
equating the expressions (IV) and (V)
\[
\begin{gather}
\omega r=\sqrt{2gh} \tag{VI}
\end{gather}
\]
writing
\( \omega =\dot{\theta} \)
and lets r = L
\[
\begin{gather}
\dot{\theta}L=\sqrt{2gh}\\
\dot{\theta}=\frac{\sqrt{2gh}}{L} \tag{VII}
\end{gather}
\]
The difference in height, h, between two instants of the oscillation will be (Figure 1)
\[
\begin{gather}
h=y-y_{0} \tag{VIII}
\end{gather}
\]
Using Trigonometry, we can write the distances, y0 and y, as a function of
L, θ, and θ0 (Figure 1)
\[
\begin{gather}
y_{0}=L\cos \theta _{0} \tag{IX-a}\\[10pt]
y=L\cos \theta \tag{IX-b}
\end{gather}
\]
substituting the expressions (IX-a) and (IX-b) into expression (VIII)
\[
\begin{gather}
h=L\cos \theta -L\cos \theta _{0}\\[5pt]
h=L\left(\cos \theta-\cos \theta _{0}\right) \tag{X}
\end{gather}
\]
substituting the expression (X) into expression (VII)
\[
\begin{gather}
\dot{\theta}=\frac{\sqrt{2gL\left(\cos \theta -\cos\theta _{0}\right)}}{L}\\[5pt]
\dot{\theta}=\sqrt{\frac{2gL}{L^{2}}\left(\cos \theta -\cos \theta_{0}\right)}\\[5pt]
\dot{\theta}=\sqrt{\frac{2g}{L}\left(\cos \theta -\cos\theta _{0}\right)} \tag{XI}
\end{gather}
\]
To find the equation of motion, we differentiate the expression (XI) as a function of time
\[
\frac{d}{dt}\dot{\theta}=\frac{d}{dt}\sqrt{\frac{2g}{L}\left(\cos \theta -\cos \theta_{0}\right)}
\]
Differentiation of \( \displaystyle \left[\frac{2g}{L}\left(\cos \theta -\cos \theta_{0}\right)\right]^{1/2} \)
this is a composite function whose derivative is given by the chain rule
this is a composite function whose derivative is given by the chain rule
\[
\frac{dv[(w)]}{dt}=\frac{dv}{dw}\frac{dw}{dt}
\]
with
\( v(w)=w^{1/2} \)
and
\( w(t)=\frac{g}{L}\left(\cos \theta -\cos \theta _{0}\right) \),
so the derivative will be
\[
\begin{gather}
\frac{dv[(w)]}{dt}=\frac{1}{2}w^{-1/2}\left(\frac{dw}{dt}\right)\left(\frac{d\theta}{dt}\right)\\
\frac{dv[(w)]}{dt}=\frac{1}{2}\left[\frac{2g}{L}\left(\cos\theta -\cos \theta_{0}\right)\right]^{-1/2}\left(-{\frac{2g}{L}}\operatorname{sen}\theta\right)\frac{d\theta}{dt}
\end{gather}
\]
Note: The variable θ is a function of time, so it must be differentiated, the last
term of the derivative, and θ0 is a constant whose derivative is zero.
\[
\ddot{\theta}=\frac{1}{2}\left[\frac{2g}{L}\left(\cos \theta -\cos\theta_{0}\right)\right]^{-1/2}\left(-{\frac{2g}{L}}\operatorname{sen}\theta\right)\dot{\theta}
\]
substituting with the expression (XI)
\[
\begin{gather}
\ddot{\theta}=\frac{1}{\cancel{2}}\cancel{\left[\frac{2g}{L}\left(\cos\theta -\cos \theta_{0}\right)\right]^{-1/2}}\left(-{\frac{\cancel{2}g}{L}}\operatorname{sen}\theta\right)\cancel{\left[\frac{2g}{L}\left(\cos \theta -\cos \theta_{0}\right)\right]^{1/2}}\\[5pt]
\ddot{\theta}=-{\frac{g}{L}}\operatorname{sen}\theta \\[5pt]
\ddot{\theta}+\frac{g}{L}\operatorname{sen}\theta =0
\end{gather}
\]
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, we have
\( \displaystyle \frac{f^{0}(0)}{0!}\theta ^{0}=\frac{\sin 0}{1}.1=0 \)
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%.
\[
\ddot{\theta}+\frac{g}{L}\theta =0
\]
Solution of the differential equation \( \displaystyle \ddot{\theta}+\frac{g}{L}\theta =0 \)
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
\[
\begin{gather}
\lambda ^{2}\operatorname{e}^{\lambda t}+\frac{g}{L}\operatorname{e}^{\lambda t}=0\\[5pt]
\lambda^{2}+\frac{g}{L}=0\\[5pt]
\lambda ^{2}=-{\frac{g}{L}}\\[5pt]
\lambda =\pm i\sqrt{\frac{g}{L}}
\end{gather}
\]
setting
\( \omega_{0}^{2}=\frac{g}{L} \)
the solution is as follows, where C1 and C2 are constant
\[
\theta (t)=C_{1}\operatorname{e}^{i\omega_{0}t}+C_{2}\operatorname{e}^{-i\omega_{0}t}
\]
using Euler's formula
\( \operatorname{e}^{ix}=\cos x+i\sin x \)
\[
\begin{gather}
\theta (t)=C_{1}\left(\cos \omega_{0}t+i\sin \omega_{0}t\right)+C_{2}\left(\cos \omega_{0}t-i\sin \omega_{0}t\right)\\[5pt]
\theta(t)=\left(C_{1}+C_{2}\right)\cos \omega_{0}t-i\left(C_{2}-C_{1}\right)\sin \omega_{0}t
\end{gather}
\]
defining the following constants
\[
A=C_{1}+C_{2}\quad ,\quad B=i(C_{2}-C_{1})
\]
\[
\theta (t)=A\cos \omega_{0}t+B\sin \omega_{0}t
\]
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
\[
\begin{gather}
\theta (t)=\left(A\cos \omega_{0}t-B\sin \omega_{0}t\right)\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{A^{2}+B^{2}}}\\[5pt]
\theta(t)=\sqrt{A^{2}+B^{2}}\left(\frac{A}{\sqrt{A^{2}+B^{2}}}\cos \omega_{0}t-\frac{B}{\sqrt{A^{2}+B^{2}}}\sin \omega_{0}t\right)\\[5pt]
\theta (t)=\theta_{0}\left(\cos \phi \cos \omega_{0}t-\sin \phi \sin\omega_{0}t\right)
\end{gather}
\]
The equation of motion will be
\[ \bbox[#FFCCCC,10px]
{\theta (t)=\theta _{0}\cos \left(\omega_{0}t+\phi \right)}
\]
The period of oscillations is given by
\[ \bbox[#99CCFF,10px]
{T=\frac{2\pi }{\omega_{0}}}
\]
substituting the definition of ω0 made above
\[
T=\frac{2\pi }{\sqrt{\frac{g}{L}}}
\]
\[ \bbox[#FFCCCC,10px]
{T=2\pi \sqrt{\frac{L}{g}}}
\]
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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 .