Solved Problem on Harmonic Oscillations
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A block of mass m is attached to a spring, the spring constant is equal to k and a shock
absorber of damping coefficient b. The block is moved from its equilibrium position O to
a point x0 and released from rest. Determine:
a) The differential equation of motion;
b) The equation solution for the system in the case of underdamped oscillations and the angular frequency of the oscillations.
a) The differential equation of motion;
b) The equation solution for the system in the case of underdamped oscillations and the angular frequency of the oscillations.
Problem data:
- Mass of body: m;
- Spring constant: k;
- Damping coefficient: b;
- Initial position (t = 0): x0;
- Initial speed (t = 0): v0 = 0.
Problem diagram:
We choose a reference frame with a positive direction to the right. The block is moved to position
x0 and released with an initial velocity equal to zero. When released, the spring
force will act to restore the equilibrium position (Figure 1). With this, we write the
Initial Conditions of the problem
\[
\begin{align}
& x(0)=x_{0}\\[10pt]
& v_{0}=\frac{dx(0)}{dt}=0
\end{align}
\]
Solution
a) Applying Newton's Second Law (Figure 1)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F=m\frac{d^{2}x}{dt^{2}}} \tag{I}
\end{gather}
\]
the forces acting on the block are the spring force
\( {\vec{F}}_{S} \)
and the damping force
\( {\vec{F}}_{D} \)
given, in magnitude, by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{S}=-kx} \tag{II-a}
\end{gather}
\]
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{D}=-bv=-b\frac{dx}{dt}} \tag{II-b}
\end{gather}
\]
the minus sign in the spring force means that it acts in the opoosite direction of the displacement
of the block (acts in the direction of restoring equilibrium), in the damping force means that it acts
in the opposite direction of the velocity (acts in the direction of braking the movement).
Substituting the expressions (II-a) and (II-b) into the expression (I)
\[
\begin{gather}
-kx-b\frac{dx}{dt}=m\frac{d^{2}x}{dt^{2}}\\[5pt]
m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0
\end{gather}
\]
this is a Second Order Homogeneous Differential Equation. Dividing the equation by the mass m
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{d^{2}x}{dt^{2}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0}
\end{gather}
\]
b) In the equation of the previous item we will make the following definitions
\[
\begin{gather}
2\gamma \equiv \frac{b}{m} \tag{III-a}\\[10pt]
\omega_{0}^{2}\equiv\frac{k}{m} \tag{III-b}
\end{gather}
\]
\[
\begin{gather}
\frac{d^{2}x}{dt^{2}}+2\gamma \frac{dx}{dt}+\omega_{0}^{2}x=0
\end{gather}
\]
Solution of \( \displaystyle \frac{d^{2}x}{dt^{2}}+2\gamma \frac{dx}{dt}+\omega_{0}^{2}x=0 \)
The solution to this type of equation is found substituting
The angular frequency ω is given by
The solution to this type of equation is found substituting
\[
\begin{align}
& x=\operatorname{e}^{\lambda t}\\[10pt]
& \frac{dx}{dt}=\lambda \operatorname{e}^{\lambda t}\\[10pt]
& \frac{d^{2}x}{dt^{2}}=\lambda ^{2}\operatorname{e}^{\lambda t}
\end{align}
\]
substituting these values into the differential equation
\[
\begin{gather}
\lambda ^{2}\operatorname{e}^{\lambda t}+2\gamma \lambda\operatorname{e}^{\lambda t}+\omega_{0}^{2}\operatorname{e}^{\lambda t}=0\\[5pt]
\operatorname{e}^{\lambda t}\left(\lambda ^{2}+2\gamma \lambda+\omega_{0}^{2}\right)=0\\[5pt]
\lambda ^{2}+2\gamma \lambda +\omega_{0}^{2}=\frac{0}{\operatorname{e}^{\lambda t}}\\[5pt]
\lambda ^{2}+2\gamma\lambda +\omega_{0}^{2}=0
\end{gather}
\]
this is the Characteristic Equation that has a solution
\[
\begin{gather}
\Delta =b^{2}-4ac=\left(2\gamma \right)^{2}-4\times 1\times \omega_{0}^{2}=4\gamma^{2}-4\omega_{0}^{2}=4\left(\gamma^{2}-\omega_{0}^{2}\right)\\[10pt]
\lambda_{1}=\frac{-b+\sqrt{\Delta\;}}{2a}=\frac{-2\gamma +\sqrt{4\left(\gamma^{2}-\omega_{0}^{2}\right)\;}}{2\times 1}=-{\frac{2\gamma}{2}}+\frac{2\sqrt{\gamma^{2}-\omega_{0}^{2}\;}}{2}=-\gamma +\sqrt{\gamma^{2}-\omega_{0}^{2}\;}\\[5pt]
\lambda_{2}=\frac{-b-\sqrt{\Delta\;}}{2a}=\frac{-2\gamma -\sqrt{4\left(\gamma^{2}-\omega_{0}^{2}\right)\;}}{2\times 1}=-{\frac{2\gamma}{2}}-\frac{2\sqrt{\gamma^{2}-\omega_{0}^{2}\;}}{2}=-\gamma -\sqrt{\gamma^{2}-\omega_{0}^{2}\;}
\end{gather}
\]
For the system to oscillate with underdamped oscillations we must have
ω02>γ2, the term in the square root will be
\[
\begin{gather}
\sqrt{-1\times \left(\omega_{0}^{2}-\gamma^{2}\right)\;}=\sqrt{-1}\times \sqrt{\left(\omega_{0}^{2}-\gamma^{2}\right)\;}=\mathsf{i}\sqrt{\left(\omega_{0}^{2}-\gamma^{2}\right)\;}
\end{gather}
\]
where \( \mathsf{i}=\sqrt{-1\;} \).The angular frequency ω is given by
\[
\begin{gather}
\omega =\sqrt{\omega_{0}^{2}-\gamma^{2}\;}
\end{gather}
\]
Using the definitions made in (III-a) and (III-b) for ω02 and γ
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\omega =\sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^{2}\;}}
\end{gather}
\]
The solution of the differential equation will be
Differentiation of the expression (IV) with respect to time
Substituting the value of φ into expression (VII)
\[
\begin{gather}
x=C_{1}\operatorname{e}^{\lambda_{1}t}+C_{2}\operatorname{e}^{\lambda_{2}t}\\[5pt]
x=C_{1}\operatorname{e}^{\left(-\gamma +\mathsf{i}\sqrt{\omega_{0}^{2}-\gamma^{2}\;}\right)t}+C_{2}\operatorname{e}^{\left(-\gamma-\mathsf{i}\sqrt{\omega_{0}^{2}-\gamma^{2}\;}\right)t}\\[5pt]
x=C_{1}\operatorname{e}^{\left(-\gamma t+\mathsf{i}\sqrt{\omega_{0}^{2}-\gamma^{2}\;}t\right)}+C_{2}\operatorname{e}^{\left(-\gamma t-\mathsf{i}\sqrt{\omega_{0}^{2}-\gamma^{2}\;}t\right)}\\[5pt]
x=C_{1}\operatorname{e}^{-\gamma t}\operatorname{e}^{\mathsf{i}\omega \;t}+C_{2}\operatorname{e}^{-\gamma t}\operatorname{e}^{-\mathsf{i}\omega t}\\[5pt]
x=\operatorname{e}^{-\gamma t}\left(C_{1}\operatorname{e}^{\mathsf{i}\omega t}+C_{2}\operatorname{e}^{-\mathsf{i}\omega t}\right)
\end{gather}
\]
where C1 and C2 are constants of integration, using
Euler's Formula
\( \operatorname{e}^{\mathsf{i}\theta }=\cos \theta +\mathsf{i}\sin \theta \)
\[
\begin{gather}
x=\operatorname{e}^{-\gamma t}\left[C_{1}\left(\cos \omega t+\mathsf{i}\sin \omega t\right)+C_{2}\left(\cos \omega t-\mathsf{i}\sin \omega t\right)\right]\\[5pt]
x=\operatorname{e}^{-\gamma t}\left(C_{1}\cos \omega t+\mathsf{i}C_{1}\sin \omega t+C_{2}\cos \omega t-\mathsf{i}C_{2}\sin \omega t\right)\\[5pt]
x=\operatorname{e}^{-\gamma t}\left[\left(C_{1}+C_{2}\right)\cos \omega t+\mathsf{i}\left(C_{1}-C_{2}\right)\sin \omega t\right]
\end{gather}
\]
defining two new constants α and β in terms of C1 and
C2
\[
\begin{gather}
\alpha \equiv C_{1}+C_{2}\\[5pt]
\text{e}\\[5pt]
\beta \equiv \mathsf{i}(C_{1}-C_{2})
\end{gather}
\]
\[
\begin{gather}
x=\operatorname{e}^{-\gamma t}\left(\alpha \cos \omega t+\beta\sin \omega t\right)
\end{gather}
\]
multiplying and dividing this expression by
\( \sqrt{\alpha^{2}+\beta^{2}\;} \)
\[
\begin{gather}
x=\operatorname{e}^{-\gamma t}\left(\alpha \cos \omega t+\beta \sin \omega t\right)\frac{\sqrt{\alpha^{2}+\beta^{2}\;}}{\sqrt{\alpha^{2}+\beta^{2}\;}}\\[5pt]
x=\sqrt{\alpha^{2}+\beta^{2}\;}\operatorname{e}^{-\gamma t}\left(\frac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}\;}}\cos \omega t+\frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}\;}}\sin \omega t\right)
\end{gather}
\]
setting
\[
\begin{gather}
A\equiv \sqrt{\alpha^{2}+\beta^{2}\;} \\[5pt]
\cos \varphi \equiv \frac{\alpha }{\sqrt{\alpha^{2}+\beta^{2}\;}}\\[5pt]
\sin \varphi \equiv \frac{\beta }{\sqrt{\alpha^{2}+\beta^{2}\;}}
\end{gather}
\]
\[
\begin{gather}
x=A\operatorname{e}^{-\gamma t}\left(\cos \varphi \cos \omega t+\sin \varphi \sin \omega t\right)
\end{gather}
\]
From the trigonometric identity
\( \cos (a-b)=\cos a\cos b+\sin a\sin b \)
\[ \cos (a-b)=\cos a\cos b+\sin a\sin b \]
\[
\begin{gather}
{x(t)=A\operatorname{e}^{-\gamma t}\cos \left(\omega t-\varphi \right)} \tag{IV}
\end{gather}
\]
where A and φ are constants determined by the Initial Conditions.
Differentiation of the expression (IV) with respect to time
\[
\begin{gather}
x=\underbrace{A\operatorname{e}^{-\gamma t}}_{u}\underbrace{\cos (\omega t-\varphi)}_{v}
\end{gather}
\]
using the Product Rule for the differentiation of functions
\[
\begin{gather}
(uv)'=u'v+uv'
\end{gather}
\]
where
\( u=A\operatorname{e}^{-\gamma t} \)
and
\( v=\cos (\omega t-\varphi) \), the function v is a composite function, using the
Chain Rule
\[
\begin{gather}
\frac{dv[w(t)]}{dt}=\frac{dv}{dw}\frac{dw}{dt}
\end{gather}
\]
with
\( v=\cos w \)
and
\( w=\omega t-\varphi \)
\[
\begin{gather}
\frac{dx}{dt}=\frac{du}{dt}v+u\frac{dv}{dt}\\[5pt]
\frac{dx}{dt}=\frac{du}{dt}v+u\frac{dv}{dw}\frac{dw}{dt}\\[5pt]
\frac{dx}{dt}=\frac{d\left(A\operatorname{e}^{-\gamma t}\right)}{dt}\left[\cos{\left(\omega t-\varphi\right)}\right]+\left(A\operatorname{e}^{-\gamma t}\right)\frac{d(\cos w)}{dw}\frac{d\left(\omega t-\varphi \right)}{dt}\\[5pt]
\frac{dx}{dt}=-\gamma A\operatorname{e}^{-\gamma t}\cos{\left(\omega t-\varphi\right)}+\left(A\operatorname{e}^{-\gamma t}\right)(-\sin w)(\omega)\\[5pt]
\frac{dx}{dt}=-\gamma A\operatorname{e}^{-\gamma t}\cos{\left(\omega t-\varphi\right)}-\omega A\operatorname{e}^{-\gamma t}\sin \left(\omega t-\varphi \right)\\[5pt]
\frac{dx}{dt}=-A\operatorname{e}^{-\gamma t}\left[\gamma \cos{\left(\omega t-\varphi\right)}-\omega\sin \left(\omega t-\varphi \right)\right] \tag{V}
\end{gather}
\]
Substituting Initial Conditions into expressions (IV) amd (V)
\[
\begin{gather}
x(0)=x_{0}=A\operatorname{e}^{-\gamma \times 0}\cos (\omega\times 0-\varphi)\\[5pt]
x_{0}=A\cos (-\varphi)
\end{gather}
\]
since cosine is an even function
\( \cos \varphi =\cos (-\varphi) \)
\[
\begin{gather}
x_{0}=A\cos \varphi\\[5pt]
A=\frac{x_{0}}{\cos \varphi} \tag{VI}
\end{gather}
\]
\[
\begin{gather}
\frac{dx(0)}{dt}=0=-A\operatorname{e}^{-\gamma\times 0}\left[\gamma \cos (\omega \times 0-\varphi)+\omega\sin (\omega \times 0-\varphi)\right]\\[5pt]
0=-A\left[\gamma \cos(-\varphi)+\omega \sin (-\varphi)\right]\\[5pt]
0=-A\gamma\cos (-\varphi)+A\omega \sin (-\varphi)
\end{gather}
\]
as cosine is an even function and sine is an odd function
\( \sin \varphi =-\sin (-\varphi) \)
\[
\begin{gather}
0=-A\gamma \cos \varphi -A\omega \sin \varphi \tag{VII}
\end{gather}
\]
and substituting the expression (VI) into expression (VII)
\[
\begin{gather}
0=-{\frac{x_{0}}{\cancel{\cos \varphi}}}\gamma \cancel{\cos \varphi}-\frac{x_{0}}{\cos \varphi}\omega \sin \varphi\\[5pt]
0=-\gamma x_{0}-\omega x_{0}\operatorname{tg}\varphi\\[5pt]
\omega c_{0}\operatorname{tg}\varphi =-\gamma x_{0}\\[5pt]
\operatorname{tg}\varphi=-{\frac{\gamma}{\omega}}\\[5pt]
\varphi=\operatorname{arctg}\left(-{\frac{\gamma}{\omega}}\right)
\end{gather}
\]
since the arctangent is an odd function, we have
\( \operatorname{arctg}(-x)=-\operatorname{arctg}(x) \).Substituting the value of φ into expression (VII)
\[
\begin{gather}
A=\frac{x_{0}}{\cos\left[-\operatorname{arctg}\left(\frac{\gamma}{\omega}\right)\right]}\\[5pt]
A=\frac{x_{0}}{\cos\left[\operatorname{arctg}\left(\frac{\gamma}{\omega}\right)\right]}
\end{gather}
\]
From trigonometric identity
\[
\begin{gather}
\cos \left(\operatorname{arctg}x\right)=\frac{1}{\sqrt{x^{2}+1\;}}
\end{gather}
\]
\[
\begin{gather}
A=\frac{x_{0}}{\frac{1}{\sqrt{\left(\frac{\gamma}{\omega}\right)^{2}+1\;}}}\\[5pt]
A=\frac{x_{0}}{\frac{1}{\sqrt{\frac{\gamma^{2}}{\omega^{2}}+1\;}}}\\[5pt]
A=\frac{x_{0}}{\frac{1}{\sqrt{\frac{\gamma^{2}+\left(\sqrt{\omega_{0}^{2}-\gamma^{2}\;} \right)^{2}}{\omega^{2}}\;}}}\\[5pt]
A=\frac{x_{0}}{\frac{1}{\sqrt{\frac{\gamma^{2}+\omega_{0}^{2}-\gamma^{2}}{\omega^{2}}\;}}}\\[5pt]
A=\frac{x_{0}}{\frac{1}{\sqrt{\frac{\omega_{0}^{2}}{\omega^{2}}\;}}}\\[5pt]
A=x_{0}\frac{\omega}{\omega_{0}}
\end{gather}
\]
substituting the constants A and φ into expression (VI)
\[
\begin{gather}
x=x_{0}\frac{\omega}{\omega_{0}}\operatorname{e}^{-\gamma t}\cos \left[\omega t-\operatorname{arctg}\left(\frac{\gamma}{\omega}\right)\right]
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{x(t)=x_{0}\frac{\omega_{0}}{\omega}\operatorname{e}^{-\gamma t}\cos \left[\omega t-\operatorname{arctg}\left(\frac{\gamma}{\omega}\right)\right]}
\end{gather}
\]
Note: The trigonometric identity
\( \cos \left(\arctan x\right)=\frac{x}{\sqrt{x^{2}+1}\;} \)
is obtained by taking a right triangle with hypotenuse a and legs b and c
with angle θ (Figure 2).
\[
\begin{gather}
\tan\theta =\frac{b}{c} \tag{VIII}
\end{gather}
\]
\[
\begin{gather}
\tan\theta=\tan\left(\arctan x\right)=x=\frac{x}{1} \tag{IX}
\end{gather}
\]
equating the expressions (VIII) and (IX)
\[
\begin{gather}
\frac{b}{c}=\frac{x}{1}
\end{gather}
\]
we determine the legs b and c
\[
\begin{gather}
b=x\\[10pt]
c=1
\end{gather}
\]
The hypotenuse is found using the Pythagorean Theorem (Figure 3)
\[
\begin{gather}
a^{2}=b^{2}+c^{2}\\[5pt]
a^{2}=x^{2}+1^{2}\\[5pt]
a=\sqrt{x^{2}+1^{2}\;}
\end{gather}
\]
\[
\begin{gather}
\cos \theta =\frac{1}{\sqrt{x^{2}+1\;}}
\end{gather}
\]
Using the expression (IX)
\[
\begin{gather}
\theta =\arctan x
\end{gather}
\]
Therefore
\[
\begin{gather}
\cos \left(\arctan x\right)=\frac{1}{\sqrt{x^{2}+1\;}}
\end{gather}
\]
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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 .