Solved Problem on Harmonic Oscillations
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A block of mass m = 0.25 kg is connected to a spring of constant k = 1 N/m. The block is
displaced from its equilibrium position O to a point P to 0.5 m and released from rest.
Determine:
a) Equation of displacement as a function of time;
b) The speed of the block;
c) Calculate the mechanical energy of the oscillator; d) The graph of position x as a function of time t.
a) Equation of displacement as a function of time;
b) The speed of the block;
c) Calculate the mechanical energy of the oscillator; d) The graph of position x as a function of time t.
Problem data:
- Mass of the body: m = 0.25 kg;
- Spring constant: k = 1 N/m;
- Initial position (t = 0): x0 = 0.5 m;
- Initial velocity (t = 0): v0 = 0.
We choose a frame of reference with a positive direction to the right. The block is displaced to the
position x0 = 0.5 m, and with an initial velocity equal to zero,
v0 = 0. When the block is released, the spring force will return the block to the
equilibrium position, and the velocity will be pointing in the opposite direction of the reference frame
and increase in magnitude towards the equilibrium position (Figure 1). Writing the
Initial Conditions of the problem
\[
\begin{array}{l}
x(0)=0.5\;\text{m}\\[10pt]
v_{0}=\dfrac{dx(0)}{dt}=0
\end{array}
\]
Solution
a) Applying Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F=m\frac{d^{2}x}{dt^{2}}} \tag{I}
\end{gather}
\]
the only force acting on the block is the spring force,
\( {\vec{F}}_{S} \),
dada por
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{S}=-kx} \tag{II}
\end{gather}
\]
the negative sign in spring force represents that it acts in the opposite direction of block
displacement (acts to restore the equilibrium). Substituting the expression (II) into expression (I)
\[
\begin{gather}
-kx=m\frac{d^{2}x}{dt^{2}}\\[5pt]
m\frac{d^{2}x}{dt^{2}}+kx=0
\end{gather}
\]
that is a Second-Order Linear Homogeneous Differential Equation. Dividing the equation by the
mass m
\[
\begin{gather}
\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x=0
\end{gather}
\]
substituting the values given in the problem
\[
\begin{gather}
\frac{d^{2}x}{dt^{2}}+\frac{1}{0.25}x=0\\[5pt]
\frac{d^{2}x}{dt^{2}}+4x=0 \tag{III}
\end{gather}
\]
Solution of \( \displaystyle \frac{d^{2}x}{dt^{2}}+4x=0 \)
The solution to this type of equation is found by substituting
Differentiation of the expression (IV) with respect to time, the function x(t) is a composite function, using the Chain Rule
The solution to this type of equation is found by substituting
\[
\begin{array}{l}
x=\operatorname{e}^{\lambda t}\\[10pt]
\dfrac{dx}{dt}=\lambda\operatorname{e}^{\lambda t}\\[10pt]
\dfrac{d^{2}x}{dt^{2}}=\lambda^{2}\operatorname{e}^{\lambda t}
\end{array}
\]
substituting these values into the differential equation
\[
\begin{gather}
\lambda ^{2}\operatorname{e}^{\lambda t}+4\operatorname{e}^{\lambda t}=0\\[5pt]
\operatorname{e}^{\lambda t}\left(\lambda^{2}+4\right)=0\\[5pt]
\lambda^{2}+4=\frac{0}{\operatorname{e}^{\lambda t}}\\[5pt]
\lambda ^{2}+4=0
\end{gather}
\]
that is the Characteristic Equation that has as its solution
\[
\begin{gather}
\Delta =b^{2}-4ac=0^{2}-4\times 1\times 4=0-18=-16
\end{gather}
\]
as Δ < 0 the roots are complex in the form a+bi, where
\( \mathrm{i}=\sqrt{-1\;} \)
\[
\begin{gather}
\lambda^{2}=-4\\[5pt]
\lambda =\sqrt{-4\;}\\[5pt]
\lambda_{1}=2\mathrm{i}\qquad \mathrm{or} \qquad \lambda_{2}=-2\mathrm{i}
\end{gather}
\]
The solution of the differential equation will be
\[
\begin{gather}
x=C_{1}\operatorname{e}^{\lambda_{1}t}+C_{2}\operatorname{e}^{\lambda_{2}t}\\[5pt]
x=C_{1}\operatorname{e}^{2\mathrm{i}t}+C_{2}\operatorname{e}^{-2\mathrm{i}t}
\end{gather}
\]
where C1 and C2 are constants of integration, using
Euler's Formula
\( \operatorname{e}^{\mathrm{i}\theta}=\cos \theta +\mathrm{i}\sin \theta \)
\[
\begin{gather}
x=C_{1}\left(\cos 2t+\mathrm{i}\sin 2t\right)+C_{2}\left(\cos 2t-\mathrm{i}\sin 2t\right)\\[5pt]
x=C_{1}\cos 2t+\mathrm{i}C_{1}\sin 2t+C_{2}\cos 2t-\mathrm{i}C_{2}\sin 2t\\[5pt]
x=\left(C_{1}+C_{2}\right)\cos 2t+\mathrm{i}\left(C_{1}-C_{2}\right)\sin 2t
\end{gather}
\]
defining two new constants α and β in terms of C1 and
C2
\[
\begin{gather}
\alpha \equiv C_{1}+C_{2}\\
\mathrm{e}\\
\beta \equiv \mathrm{i}(C_{1}-C_{2})
\end{gather}
\]
\[
\begin{gather}
x=\alpha \cos 2t+\beta \sin 2t
\end{gather}
\]
multiplying and dividing this expression by
\( \sqrt{\alpha^{2}+\beta^{2}\;} \)
\[
\begin{gather}
x=\left(\alpha \cos 2t+\beta\sin 2t\right)\frac{\sqrt{\alpha^{2}+\beta ^{2}}}{\sqrt{\alpha^{2}+\beta^{2}\;}}\\[5pt]
x=\sqrt{\alpha^{2}+\beta^{2}\;}\left(\frac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}\;}}\cos 2t+\frac{\beta }{\sqrt{\alpha^{2}+\beta^{2}\;}}\sin 2t\right)
\end{gather}
\]
setting
\[
\begin{array}{l}
A\equiv \sqrt{\alpha ^{2}+\beta^{2}\;}\\[10pt]
\cos \varphi \equiv \dfrac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}\;}}\\[10pt]
\sin \varphi \equiv \dfrac{\beta}{\sqrt{\alpha^{2}+\beta^{2}\;}}
\end{array}
\]
\[
\begin{gather}
x=A(\cos \varphi \cos 2t+\sin \varphi \sin 2t)
\end{gather}
\]
From the trigonometric identity
\( cos(a-b)=\cos a\cos b+\sin a\sin b \)
\[ cos(a-b)=\cos a\cos b+\sin a\sin b \]
\[
\begin{gather}
x=A\cos (2t-\varphi) \tag{IV}
\end{gather}
\]
where A and φ are constants determined by the Initial Conditions.Differentiation of the expression (IV) with respect to time, the function x(t) is a composite function, using the Chain Rule
\[
\begin{gather}
\frac{dx[v(t)]}{dt}=\frac{dx}{dv}\frac{dv}{dt}
\end{gather}
\]
with
\( x(v)=A\cos v \)
and
\( v(t)=(2t-\varphi) \)
\[
\begin{gather}
\frac{dx}{dt}=\frac{dx}{dv}\frac{dv}{dt}\\[5pt]
\frac{dx}{dt}=\frac{d(A\cos v)}{dv}\frac{d(2t-\varphi)}{dt}\\[5pt]
\frac{dx}{dt}=A(-\sin v)(2)\\[5pt]
\frac{dx}{dt}=-2 A\sin (2t-\varphi) \tag{V}
\end{gather}
\]
Substituting the Initial Conditions into expressions (IV) and (V)
\[
\begin{gather}
x(0)=0.5=A\cos (2\times 0-\varphi)\\[5pt]
0.5=A\cos (-\varphi)
\end{gather}
\]
since cosine is an even function, we have
\( \cos \varphi=\cos (-\varphi) \)
\[
\begin{gather}
0.5=A\cos \varphi \tag{VI}
\end{gather}
\]
\[
\begin{gather}
\frac{dx(0)}{dt}=0=-2A\sin (2\times 0-\varphi)\\[5pt]
0=-2A\sin (-\varphi)
\end{gather}
\]
since sine is an odd function, we have
\( \sin \varphi=-\sin(-\varphi) \)
\[
\begin{gather}
0=2A\sin \varphi \tag{VII}
\end{gather}
\]
solving the equation (VI) for A
\[
\begin{gather}
A=\frac{0.5}{\cos \varphi} \tag{VIII}
\end{gather}
\]
and substituting into the expression (VII)
\[
\begin{gather}
0=2\times \frac{0.5}{\cos \varphi}\times\sin \varphi\\[5pt]
0=\tan\varphi\\[5pt]
\varphi=\arctan(0)\\[5pt]
\varphi=0
\end{gather}
\]
substituting the values of φ into expression (VIII)
\[
\begin{gather}
A=\frac{0.5}{\cos 0}\\[5pt]
A=\frac{0.5}{1}\\[5pt]
A=0.5
\end{gather}
\]
substituting the constants A and φ into expression (IV)
\[
\begin{gather}
x(t)=0.5\cos 2t
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{x(t)=0.5\cos 2t}
\end{gather}
\]
b) The velocity is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\end{gather}
\]
the derivative is given by the expression (V), substituting the constants obtained above
\[
\begin{gather}
v(t)=-2\times 0.5\sin (2t-0)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v(t)=-\sin 2t}
\end{gather}
\]
c) The mechanical energy of a simple harmonic oscillator is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{E=\frac{1}{2}kA^{2}}
\end{gather}
\]
substituting the spring constant given in the problem and the amplitude calculated above
\[
\begin{gather}
E=\frac{1}{2}\times 1\times 0.5^{2}\\[5pt]
E=\frac{1}{2}\times 0.25
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{E=0.125\;\text{J}}
\end{gather}
\]
d) Plotting the graph of
\[
\begin{gather}
x(t)=0.5\cos 2t \tag{IX}
\end{gather}
\]
Setting x(t) = 0 in the expression (IX), we find the roots of the equation
\[
\begin{gather}
x(t)=0.5\cos 2t=0\\[5pt]
\cos 2t=\frac{0}{0.5}\\[5pt]
\cos 2t=0
\end{gather}
\]
the function cosine is equal to zero when its argument (2t) is equal to
\( \dfrac{\pi}{2} \),
\( \dfrac{3\pi}{2} \),
\( \dfrac{5\pi}{2} \),...,
\( \dfrac{(2n+1)\pi}{2} \),
with n = 0, 1, 2, 3,..., thus we have
\[
\begin{gather}
2t=\frac{(2n+1)\pi}{2}\\[5pt]
t=\frac{(2n+1)\pi}{2.2}\\[5pt]
t=\frac{(2n+1)\pi}{4}
\end{gather}
\]
For these values of t, we have the roots of the cosine function, the first four values will be, for n = 0, 1, 2 e 3, respectively, t = \( \dfrac{\pi}{4} \), \( \dfrac{3\pi}{4} \), \( \dfrac{5\pi}{4} \), \( \dfrac{7\pi}{4} \). These values are shown in Graph 1.
The function oscillates between +0.5 e − 0.5 of the amplitude.
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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 .