Natural frequency: Two objects with a spring

Question:
Two objects which have equal masses, $m$, are connected with a spring whose force constant is $k$. The objects are placed on a frictionless surface and oscillating without external forces. Find the natural frequency of this motion.

Answer:
$x_1$ and $x_2$ are the displacements of each object from the natural length. The displacement, $x$, is the one for the spring, so $x=x_1-x_2$ for object 1 and $-x=x_2-x_1$ for object 2. Set up the equations of motion:
\begin{eqnarray}
mx_1'' &=& -kx  \\
mx_2'' &=& kx
\end{eqnarray}
Subtract (2) from (1).
\begin{equation}
m(x_1''-x_2'') = -2kx
\end{equation}
Since $x'' = x_1''-x_2''$, we have
\begin{equation}
mx'' + 2kx = 0 \quad \rightarrow \quad x'' + \frac{2k}{m} = 0
\end{equation}
The solution of this differential equation is
\begin{equation}
x = A\sin\left(\sqrt{\frac{2k}{m}} t\right) + B\cos\left(\sqrt{\frac{2k}{m}} t\right)
\end{equation}
where $\sqrt{\frac{2k}{m}}$ is the natural frequency of this spring motion.

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Horizontal spring motion: Wave equation

Question:
An object attached to a spring moves on a horizontal frictionless surface. This is a harmonic motion with amplitude of 0.16 m and period of 2.0 s. The mass is released from rest at $t=0$ s and $x=-0.16$ m. Find the displacement as a function of time.

Answer:
The simple harmonic motion is given as
\[
x = A\cos(\omega t + \delta)
\]
The amplitude, $A$, is 0.16 m. Since the period $T=2.0$ s, the frequency is calculated as
\[
f = \frac{1}{2.0}=0.50 \ \mathrm{Hz}
\]
Then, the angular frequency is
\[
\omega = 2\pi f = \pi \ \mathrm{rad/s}
\]
In order to find the phase, $\delta$, we need to check the initial state. When $t=0$, the wave equation becomes
\begin{eqnarray*}
-0.16 &=& 0.16\cos(\delta)  \\
\cos(\delta) &=& -1.0  \\
\delta &=& \cos^{-1}(-1.0)  \\
\delta &=& \pi
\end{eqnarray*}
Thus, we have the final form of the equation.
\[
x = 0.16\cos(\pi t + \pi)
\]

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Damped oscillator: Spring motion

Question:
A 0.400-kg object is attached to a spring of its force constant $k=200$ N/m and this motion is subject to a resistive force, $-bv$, which is proportional to the velocity. If the damped frequency is 99.5 % of the undamped frequency, find the value of $b$.

Answer:
The equation of motion gives:
\[
\sum F = -kx - b\frac{dx}{dt} = m\frac{d^2x}{dt^2}
\]
Let us put $\frac{dx}{dt}=x'\equiv Dx$ and $\frac{d^2x}{dt^2}=x''\equiv D^2x$. The differential equation becomes
\begin{eqnarray}
x''+\frac{b}{m}x'+\frac{k}{m}x &=& 0  \quad \mathrm{or} \\
\left(D^2+\frac{b}{m}D+\frac{k}{m}\right)x &=& 0
\end{eqnarray}
Let us also put $\omega_0 = \sqrt{\frac{k}{m}}$, which is called the rational frequency. This system is a damped oscillation, so the discriminant must be negative. Namely,
\begin{equation}
\left(\frac{b}{m}\right)^2 - 4\omega^2 < 0 \ \rightarrow \ \left(\frac{b}{m}\right)^2 < 4\omega^2
\end{equation}
The solution should be
\begin{equation}
D = -\frac{b}{2m} \pm \sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}i
\end{equation}
Therefore,
\begin{equation}
x = Ae^{-\frac{m}{2m}t}\cos(\omega t + \phi)
\end{equation}
where $\omega=\sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}$, which is called the damped frequency. This frequency is 99.5% of the undamped, so
\begin{eqnarray}
& & 0.995\omega_0 = \sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}  \\
& & 0.010\omega^2_0 = \frac{1}{4}\frac{b^2}{m^2}  \\
& & b = \sqrt{0.010 \times 4 \frac{k}{m} m^2}   \\
& & b = \sqrt{0.040 km}
\end{eqnarray}
Therefore,
\[
b = 1.789 \ \mathrm{kg/s}
\]

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