Solved Problem on Harmonic Oscillations
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A body of mass m is resting on a wooden support. The support starts to oscillate with simple harmonic motion, increasing the frequency of oscillations until the body begins to slide over the wood.
a) Calculate the coefficient of friction between the body and the wood as a function of the period T and the amplitude A of the oscillations;
b) For period T = 3 s and amplitude A = 0.4 m, what is the coefficient of friction?


Problem data:
  • Body mass:    m;
  • Period of oscillations:    T;
  • Amplitude of oscillations:    A.
Problem diagram:

The forces acting on the system are the gravitational force \( {\vec F}_{g} \), the reaction force \( \vec{N} \), the force of friction \( {\vec{F}}_{f} \), and the external force \( \vec{F} \) (Figure 1).
Figure 1

The system is initially at rest relative to position O (Figure 2-A).
An oscillating force begins to act on the wooden support pushing it back and forth. The body on the support remains in the same position relative to the support, but not relative to position O, it moves along with the support (Figures 2-B, 2-C, and 2-D).
The external force \( \vec{F} \) increases, and as long as it is smaller than the force of friction \( {\vec{F}}_{f} \), the body remains in the same position on the support (Figures 2-B, 2-C, and 2-D)..
This situation continues until the external force is greater than the force of friction. At that moment, the support moves under the body and the force of friction is not enough to make the body remain in the same position on the support (Figure 2-E).
Figure 2

Solution

a) Applying Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]
  • Direction x:
We have the external oscillation force \( \vec{F} \) and the force friction \( {\vec{F}}_{f} \), and as long as the external force is less than the force of friction, the body remains at rest relative to the wooden support. In simple harmonic motion, the acceleration is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {a=-\omega ^{2}A\cos \omega t} \end{gather} \]
the acceleration is maximum when   \( \cos \omega t=1 \), in magnitude
\[ \begin{gather} a_{max}=\omega ^{2}A \tag{II} \end{gather} \]
Substituting expression (II) into expression (I), the maximum external force will be
\[ \begin{gather} F=m\omega ^{2}A \tag{III} \end{gather} \]
The frequency of oscillations is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\omega =\frac{2\pi}{T}} \tag{IV} \end{gather} \]
substituting expression (IV) into expression (III)
\[ \begin{gather} F=mA\left(\frac{2\pi}{T}\right)^{2} \tag{V} \end{gather} \]
The force of friction is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{f}=\mu N} \tag{VI} \end{gather} \]
  • Direction y:
We have the gravitational force \( {\vec F}_{g} \) and the reaction force \( \vec{N} \), and as there is no motion in that direction, the two forces cancel each other out
\[ \begin{gather} F_{g}=N \tag{VII} \end{gather} \]
the gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{VIII} \end{gather} \]
substituting expression (VIII) into expression (VII)
\[ \begin{gather} N=mg \tag{IX} \end{gather} \]
Substituting expression (IX) into expression (VI)
\[ \begin{gather} F_{f}=\mu mg \tag{X} \end{gather} \]
Equating expressions (V) and (X)
\[ \begin{gather} \mu \cancel{m}g=\cancel{m}A\left(\frac{2\pi}{T}\right)^{2} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\mu =\frac{A}{g}\left(\frac{2\pi}{T}\right)^{2}} \end{gather} \]

b) For T = 3 s and A = 0.4 m, and assuming π =3.14 and g = 9.81 m/s2
\[ \begin{gather} \mu =\frac{0.4}{9.81}\times\left(\frac{2\times 3.14}{3}\right)^{2} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\mu =0.1} \end{gather} \]
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