A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the block and the surface it moves on are respectively \(\mu_{s}=0.12\) and \(\mu_{k}=0.10\), determine the drop in amplitude between successive periods during free vibration. What is the frequency of the oscillations?

The natural frequency (\(\omega_n\)) can be calculated as the square root of the stiffness divided by the mass. In this case, \(\omega_n = \sqrt{4/2}\). $$ \omega_n = \sqrt{\frac{4 \, \mathrm{N/m}}{2 \, \mathrm{kg}}} = 1 \, \mathrm{rad/s} $$

To determine the total friction force, we first find the maximum static friction (\(F_s\)), which can be calculated as \(F_s = \mu_s \cdot m \cdot g\), where \(\mu_s\) is the static friction coefficient, \(m\) is the mass, and \(g\) is the acceleration due to gravity (approximately \(9.81 \, \mathrm{m/s^2}\)). Next, we find the maximum kinetic friction (\(F_k\)), which can be calculated as \(F_k = \mu_k \cdot m \cdot g\), where \(\mu_k\) is the kinetic friction coefficient. $$ F_s = 0.12 \cdot 2 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^2} = 2.36 \, \mathrm{N} $$ $$ F_k = 0.10 \cdot 2 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^2} = 1.96 \, \mathrm{N} $$ Since the system is in free vibration, the maximum friction will always be the kinetic friction force.

The energy dissipated per cycle due to friction can be calculated as the work done by friction throughout one full oscillation. Assuming a simple harmonic motion for the system, this can be calculated as the product of the total friction force (\(F_k\)), the amplitude of oscillation (\(A\)), and the number of oscillation cycles (\(N\)). The energy, after \(N\) cycles, can be expressed as: $$ E_d = F_k \cdot A \cdot N $$

Given the energy dissipation per cycle calculated in step 3, we can find the drop in amplitude between successive periods, denoted as \(\Delta A\). To do this, we set the energy dissipation equal to the change in potential energy of the system, which is: $$ \frac{1}{2} k (\Delta A)^2 = E_d $$ Solving for \(\Delta A\): $$ \Delta A = \sqrt{\frac{2 E_d}{k}} = \sqrt{\frac{2 F_k \cdot A \cdot N}{4 \mathrm{~N} / \mathrm{m}}} = \sqrt{\frac{2 \cdot 1.96 \mathrm{~N} \cdot A \cdot N}{4 \mathrm{~N} / \mathrm{m}}} $$

The frequency of the oscillations can be calculated as the natural frequency divided by \(2\pi\): $$ f = \frac{\omega_n}{2 \pi} = \frac{1 \, \mathrm{rad/s}}{2 \pi} = 0.159 \, \mathrm{Hz} $$ In conclusion, the drop in amplitude between successive periods during free vibration is given by the expression: $$ \Delta A = \sqrt{\frac{3.92 \mathrm{~N} \cdot A \cdot N}{2 \mathrm{~N} / \mathrm{m}}} $$ The frequency of the oscillations is \(0.159 \, \mathrm{Hz}\).