Question

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?

Short answer

Answer: The drop in amplitude between successive periods is $A_n(1-e^{-0.512})$, and the frequency of oscillations is 0.225 Hz.

Step-by-step solution

Calculate the damping constant

To find the damping constant, we need to use the kinetic friction coefficient, mass of the block, and gravitational acceleration (9.81 m/s²). The damping constant (c) can be calculated using the formula: $c={\mu }_k\ast m\ast g$ Where: $m=2\mathrm{kg}$ (mass of the block) ${\mu }_k=0.10$ (kinetic friction coefficient) $g=9.81\mathrm{m}/{\mathrm{s}}^2$ (gravitational acceleration) $c=0.10\ast 2\ast 9.81=1.962\mathrm{N}\cdot \mathrm{s}/\mathrm{m}$

Calculate the natural frequency

To find the natural frequency (wn) of the system, we will use the formula: $w_n=\sqrt{\frac{k}{m}}$ Where: $k=4\mathrm{N}/\mathrm{m}$ (spring stiffness) $m=2\mathrm{kg}$ (mass of the block) $w_n=\sqrt{\frac{4}{2}}=\sqrt{2}\mathrm{rad}/\mathrm{s}$

Calculate the damping ratio

To find the damping ratio (ζ), we will use the formula: $\zeta =\frac{c}{2\sqrt{mk}}$ Where: $c=1.962\mathrm{N}\cdot \mathrm{s}/\mathrm{m}$ (damping constant) $m=2\mathrm{kg}$ (mass of the block) $k=4\mathrm{N}/\mathrm{m}$ (spring stiffness) $\zeta =\frac{1.962}{2\sqrt{2\cdot 4}}=0.245$

Calculate the logarithmic decrement

To find the logarithmic decrement (δ) in amplitude, we will use the formula: $\delta =2\pi \frac{\zeta }{\sqrt{1-{\zeta }^2}}$ Where: $\zeta =0.245$ (damping ratio) $\delta =2\pi \frac{0.245}{\sqrt{1-{0.245}^2}}=0.512$

Calculate the drop in amplitude between successive periods

To find the drop in amplitude between successive periods, we will use the formula: $A_{n+1}=A_n{e}^{-\delta }$ Where: $A_n$ is the amplitude at period n $A_{n+1}$ is the amplitude at period n+1 $\delta =0.512$ (logarithmic decrement) $A_{n+1}=A_n{e}^{-0.512}$ The drop in amplitude between successive periods is equal to $A_n-A_{n+1}=A_n(1-e^{-0.512})$.

Calculate the frequency of oscillations

To find the frequency of oscillations, we will use the formula: $f=\frac{w_n}{2\pi }$ Where: $w_n=\sqrt{2}\mathrm{rad}/\mathrm{s}$ (natural frequency) $f=\frac{\sqrt{2}}{2\pi }=0.225\mathrm{Hz}$ The frequency of oscillations is 0.225 Hz. In conclusion, the drop in amplitude between successive periods is $A_n(1-e^{-0.512})$, and the frequency of oscillations is 0.225 Hz.