Question

A mass of 1 kg is attached to a spring having a stiffness of 3920 N/m. The mass slides on a horizontal surface, the co-efficient of friction between mass and surface being 0.1. Determine the frequency of vibration of the system and amplitude after one cycle if the initial amplitude is 0.25 cm.

Answer 1

Question #81301

A mass of $1\,\mathrm{kg}$ is attached to a spring having a stiffness of $3920\,\mathrm{N/m}$. The mass slides on a horizontal surface, the co-efficient of friction between mass and surface being 0.1. Determine the frequency of vibration of the system and amplitude after one cycle if the initial amplitude is $0.25\,\mathrm{cm}$.

Answer:

The frequency of vibration is given by:

$$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}},$$

where $k$ is spring stiffness and $m$ is the mass. Thus

$$f = \frac{1}{2\pi} \sqrt{\frac{3920}{1}} = 9.965\,\mathrm{Hz}.$$

Based on the coefficient of friction $\mu = 0.1$, we can determine the frictional force as follows

$$F = \mu m g = 0.1 \cdot 1 \cdot 9.81 = 0.981\,\mathrm{N}.$$

Since, the reduction in amplitude per cycle is given by

$$\Delta A = \frac{4F}{k} = \frac{4 \cdot 0.981}{3920} = 0.0010\,\mathrm{m} = 0.1\,\mathrm{cm},$$

the amplitude after one cycle equal to $0.25 - 0.1 = 0.15\,\mathrm{cm}$.