Question 1.4.4: Model the mass of the spring in the system shown in Figure 1......

One approach to considering the mass of the spring in analyzing the system vibration response is to calculate the kinetic energy of the spring. Consider the kinetic energy of the element dy of the spring. If m_{s} is the total mass of the spring, then \frac{m_{s}}{l} dy, is the mass of the element dy. The velocity of this element, denoted by ν_{dy}, may be approximated by assuming that the velocity at any point varies linearly over the length of the spring:

ν_{dy} = \frac{y}{l} \dot{x}(t)

The total kinetic energy of the spring is the kinetic energy of the element dy integrated over the length of the spring:

T_{spring} = \frac{1}{2} \int_{0}^{l}{\frac{m_{s}}{l} \left[\frac{y}{l} \dot{x}\right]²} dy

= \frac{1}{2} \left(\frac{m_{s}}{3}\right) \dot{x}²

From the form of this expression, the effective mass of the spring is \frac{m_{s}}{3}, or one-third of that of the spring. Following the energy method, the maximum kinetic energy of the system is thus

T_{max} = \frac{1}{2} \left(m  +  \frac{m_{s}}{3}\right) w²_{n}A²

Equating this to the maximum potential energy, \frac{1}{2}kA² yields the fact that the natural frequency of the system is

w_{n} = \sqrt{\frac{k}{m  +  m_{s}/3}}

Thus, including the effects of the mass of the spring in the system decreases the natural frequency. Note that if the mass of the spring is much smaller than the system mass m, the effect of the spring’s mass on the natural frequency is negligible.