Engineering Acoustics. Malcolm J. Crocker

f_n increases. If M increases with K constant, f_n decreases. These are the results we also find in practice.

      Example 2.2

      A machine of mass 600 kg is mounted on four springs of stiffness 2 × 10⁵ N/m each. Determine the natural frequency of the system

      Solution

      We model the system as a hanging spring‐mass system (see Figure 2.5). Equation (2.9) governs the displacement of the machine from its static‐equilibrium position. Since we have four equal springs, the equivalent stiffness is 4 × 2 × 10⁵ = 8 × 10⁵ N/m. The natural frequency is then determined using Eq. (2.10) as

      We have seen that a solution to Eq. (2.9) is y = A sin(ωt + ϕ) or the same as Eq. (2.3). Hence we know that any system that has a restoring force that is proportional to the displacement will have a displacement that is simple harmonic. This is an alternative definition to that given in Section 2.2 for simple harmonic motion.

b) Free Vibration – Damped

Many mechanical systems can be adequately described by the simple mass–spring system just discussed above. However, for some purposes it is necessary to include the effects of losses (sometimes called damping). This is normally done by including a viscous damper in the system (see Figure 2.6). See Refs. [8, 9] for further discussion on passive damping. With viscous or “coulomb” damping the friction or damping force F_d is assumed to be proportional to the velocity, dy/dt. If the constant of proportionality is R, then the damping force F_d on the mass is

      (2.11)

      and Eq. (2.9) becomes

      (2.12)

      or equivalently

(2.13)

      where the dots represent single and double differentiation with respect to time.

Figure 2.6 Movement of damped simple system.

The solution of Eq. (2.13) is most conveniently found by assuming a solution of the form: y is the real part of A e^jλt where A is a complex number and λ is an arbitrary constant to be determined. By substituting y = A e^jλt into Eq. (2.13) and assuming that the damping constant R is small, R < (4MK)^1/2 (which is true in most engineering applications), the solution is found that:

(2.14)

      Here ω_d is known as the damped “natural” angular frequency:

      (2.15)

where ω_n is the undamped natural frequency . The motion described by Eq. (2.14) is plotted in Figure 2.7.

Figure 2.7 Motion of a damped mass–spring system, R < (4MK)^1/2.

      The amplitude of the motion decreases with time unlike that for undamped motion (see Figure 2.3). If the damping is increased until R equals (4MK)^1/2, the damping is then called critical, R_crit = (4MK)^1/2. In this case, if the mass in Figure 2.6 is displaced, it gradually returns to its equilibrium position and the displacement never becomes negative. In other words, there is no oscillation or vibration. If R > (4MK)^1/2, the system is said to be overdamped.

      The ratio of the damping constant R to the critical damping constant R_crit is called the damping ratio δ:

      (2.16)

      In most engineering cases, the damping ratio, δ, in a structure is hard to predict and is of the order of 0.01–0.1. There are, however, several ways to measure damping experimentally [8, 9].

      Example 2.3

      A 600‐kg machine is mounted on springs such that its static deflection is 2 mm. Determine the damping constant of a viscous damper to be added to the system in parallel with the springs, such that the system is critically damped.

      Solution

The static deflection is given by Eq. (2.8a) as d = Mg/K. Therefore K = Mg/d = 600(9.8)/2 × 10⁻³ = 294 × 10⁴ N/m. The system is critically damped when the damped constant R_crit = (4MK)^1/2 = (4 × 600 × 294 × 10⁴)^1/2 = 84 000 Ns/m.

       (c) Forced Vibration – Damped

If a damped spring–mass system is excited by a simple harmonic force at some arbitrary angular forcing frequency ω (see Figure 2.8), we now obtain the equation of motion Eq. (2.17):

(2.17)