Engineering Acoustics. Malcolm J. Crocker

with time. Thus, using simple calculus:

(2.4)

      and

(2.5)

Equations are plotted in Figure 2.4.

Figure 2.4 Displacement, velocity, and acceleration.

Note, by trigonometric manipulation we can rewrite Eqs. (2.4) and (2.5) as (2.6) and (2.7):

(2.6)

      and

(2.7)

and from Eq. (2.3) we see that a = −ω² y.

      Equations tell us that for simple harmonic motion the amplitude of the velocity is ω or 2πf greater than the amplitude of the displacement, while the amplitude of the acceleration is ω² or (2πf)² greater. The phase of the velocity is π/2 or 90° ahead of the displacement, while the acceleration is π or 180° ahead of the displacement.

      Note we could have come to the same conclusions and much more quickly if we had used the complex exponential notation. Writing

      then

      and

      Example 2.1

      In a simple harmonic motion of frequency 10 Hz, the displacement amplitude is 2 mm. Calculate the maximum velocity amplitude and maximum acceleration amplitude.

      Solution

      Since ω = 2πf = 2π(10) = 62.83 rad/s. The velocity amplitude is calculated as

      ν = ω × 2 = 62.83 × 2 = 125.7 mm/s and the acceleration amplitude is a = ω² × 2 = (62.83)² × 2 = 7896 mm/s².

2.3 Vibrating Systems

      2.3.1 Mass–Spring System

       a) Free Vibration – Undamped

Suppose a mass of M kilogram is placed on a spring of stiffness K newton‐metre (see Figure 2.5a), and the mass is allowed to sink down a distance d metres to its equilibrium position under its own weight Mg newtons, where g is the acceleration of gravity 9.81 m/s². Taking forces and deflections to be positive upward gives

      (2.8)

Figure 2.5 Movement of mass on a spring: (a) static deflection due to gravity and (b) oscillation due to initial displacement y₀.

      Thus the static deflection d of the mass is

(2.8a)

      The distance d is normally called the static deflection of the mass; we define a new displacement coordinate system, where Y = 0 is the location of the mass after the gravity force is allowed to compress the spring.

      Suppose now we displace the mass a distance y from its equilibrium position and release it; then it will oscillate about this position. We will measure the deflection from the equilibrium position of the mass (see Figure 2.5b). Newton's law states that force is equal to mass × acceleration. Forces and deflections are again assumed positive upward, and thus

(2.9)

Let us assume a solution to Eq. (2.9) of the form y = A sin(ωt + φ). Then upon substitution into Eq. (2.9) we obtain

      We see our solution satisfies Eq. (2.9) only if

      The system vibrates with free vibration at an angular frequency ω rad/s. This frequency, ω, which is generally known as the natural angular frequency, depends only on the stiffness K and mass M. We normally signify this so‐called natural frequency with the subscript n. And so

      and from Eq. (3.2)

(2.10)

The frequency, f_n hertz, is known as the natural frequency of the mass on the spring. This result, Eq. (2.10), looks physically correct since