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u(0) = u0, u'(0) = v0. (8)
It follows from Theorem 3.2.1 that these conditions give a mathematical problem that has a unique solution. This is consistent with our physical intuition that if the mass is set in motion with a given initial displacement and velocity, then its position will be determined uniquely at all future times. The position of the mass is given (approximately) by the solution of Eq. (7) subject to the prescribed initial conditions (8).
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is displaced an additional 6 in. in the positive direction and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/sec. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
The required initial value problem consists of the differential equation (7) and initial conditions (8), so our task is to determine the various constants that appear in these equations. The first step is to choose the units of measurement. Based on the statement of the problem, it is natural to use the English rather than the metric system of units. The
3.8 Mechanical and Electrical Vibrations
only time unit mentioned is the second, so we will measure t in seconds. On the other hand, both the foot and inch appear in the statement as units of length. It is immaterial
which one we use, but having made a choice, it is important to be consistent. To be
definite, let us measure the displacement u in feet.
Since nothing is said in the statement of the problem about an external force, we assume that F(t) = 0. To determine m note that
w 4 lb 1 lb-sec2 g 32 ft/ sec2 8 ft
The damping coefficient y is determined from the statement that y u' is equal to 6 lb when u' is 3 ft/sec. Therefore
3 ft/sec ft
The spring constant k is found from the statement that the mass stretches the spring by 2 in., or by 1/6 ft. Thus
4 lb „ËÜ
1/6 ft ft
Consequently, Eq. (7) becomes
8 u" + 2U + 24u = 0,
The initial conditions are
u" + 16u' + 192u = 0.
u (0) = 2, u '(0) = 0.
The second initial condition is implied by the word “released” in the statement of the problem, which we interpret to mean that the mass is set in motion with no initial velocity.
Undamped Free Vibrations. If there is no external force, then F (t) = 0 in Eq. (7). Let us also suppose that there is no damping, so that y = 0; this is an idealized configuration of the system, seldom (if ever) completely attainable in practice. However, if the actual damping is very small, the assumption of no damping may yield satisfactory results over short to moderate time intervals. Then the equation of motion (7) reduces to
mu" + ku = 0. (11)
The general solution of Eq. (11) is
u = A cos m0t + Bsin rnQt, (12)
= k/ m. (13)
The arbitrary constants A and B can be determined if initial conditions of the form (8) are given.
Chapter 3. Second Order Linear Equations
In discussing the solution of Eq. (11) it is convenient to rewrite Eq. (12) in the form
u = R cos(m0t — S), (14)
u = Rcos S cos m0t + Rsin S sinm0t. (15)
By comparing Eq. (15) with Eq. (12), we find that A, B, R, and S are related by the equations
A = Rcos S, B = Rsin S. (16)
R = y/ A2 + B2, tan S = B/ A. (17)
In calculating S care must be taken to choose the correct quadrant; this can be done by checking the signs of cos S and sin S in Eqs. (16).
The graph of Eq. (14), or the equivalent Eq. (12), for a typical set of initial conditions is shown in Figure 3.8.3. The graph is a displaced cosine wave that describes a periodic, or simple harmonic, motion of the mass. The period of the motion is
^ 2n / m \1/2
T = — = 2n (-) . (18)
m0 V k /
The circular frequency m0 = Vk/ m, measured in radians per unit time, is called the natural frequency of the vibration. The maximum displacement R of the mass from equilibrium is the amplitude of the motion. The dimensionless parameter S is called the phase, or phase angle, and measures the displacement of the wave from its normal position corresponding to S = 0.
Note that the motion described by Eq. (14) has a constant amplitude that does not diminish with time. This reflects the fact that in the absence of damping there is no way for the system to dissipate the energy imparted to it by the initial displacement and velocity. Further, for a given mass m and spring constant k, the system always vibrates at the same frequency m0, regardless of the initial conditions. However, the initial conditions do help to determine the amplitude of the motion. Finally, observe from Eq. (18) that T increases as m increases, so larger masses vibrate more slowly. On the other hand, T decreases as k increases, which means that stiffer springs cause the system to vibrate more rapidly.
FIGURE 3.8.3 Simple harmonic motion; u = Rcos(«0f — S).
3.8 Mechanical and Electrical Vibrations