Chapter 2: Theoretical Background
2.6 Mass Spring Damper
Existing mass spring damper or vibration theory is directly applicable to point absorbers devices. Using standard notation, the governing equation is given by
ππ§Μ + ππ§Μ + ππ§ = 0 (2.87)
Dividing by π yields,
π§Μ + 2ππππ§Μ + ππ2π§ = 0 (2.88)
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π = π 2πππ (2.89) ππ= β π π (2.90)Here ππ is the natural frequency (in rad/s) of the system and π is a measure of damping
present in the system relative to the critical level of damping. In an actual vibration, there will always be some damping present. The solution of equation (2.88) must be a function which has the property that repeated differentiations do not change its form since the function and its first and second derivatives must be added together to give zero (Housner & Hudson, 1950). Assume a solution of the form;
π§(π‘) = π΄ππΌπ‘ (2.91)
Substituting equation (2.91) into equation (2.88) yields the characteristic equation,
πΌ2+ 2ππππΌ + ππ2= 0 (2.92)
An equation, whose roots are
πΌ1 = ππ(βπ + βπ2β 1) , πΌ2= ππ(βπ β βπ2β 1) (2.93)
The general solution of the governing equation is therefore
π§(π‘) = π΄1ππΌ1π‘+ π΄2ππΌ2π‘ =π΄1π(βπ+βπ 2β1)π ππ‘+ π΄ 2π(βπββπ 2β1)π ππ‘ (2.94)
The physical significance of the solution depends on the amount of damping present. There are three distinct scenarios to be considered:
π > 1: The system is said to be overdamped; the roots πΌ1, πΌ2 are real negative numbers.
When an overdamped system is given an initial displacement, the damping is sufficiently large to ensure the mass never oscillates past the static equilibrium position but rather exponentially subsides.
π = 0: The system is said to be critically damped with πΌ1 = πΌ2 = βππ. A critically damped
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π < 0: The system is underdamped. Such a system, when given a displacement, will oscillate about the equilibrium before eventually coming to rest. The frequency of oscillation is known as the damped angular frequency ππ , and is given byππ= ππ(β1 β π2) (2.95)
Rearranging equation (2.94) gives
z(t) = { π΄1ππ(β1βπ 2)π ππ‘+ π΄ 2πβπ(β1βπ 2)π ππ‘} πβππππ‘ = { π΄1πππππ‘+ π΄2πβππππ‘}πβππππ‘ (2.96)
Eulerβs identity πππ= cos(π) + π sin(π), allows the displacement to be written
π§(π‘) = πβππππ‘{ (π΄ 1+ π΄2) cos(πππ‘) + π( π΄1β π΄2)π ππ(πππ‘)} =πβππππ‘{π΄ 1 β² cos(π ππ‘) + π΄2β²π ππ(πππ‘)} (2.97)
Alternatively, equation (2.97) can be written in amplitude-phase form as
π§(π‘) = πΆπβππππ‘cos(π
ππ‘ + π) (2.98)
In an effort to determine the hydrodynamic parameters of added mass and damping, it is usual to carry out a decay test. A decay test or extinction test is one where the buoy is initially displaced a distance π§0 from its equilibrium position and then released. The buoy then
undergoes a damped free oscillation; parameters such as the natural frequency and hydrodynamic coefficients of added mass and damping can then be inferred from the recorded response of the buoy as it comes to rest. The unforced solution to a pure mass spring damper system is that of Equation (2.98) where the parameters πΆ and π are determined by the initial conditions. If the buoy is initially displaced a distance π§0, with an
accompanying initial velocity π£0= 0 , then
πΆ = π§0
β1 β π2 (2.99)
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π = tanβ1( βπβ1 β π2) (2.100)
A typical response of a damped free oscillation is shown in Figure 2:16(a). The envelope of decay is determined by the πΆπβππππ‘ term, whilst the trigonometric term determines the
frequency of oscillation.
Figure 2:16: Typical free decay curve of (a) a viscously damped system; (b) Coulomb damping
According to (Feeny & Liang, 1996):
βIn his Theory of Sound, Lord Rayleigh noted that, for the free vibration of a linear damped oscillator, βthe difference in the logarithms of successive extreme excursions is nearly constant, and is called the logarithmic decrementβ. In fact, the idea goes back to Hermann Helmhotz, who in 1863 applied the logarithmic decrement to determine frequency information in musical tones given a known damping coefficientβ
Therefore, the logarithmic ratio of successive peaks (called the logarithmic decrement Ξ΄) can be used to estimate the damping experimentally
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πΏ = ln (π§1 π§2 ) = πππππ= 2ππ β1 β π2 (2.101)The influence of mechanical friction in any test set up cannot be ignored. If the energy dissipation in a system is achieved purely by friction, as is the case in the free response of a mass and spring with Coulomb friction, then the amplitude of successive peaks decays linearly with time as shown in Figure 2:16(b). Thus, to truly encapsulate all the dynamics of the experimental system both viscous and Coulomb damping must be considered. A method of identifying Coulomb and viscous friction from free vibration decrements is described in (Liang and Feeny 1998).
The equation of motion of a single degree of freedom SDOF mass spring damper with dry friction is
ππ§Μ + ππ§Μ + ππ§ + π(π§Μ) = 0 (2.102)
The dry friction will always oppose the direction of motion. Hence
π(π§Μ) = πππ πππ(π§Μ)
π§Μ β 0, βππ β€ π(0) β€ ππ
(2.103)
Where ππ, is the kinetic friction force and ππ is the static friction force. The equilibrium
solution of Equation (2.102) can be obtained by letting π§Μ = π§Μ = 0. A physical consequence of the friction is that the system doesnβt have a single equilibrium point but rather a locus of equilibria βπ§π β€ π§ β€ π§π where π§π = ππ β . π
If follows from the above that if the buoy where to be submerged by some small distance π§ < π§π , then the restoring buoyancy force would not be able to overcome the stiction force
so the buoy would remain stationary.
Equation (2.102) is piecewise solvable, when written as
π§Μ + 2ππππ§Μ + ππ2π§ = βππ2π§π π§Μ > 0 (2.104)
π§Μ + 2ππππ§Μ + ππ2π§ = +ππ2π§π π§Μ < 0 (2.105)
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In a decay test the buoy is initially displaced π§(π‘0) = π0< βπ§π with π§Μ(π‘0) = 0. Then themotion starts with π§Μ > 0, and the response to equation (2.104) has the form
π§(π‘) = (π0+ π§π)πβπππ(π‘βπ‘0){πππ (ππ(π‘ β π‘0)) + π½π ππ(ππ(π‘ β π‘0))} β π§π (2.106)
Where π½ = π
β1βπ2.
This equation is valid until the velocity vanishes after time π‘ = π‘1= π‘0+ π πβ π. If the next
peak excursion point is denoted π1, then
π1= π§(π‘1) = βπβπ½ππ0+ (πβπ½π+ 1)π§π (2.107)
If π1> π§π then the mass will reverse direction and continue moving only with π§Μ < 0, so that
the governing equation is now that given in equation (2.105). The solution for this interval is
π§(π‘) = (π1β π§π)πβπππ(π‘βπ‘1){πππ (ππ(π‘ β π‘1)) + π½π ππ(ππ(π‘ β π‘1))} + π§π (2.108)
This is valid until π§Μ = 0 after time π‘ = π‘2= π‘1+ π πβ π. The next peak excursion point π2 =
π§(π‘2) = βπβπ½ππ1β (πβπ½π+ 1)π§π. If π2< βπ§π , motion will continue again. This process will
repeat itself until βπ§π β€ ππβ€ π§π , at which time the motion stops. This iterated process leads
to a recursive relation for the successive peaks and valleys in the oscillatory response:
ππ = βπβπ½πππβ1+ (β1)π(πβπ½π+ 1)π§π, i=1,2,... ,n. (2.109)
It is therefore possible to utilise successive maxima and minima points to isolate the viscous effect and then calculate the Coulomb effect. Taking the sum of extreme displacement values cancels out the friction contribution
ππ+ ππ+1
ππβ1+ ππ
= βπβπ½π (2.110)
Taking a logarithmic decrement gives the viscous dependence
ln (ππ+ ππ+1 ππβ1+ ππ