Chapter 3 Notes - DAMPED HARMONIC MOTION PDF

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DAMPED HARMONIC MOTION...

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Chapter 3

DAMPED H ARMON IC MOTION SPECIFIC AIM S ➢ To show a method for solving linear DE’s with constant coefficients. ➢ To understand the physics of damped simple harmonic motion. CONTENT OF CHAPTER - Introduction - Damped Harmonic Motion - Overdamped - Critically Damped - Underdamped - Logarithmic Decrement - Energy Dissipation

INTRODUCTION Background reading Read the section in the Oscillations chapter of Halliday/Resnick on damped HM. Notice that damped HM is more common in our everyday experience than that of (pure) SHM. DAM PED HARMONIC MOTION (DAMPED HM)

The frictional force opposing HM is often directly proportional to velocity of the oscillating body. From the sketch above it follows from Newton II: 𝐹 = 𝑚𝑥󰇘 󰆹i = 𝐹𝑆 + 𝐹𝐷 = −𝑘𝑥󰆹i − 𝑏𝑣i󰆹 or 𝑚𝑥󰇘 + 𝑏𝑥󰇗 + 𝑘𝑥 = 0 . . . (1)

where b is called the damping constant of the system. Let us consider (1) a bit more carefully. Firstly, let’s write it in a form that will be easier to work with: where

𝑥󰇘 + 2𝑝𝑥󰇗 + 𝜔 2 𝑥 = 0 𝑘

𝜔 = √ 𝑚 and 𝑝 =

. . . (2)

𝑏 . 2𝑚

That is, 𝜔 is the angular frequency of the associated SHM (i.e. if there were no friction). From general DE theory, the general solution of any homogenous linear DE with constant coefficients, such as (2) (2 constant coefficients are 1,2𝑝, ω2 ), can be found by substituting 𝑒 𝜆𝑡 and finding the roots of the resulting polynomial i.t.o. 𝜆. By the fundamental theorem of algebra, up to algebraic multiplicity, there will be 𝑛 roots (some complex and/or repeating) which can be used to express a general solution for the DE. We will illustrate this process in the case of (2). Substitute 𝑥 = 𝑒 𝜆𝑡 into (2). This yields: so

𝜆2 + 2𝑝𝜆 + 𝜔 2 = 0 𝜆 = −𝑝 ± √𝑝2 − 𝜔 2

… (3)

Thus (3) has two distinct real roots if

𝑝2 > 𝜔 2 or, equivalently, 𝑏 2 > 4𝑘𝑚

(3) has a repeating real root if

𝑝2 = 𝜔 2 or, equivalently, 𝑏 2 = 4𝑘𝑚

(3) has a conjugate pair of complex roots if

𝑝2 < 𝜔 2 or, equivalently, 𝑏 2 < 4𝑘𝑚

Case 1: 𝒑𝟐 > 𝝎𝟐 (two distinct real roots) - overdamped oscillatory motion (3) has two distinct real roots 𝜆1 = −𝑝 − √𝑝 2 − 𝜔 2 and 𝜆2 = −𝑝 + √𝑝 2 − 𝜔 2 . We thus

have two linearly independent solutions, 𝑥 = 𝑒 𝜆1 𝑡 and 𝑥 = 𝑒 𝜆2 𝑡 of the 2nd order linear DE (2). By general DE theory, the general solution of (2) is therefore where 𝐴, 𝐵 ∈ ℝ are arbitrary.

𝑥 = 𝐴𝑒 𝜆1 𝑡 + 𝐵𝑒 𝜆2 𝑡

In this case the damping force is large compared to the restoring force. This results in a displaced mass slowly returning to the equilibrium position with no oscillation. The system is said to be overdamped. Case 2: 𝒑𝟐 = 𝝎𝟐 (a repeating real root) – critically damped oscillatory motion (3) has one repeating real root 𝜆 = −𝑝. So 𝑥 = 𝑒 −𝑝𝑡 is one solution of (2), but by general DE theory we need another solution linearly independent from 𝑥 = 𝑒 −𝑝𝑡 to be able to write down an expression for the general solution. Check (homework) that in this case 𝑥 = 𝑡𝑒 −𝑝𝑡 is another solution of (2). Thus the general solution of (2) is therefore where 𝐴, 𝐵 ∈ ℝ are arbitrary.

𝑥 = 𝐴𝑒 −𝑝𝑡 + 𝐵𝑡𝑒 −𝑝𝑡 = 𝑒 −𝑝𝑡 (𝐴 + 𝐵𝑡)

In this case the damping force is just large enough to damp out any oscillations. The system is said to be critically damped, but even a slight reduction in resistance will bring us to the 3rd case, which shows the more interesting behavior.

Case 3: 𝒑𝟐 < 𝝎𝟐 (conjugate pair of complex roots) – lightly-/under damped oscillatory motion (3) has a pair of conjugate complex roots 𝜆1 = −𝑝 + 𝑖√ω2 − 𝑝2 and 𝜆2 = −𝑝 − 𝑖√ω2 − 𝑝 2 . We thus have two complex solutions of (2):

𝑥 = 𝑒 −𝑝𝑡 (cos ω′𝑡 + 𝑖 𝑠𝑖𝑛 𝜔′𝑡) and 𝑥 = 𝑒 −𝑝𝑡 (cos 𝜔′𝑡 − 𝑖 𝑠𝑖𝑛 𝜔′𝑡)

where ω′ = √ω2 − 𝑝 2 . However, one can easily show from this, either by checking directly or by using the linearity of (2), that

𝑥 = 𝑒 −𝑝𝑡 (cos 𝜔′𝑡) and 𝑥 = 𝑒 −𝑝𝑡 (𝑠𝑖𝑛 𝜔′𝑡) are two linearly independent real solutions of (2). Hence the general solution of (2) is given by 𝑥 = 𝐴𝑒 −𝑝𝑡 (cos ω′𝑡) + 𝐵𝑒 −𝑝𝑡 (sin ω′𝑡) = 𝑒 −𝑝𝑡 (𝐴 cos 𝜔′ 𝑡 + 𝐵 sin 𝜔 ′ 𝑡) where 𝐴, 𝐵 ∈ ℝ are arbitrary. By using the cosine addition formula this can be written as 𝑥 = 𝐶𝑒 −𝑝𝑡 (cos (ω′𝑡 + ϕ)) . . . (3) where 𝐶, ϕ ∈ ℝ are arbitrary. Here the system is lightly damped and gives oscillatory damped harmonic motion (short damped oscillatory motion)

Ex1. Consider the following mass-spring-dashpot arrangement, with 𝑚 = 20 kg, 𝑘 = 2000 N/m:

Find the displacement of the mass from equilibrium when pulled 0.5 m from its equilibrium position before being released, in the case that a) 𝑏 = 500 N.s.m−1 b) 𝑏 = 400 N.s.m−1 c) 𝑏 = 100 N.s.m−1

Ex2. Consider the following RLC circuit.

It’s an oscillator circuit if the resistance 𝑅 is low enough, meaning the current continuously changes direction at a certain frequency. For what Ohm range of 𝑅 is the above circuit an oscillator circuit if 𝐶 = 4.5 𝜇𝐹 and 𝐿 = 350 mH?

The left figure compares the underdamped case (c) of Ex1. with the associated SHM case, and the right figure compares the displacement of all three cases exhibited in (a),(b) and (c) of Ex1. Remark Though there is a strong similarity between the oscillations of the SHM and the underdamped oscillations, as can be seen on the left figure on the previous page, there are some significant differences. 1.) The exponentially decaying amplitude of the oscillations 2.) The periods are not the same, due to the term in the definition of ω′.

𝑝2 =

𝑏2 2𝑚

It can be seen from the right figure on the previous page that the critically damped HM approaches equilibrium the fastest. This property is used in some applications, such as shock absorbers in vehicles.

THE LOGARITHMIC DECREMEN T The logarithmic decrement  is defined as the natural logarithm of the ratio of two maxima a period apart, i.e. 𝐶𝑒 −𝑝𝑡 𝑐𝑜𝑠 (𝜔′𝑡 − 𝜙) 𝑥(𝑇) = ln [ −𝑝(𝑡+𝑇) ] = ln 𝑒 𝑝𝑇 = 𝑝𝑇 𝛿 = ln 𝑥(𝑡 + 𝑇) 𝐶𝑒 𝑐𝑜𝑠 (𝜔′(𝑡 + 𝑇) − 𝜙 ) Since critical damping is often desirable, there needs to be way in which one can tell how close/far a system is to critical damping. For this there is the damping ratio: 𝑏 𝜁= 𝑏crit

Hence 𝜁 < 1 means the system is underdamped, 𝜁 = 1 means the system is critically damped, and 𝜁 > 1 means the system is overdamped. Clearly the closer to 1 the damping ratio is, the closer the system is to being critically damped. Since 𝑏crit = 2√𝑘𝑚, it follows that 𝜁 =

𝑏

2𝑚𝜔

=

𝑝

𝜔

So it is an easy enough ratio to calculate, if you have the constants 𝑏, 𝑘 and 𝑚. But what are you supposed to do if you do not have any of the constants? Say you are only looking at a graph of the oscillations of an underdamped system? Then we can use the logarithmic decrement, which is the natural logarithm the ratio of the 1

two maxima a period 𝑇 apart, or of the natural logarithm of the ratio of the two maxima 𝑛 𝑛 periods apart: 1 𝑥(𝑡) 𝐶𝑒 −𝑝𝑡 𝑐𝑜𝑠(ω′ 𝑡 − ϕ) 1 1 δ = ln = ln [ −𝑝(𝑡+𝑛𝑇) ] = ln 𝑒 𝑛𝑝𝑇 = 𝑝𝑇 ′ 𝑛 𝑥(𝑡 + 𝑛𝑇) 𝑛 𝐶𝑒 𝑐𝑜𝑠(ω (𝑡 + 𝑛𝑇) − ϕ) 𝑛 This is a value you can simply read from a graph, and it’s related to the damping ratio by 1 𝜁= 2 √1 + (2π ) δ ENERGY DISSIPATION

It is obvious that the amplitude in damped oscillations decreases because the resistive force consumes energy. It is also easily proven mathematically. Consider the mechanical energy 1 1 𝐸𝑀 = 𝐸𝐾 + 𝐸𝑃 = 𝑚𝑥 2󰇗 + 𝑘𝑥 2 2 2 For constant 𝐸𝑀 , the derivative should be equal to zero, but in this case 𝑑 1 1 𝑑𝐸𝑀 = ( 𝑚𝑥 2󰇗 + 𝑘𝑥 2 ) = 𝑚𝑥𝑥󰇘󰇗 + 𝑘𝑥𝑥󰇗 = 𝑥󰇗 (𝑚𝑥󰇘 + 𝑘𝑥) 2 𝑑𝑡 𝑑𝑡 2 From our original DE we have 𝑚𝑥󰇘 + 𝑘𝑥 = −𝑏𝑥󰇗 which when substituted above reveals the rate of loss of energy as 𝑑𝐸 = −𝑏𝑥 2󰇗 < 0 𝑑𝑡 which is also the rate of doing work against the frictional force 𝐹𝐷 = −𝑏𝑣. For an undamped 1 oscillator, the total energy is constant and given by 𝐸𝑃 = 𝑘𝐶 2 where 𝐶 is the amplitude of 2 the motion. In (3) the amplitude of the oscillations when underdamped decreases by 𝑒 −𝑝𝑡, which when substituted for the SHM case gives us an estimate for the total energy of the underdamped oscillator w.r.t. to time 1 1 𝐸(𝑡) ≈ 𝑘(𝑒 −𝑝𝑡 )2 𝐶 2 = 𝑘𝐶 2 𝑒 −𝑏𝑡/𝑚 2 2 The estimate is more accurate the smaller the damping force is, i.e the smaller b is.

Tutorial problems

1) Consider a lightly damped oscillator with m = 0.010 kg and k = 36 N m-1. Calculate (a) the value of b which would make the amplitude decrease from a to ae-1 in 1 s; (b) the value of b which would produce critical damping. [Hint: Amplitude at any 𝑡 is 𝐶𝑒 −𝑝𝑇 ][Ans. 0.020 kg s-1 1.2 kg s-1] 2) For the damped oscillator system shown in the figure to the right, suppose now that the block has a mass of 1.50 kg and the spring constant is 8.00 N/m. The damping force is given by 𝑏 = 230 g/s. The block is pulled down 12.0 cm and released. (a) Calculate the time required for the amplitude of the resulting oscillations to fall to one-third of its initial value. (b) How many oscillations are made by the block in this time? [Ans. (a) 14.3 s (b) 5.27]

3*) The suspension system of a 2000 kg automobile sent in for repairs “sags” 10 cm when the vehicle is lowered to the ground, and the chassis oscillates up and down with an amplitude that decreases by 50% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 500 kg. [Hint: first use the estimate ω′ = ω, and optionally see if you can also work it out without the use of that estimate – compare the answers] [Ans. (a) 4.9x102 N/cm (b) 1.1x103 kg/s] 4) Suppose that the mass in a mass-spring-dashpot system with m = 10kg, b = 9.0 kg/s, and k = 2.0 N/m is set in motion with x(0) = 0 and x'(0) = 5.0. Find (a) the position function x(t), and (b) how far the mass moves to the right before starting back toward the origin. 1

2

[Ans. (a) 𝑥(𝑡) = −50𝑒 −2𝑡 + 50𝑒 − 5𝑡 (b) 𝑥(2.2) = 4.1 m ]

5) The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle? [Ans: 6% Hint: 𝑒 −𝑝𝑇 = 0.97 and calculate E(t+T)/E(t)] 6) A mass of 2.0 kg is attached to a horizontal spring with a spring constant of 50 N/m. There is a damping force of 12v (N) where v is the velocity of the mass. If the horizontal displacement from equilibrium function of the mass is given by x(t), then show from first principles that equation (1) hold. Then find x(t) if at time t = 0, x(0) = 0 m and v(0) = -8 m/s. π

[Ans: 𝑥(𝑡) = 2𝑒 −3𝑡 cos (4𝑡 + 2 )]

7) The graph below depicts the oscillations of a 200 g mass attached to a spring with spring constant of 0.10 N/m. Determine the damping ratio through the logarithmic increment. Optionally see if you can determine the damping constant directly from the graph, and use it to directly calculate the damping ratio. [Ans. ≈ 0.3]

Displacement y (m)

0.10

0.05

0.00

-0.05

-0.10 0

20

40

60

80

100

Time t (s) 8) What resistance 𝑅 should be connected in series to an inductance 𝐿 = 200 mH and capacitance 𝐶 = 12.0 μF for the maximum charge on the capacitor to decay to 99.0% of its initial value in 50 cycles? Assume, or rather make the simplifying assumption that ω′ = ω. [Ans: 8.3 mΩ]...