This is a document for physics based on second order differential equations and its applications.

1. Chapter 1
Applications of 2nd – Order
Differential Equation
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Chapter 1 Applications 1

2. Learning Objectives
• Solve a second-order differential equation representing simple harmonic
oscillator.
• Solve a second-order differential equation representing damped simple
harmonic oscillator.
• Solve a second-order differential equation representing forced simple harmonic
oscillator.
• Solve a second-order differential equation representing charge and current in an
RLC series circuit.
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3. I-1 Introduction
• In saw in chapter 0 that second-order linear differential equation are use to
model many situations in physics an engineering.
• In chapter, we look at how this works for systems of an object with mass
attached to a vertical or horizontal spring and an electric circuit containing a
resistor, an inductor and a capacitor in series.
• Model such as these can be used to approximate other more complicated
situation; for example bonds between atoms or molecules are often modeled
as spring that vibrate, as described by these same differential equation.
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4. I-2 Simple Harmonic Oscillator
• Consider a mass suspend from a spring
attached to a rigid support (this is
commonly call spring-mass system).
• (a) A spring in its natural position
• (b) At equilibrium with the mass attached
• (c) In oscillatory motion.
• Gravity is pulling the mass downward and
the restoring force of the spring is pulling
the mass upward.
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5. I-2 Simple Harmonic Oscillator
• When the two force are equal, the mass is said to be in equilibrium position (Fig.b).
• If the mass is displaced from equilibrium, it oscillates up and down (Fig.c).
• The behavior can be modeled by a second-order constant-coefficient differential
equation.
• Let x(t) denote the displacement of the mass from equilibrium.
• A positive displacement indicates the mass is below the equilibrium point, whereas
a negative displacement indicates the mass is above equilibrium.
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6. I-2 Simple Harmonic Oscillator
• Consider the force acting on the mass:
• Force of gravity: mg.
• According to Hooke’s law the restoring force is proportional to the displacement and acts in the
opposite direction from the displacement, so the restoring force is given by −𝑘(𝑠 + 𝑥), where k is
the spring constant.
• By Newton’s II law, we have:
𝑚𝑥′′
= 𝑚𝑔 − 𝑘(𝑠 + 𝑥)
• Note that at equilibrium, 𝑚𝑔 = 𝑘𝑠, the differential equation becomes
𝑚𝑥′′
+ 𝑘𝑥 = 0
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7. I-2 Simple Harmonic Oscillator
• It is convenient to rearrange this equation and introduce a new variable, called the
angular frequency, , letting 𝜔 =
𝑘
𝑚
, we can write the equation as:
𝑥′′
+ 𝜔2
𝑥 = 0
• Solving the differential equation, the general solution is:
𝑥 𝑡 = 𝑐1𝑐𝑜𝑠𝜔𝑡 + 𝑐2𝑠𝑖𝑛𝜔𝑡
which gives the position of the mass at any time.
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8. I-2 Simple Harmonic Oscillator
• For  = 0:
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9. I-2 Simple Harmonic Oscillator
• The motion of the mass is called simple harmonic motion.
• The period of this motion is 𝑇 =
2𝜋
𝜔
and the frequency is 𝑓 =
1
𝑇
=
𝜔
2𝜋
.
• Example:
• A 200g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released
from rest from a postion 10 cm below the equilibrium position. What is the frequency of this
motion.
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10. I-2 Simple Harmonic Oscillator
• Writing the general solution in the form 𝑥 𝑡 = 𝑐1𝑐𝑜𝑠𝜔𝑡 + 𝑐2𝑠𝑖𝑛𝜔𝑡 has some advantages.
• It is easy to see the link between the differential equation and the solution, and the period and
frequency of motion are evident.
• But the form of the function tells us little about the amplitude of the motion.
• In some situation, we may prefer to write the solution in the form
𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 .
• The period and frequency of motion are still evident. Furthermore, the amplitude of the motion,
A, is obvious in this form of the function.
• The constant  is called phase shift and has the effect of shifting the graph of the function to the
left or right.
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11. I-2 Simple Harmonic Oscillator
• To convert the solution to this form, we want to find the values of A and  such
that:
𝑐1𝑐𝑜𝑠𝜔𝑡 + 𝑐2𝑠𝑖𝑛𝜔𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙
• We first apply the trigonometric identity sin 𝛼 + 𝛽 = 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 + 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽
to get:
𝑐1cos(𝜔𝑡) + 𝑐2sin(𝜔𝑡) = 𝐴𝑠𝑖𝑛𝜙 cos(𝜔𝑡) + 𝐴𝑐𝑜𝑠𝜙(sin 𝜔𝑡 )
• Thus: 𝑐1 = 𝐴𝑠𝑖𝑛𝜙; 𝑐2 = 𝐴𝑐𝑜𝑠𝜙; If we square bot of these equation and add
them, we get: 𝐴 = 𝑐1
2
+ 𝑐2
2
.
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12. I-2 Simple Harmonic Oscillator
• Now, to find the phase shift, , go back to the equation c1 and c2, but this time divide the first
equation by the second equation to get:
𝑐1
𝑐2
= 𝑡𝑎𝑛𝜙
• We summarize this finding in the following theorem: Solution to the equation for simple harmonic
oscillator
• The function 𝑥 𝑡 = 𝑐1cos(𝜔𝑡) + 𝑐2sin(𝜔𝑡) can be written in the form 𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 , where 𝐴 =
𝑐1
2
+ 𝑐2
2
and tanϕ =
𝑐1
𝑐2
.
• Note that when using the formula tanϕ =
𝑐1
𝑐2
to find , we must take care to ensure  is in the right
quadrant.
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13. I-2 Simple Harmonic Oscillator
• Can take derivatives to find velocity and acceleration:
𝑣 𝑡 = 𝜔𝐴𝑐𝑜𝑠 𝜔𝑡 + 𝜙 ; 𝑎 𝑡 = −𝜔2
𝑥(𝑡)
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14. I-2 Simple Harmonic Oscillator
• Express the following function in the form 𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 . What is
the frequency of motion? The amplitude?
• 𝑥 𝑡 = 2cos(3𝑡) + sin(3𝑡)
• 𝑥 𝑡 = 3 cos 2𝑡 − 2sin(2𝑡)
• 𝑥 𝑡 = cos(4𝑡) + 4sin(4𝑡)
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15. I-2 Simple Harmonic Oscillator
Physical quantities in SHO:
• Period and frequency:
• The cosine function has a period of 2. The motion 𝑥 𝑡 =
𝐴𝑐𝑜𝑠(𝜔𝑡 − 𝜙) consists of a cosine function that has been
compressed by a factor of  and shifted by
𝜙
𝜔
.
• Only the compression affect the period, which is given by
𝑇 =
2𝜋
𝜔
• The frequency is the number of oscillation per unit time and is
given by:
𝑓 =
1
𝑇
=
𝜔
2𝜋
• The angular frequency is 𝜔 = 2𝜋𝑓.
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16. I-2 Simple Harmonic Oscillator
Physical quantities in SHO:
• Energy:
• The total energy of a SHO at any moment is the sum of its kinetic and
potential energy.
• Its kinetic energy is given by:
𝐸𝐾 =
1
2
𝑚𝑣2 = −𝐴𝜔sin(𝜔𝑡 − 𝜙) 2 =
1
2
𝑚𝐴2𝜔2𝑠𝑖𝑛2(𝜔𝑡 − 𝜙)
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17. I-2 Simple Harmonic Oscillator
Physical quantities in SHO: Energy
• Its potential energy is given by:
𝐸𝑃 =
1
2
𝑘𝑥2
=
1
2
𝑘 𝐴𝑐𝑜𝑠(𝜔𝑡 − 𝜙) 2
=
1
2
𝑘𝐴2
𝑐𝑜𝑠2
(𝜔𝑡 − 𝜙)
• Its total energy is therefore:
𝐸 = 𝐸𝐾 + 𝐸𝑃 =
1
2
𝑚𝐴2
𝜔2
𝑠𝑖𝑛2
𝜔𝑡 − 𝜙 +
1
2
𝑘𝐴2
𝑐𝑜𝑠2
(𝜔𝑡 − 𝜙)
• Since 𝜔2
=
𝑘
𝑚
it can be shown that the total energy is then
𝐸 =
1
2
𝑘𝐴2
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18. I-2 Simple Harmonic
Oscillator
Physical quantities in SHO: Energy
• The mechanical energy of an object oscillating on a spring
is:
𝐸 = 𝐸𝐾 + 𝐸𝑃 =
1
2
𝑚𝑣2
+
1
2
𝑘𝑥2
• Note that the object oscillates between the turning points
where the total energy line E crosses the potential energy
curve.
• You can see that the object has purely potential energy at
𝑥 = ±𝐴 and purely kinetic energy at it passes through the
equilibrium point x = 0.
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19. I-2 Simple Harmonic Oscillator
Physical quantities in SHO: Energy
• At maximum displacement, with 𝑥 = ±𝐴 and v = 0, the energy is
𝐸 𝑥 = ±𝐴 = 𝐸𝑝 =
1
2
𝑘𝑥2
• At x = 0, where 𝑣 = ±𝑣𝑚𝑎𝑥, the energy is
𝐸 𝑥 = 0 =
1
2
𝑚𝑣𝑚𝑎𝑥
2
• The system mechanical energy is conserved because of no friction and no external forces, so the energy at maximum
displacement and the energy at maximum speed must be equal:
1
2
𝑚𝑣𝑚𝑎𝑥
2
=
1
2
𝑘𝑥2
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20. I-2 Simple Harmonic Oscillator
Physical quantities in SHO: Energy
• The maximum speed is related to the amplitude:
𝑣𝑚𝑎𝑥 = 𝐴
𝑘
𝑚
= 𝐴𝜔
• Because energy is conserved, we can write:
𝐸 =
1
2
𝑚𝑣2
+
1
2
𝑘𝑥2
=
1
2
𝑘𝐴2
=
1
2
𝑚𝑣𝑚𝑎𝑥
2
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21. I-2 Simple Harmonic Oscillator
Physical quantities in SHO: Energy
• Any pair of this equation may be useful depending on the known information.
• For instance, you can use the amplitude A to find the speed at any point x:
𝑣 =
𝑘
𝑚
𝐴2 − 𝑥2 = 𝜔 𝐴2 − 𝑥2
• Similarly, you can use initial conditions to find the amplitude A;
𝐴 = 𝑥0
2
+
𝑚𝑣0
2
𝑘
= 𝑥0
2
+
𝑣0
𝜔
2
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22. I-2 Simple Harmonic Oscillator
• Example:
• A 500 g block on a spring is pulled a distance of 20 cm and released. The
subsequent oscillation are measured to have a period of 0.80 s. At what
positions is the block’s speed 1.0 m/s?
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23. I-2 Simple Harmonic Oscillator
Identifying and analyzing simple harmonic motion
• If the net force acting on a particle is a linear restoring force, the motion will be simple
harmonic motion around the equilibrium position.
• The position as a function of time is 𝑥 𝑡 = 𝐴𝑐𝑜𝑠(𝜔𝑡 + 𝜙). The velocity as a function of
time is 𝑣 𝑡 = −𝜔𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜙). The maximum speed is 𝑣𝑚𝑎𝑥 = 𝜔𝐴. The equation are
given in terms of x but they can be written in terms of y,  or some other parameters.
• The amplitude A and the phase constant  are determine by the initial conditions through
𝑥0 = 𝐴𝑐𝑜𝑠𝜙 and 𝑣0 = −𝜔𝐴𝑠𝑖𝑛𝜙.
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24. I-2 Simple Harmonic Oscillator
Identifying and analyzing simple harmonic motion
• The angular frequency  (and hence the period 𝑇 =
2𝜋
𝜔
) depends on the physics of the
particular situation. But  does not depend on A or .
• Mechanical energy is conserved. Thus:
1
2
𝑚𝑣2
+
1
2
𝑘𝑥2
=
1
2
𝑘𝐴2
=
1
2
𝑚𝑣𝑚𝑎𝑥
2
• Mechanical conservation provides a relationship between position and velocity that is
independent of time.
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25. I-3 Rotational Systems
• Consider an object suspended from a spring. Now suppose the object is twisted and released
(without changing its height) – it will then oscillate in a rotational manner!
• The rotational simple harmonic motion is very similar to what we have encountered up to now,
but there is important difference:
• Instead of forces, we must deal with torques (twisting forces)
• NII’s law, in rotational form is 𝜏 = 𝐼𝜃, where 𝜏 is the total torque, I is moment of inertia 𝜃 is the angular
acceleration.
• Hooke’s law for rotations is that if an object is twisted the restoring torque is 𝜏 = −𝑐𝜃.
• NII’s: 𝐼𝜃 = −𝑐𝜃; 𝜃 + 𝜔2𝜃 = 0; 𝜔 =
𝑐
𝐼
.
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26. I-3 Rotational Systems
Simple pendulum
• Position of mass along arc:
• Velocity along the arc:
• Tangential acceleration:
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27. I-3 Rotational Systems
Simple pendulum
• The tangential force comes from gravity (tension is always centripetal for a
pendulum):
• NII’s gives:
• This is almost a harmonic-oscillator equation, but the right-hand side has sin θ
instead of θ.
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28. I-3 Rotational Systems
Simple pendulum
• Fortunately, if θ is small, sin θ ≈ θ:
• Energy of a simple pendulum:
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29. I-3 Rotational Systems
• Example
• A simple pendulum 2 m long is suspended vertically in a region where
g=9.81m/s2. The point mass at the end is displaced from the vertical and
given a small push, so its maximum speed is 0.11m/s. What is the maximum
horizontal displacement of the mass from the vertical line it make when at
rest? Assume that all the motion takes place at small angle.
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30. I-3 Rotational Systems
The Physical Pendulum
• Any object, if suspended and then displaced so the gravitational force does
no run through the center of mass, can oscillate due to the torque.
• Also,
• And therefore
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31. I-3 Rotational Systems
Physical pendulum
• As before, sin θ can be replaced by θ if θ is small, and the motion is simple
harmonic with frequency:
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32. I-3 Rotational Systems
Physical pendulum
• Example:
• A thin, uniform rod of mass M and length L
swings from its end as a physical pendulum.
What is the period of the oscillatory motion
for small angles? Find the length ℓ of the
simple pendulum that has the same period as
the swinging rod.
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33. I-4 Damped Vibrations
• With the model just described, the motion of the mass continues indefinitely. Clearly,
this doesn’t happen in the real world.
• In the real world, there is almost always some friction in the system, which causes the
oscillation to die off slowly – an effect called damping.
• Now let’s look how to incorporate that damping force into our differential equation.
• Physical spring-mass systems almost always have some damping as a results of
friction, air resistance, or a physical damper called dashpot (a pneumatic cylinder).
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34. I-4 Damped Vibrations
• A damped oscillator
• An automobile shock absorber is a
damped harmonic oscillator.
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35. I-4 Damped Vibrations
• Let consider the motion of a mass attached to
a spring, which is free to move horizontally.
• However we assume that the table the mass
lies is not smooth, but offer resistance
proportional to the velocity of the mass and
acts in the opposite direction.
• So the damping force is given by: −𝑏𝑥′ for
some constant b > 0.
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36. I-4 Damped Vibrations
• Again applying NII’s law, the differential equation becomes:
𝑚𝑥′′ + 𝑏𝑥′ + 𝑘𝑥 = 0
• The associate characteristic equation is
𝑚𝑟2 + 𝑏𝑟 + 𝑘 = 0; divide by m define 𝜔2 =
𝑘
𝑚
and 2𝛼 =
𝑏
𝑚
• The differential equation become
𝑥′′
+ 2𝛼𝑥′
+ 𝜔2
𝑥 = 0
• Applying the quadratic formula, we have
𝑟 = −𝛼 ± 𝛼2 − 𝜔2
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37. I-4 Damped Vibrations
• Just as in second-order linear equation, we consider 3 cases, based on whether
the characteristic equation has distinct real roots, a repeated real root or
complex conjugate roots.
Case 1: 𝜶 > 𝝎
• In this case, we say the system is overdamped. The general solution has the
form:
𝑥 𝑡 = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑒𝑟2𝑡
where both r1 and r2 are less than zero.
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38. I-4 Damped Vibrations
• The expressions of r1 and r2 are:
𝑟1 = −𝛼 + 𝛾0 and 𝑟2 = −𝛼 − 𝛾0; where 𝛾0
2
= 𝛼2
− 𝜔2
• The general solution is given by:
𝑥 𝑡 = 𝑐1𝑒 −𝛼+𝛾0 𝑡 + 𝑐2𝑒 −𝛼−𝛾0 𝑡 = 𝑒−𝛼𝑡 𝑐1𝑒𝛾0𝑡 + 𝑐2𝑒−𝛾0𝑡
• Since  > , both terms decay exponentially.
• The velocity is obtained by differentiate x(t):
𝑣 𝑡 = −𝛼 + 𝛾0 𝑐1𝑒 −𝛼+𝛾0 𝑡
− (𝛼 + 𝛾0)𝑒− 𝛼+𝛾0 𝑡
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39. I-4 Damped Vibrations
• Suppose the mass is initially (t=0) released from rest (v=0) at position x=x0, find
the general solution.
• Because the exponents are negative, the displacement decays to zero over time,
usually quite quickly.
• Overdamped systems do not oscillate (no more than one change of direction),
but simply move back toward the equilibrium position.
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40. I-4 Damped Vibrations
• Behavior of an overdamped spring-mass system, with
• (a) no change in direction and
• (b) only one change in direction.
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41. I-4 Damped Vibrations
• Example:
• A 2 kg mass is attached to a spring with spring constant 24 N/m. The system is then
immersed in a medium imparting a damping for equal 16 times the instantaneous
velocity of the mass. Find the equation of motion if it is released from rest at a point
40 cm below equilibrium.
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42. I-4 Damped Vibrations
Case 2: 𝜶 = 𝝎
• In this case, we say the system is critical damped. The general solution has the form:
𝑥 𝑡 = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑡𝑒𝑟1𝑡
where r1 is less than zero. Then 𝑟1 = −𝛼 and the solution is
𝑥 𝑡 = (𝑐1+𝑐2𝑡)𝑒−𝛼𝑡
• The velocity is then: 𝑣 𝑡 = 𝑐2 − 𝛼𝑐1 − 𝛼𝑐2𝑡 𝑒−𝛼𝑡.
• Suppose the mass is initially (t=0) released from rest (v=0) at position x=x0, find the
general solution.
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43. I-4 Damped Vibrations
• The motion of a critical damped system is very similar to that of an
overdamped system. It does not oscillate.
• However, with a critical damped system, if the damping is reduced even a
little, oscillatory behavior takes place.
• From a practical perspective, physical systems are almost always either
overdamped or underdamped (case 3 we will consider next).
• It is impossible to fine-tune the characteristic of a physical system so that 
and  are exactly equal.
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44. I-4 Damped Vibrations
• Critically Damped
• (a) Behavior of a critically damped
spring-mass system.
• (b) The system graphed in part (a) has
more damping than the system
graphed in part (b)
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45. I-4 Damped Vibrations
• Example
• A 1 kg mass stretches a spring 20cm. The system is attached to a dashpot that
imparts a damping force equal to 14 times the instantaneous velocity of the mass.
Find the equation of motion if the mass is released from equilibrium with an upward
velocity of 3m/s.
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46. I-4 Damped Vibrations
Case 3: 𝛂 < 𝝎
• In this case, we say the system is underdamped.
• The solutions of the characteristic equation are: 𝑟 = −𝛼 ± 𝛼2 − 𝜔2; Let define 𝜔0
2
= 𝜔2 −
𝛼2. Then we got 𝑟 = −𝛼 ± 𝑖𝜔0
• The general solution is
𝑥 𝑡 = 𝑒−𝛼𝑡
𝑐1 cos 𝜔0𝑡 + 𝑐2sin(𝜔0𝑡) = 𝐴𝑒−𝛼𝑡
cos(𝜔0𝑡 − 𝜙)
• Underdamped systems do oscillate because of the sine and cosine terms in the solution.
• However the exponential term dominates eventually, so the amplitude of the oscillations
decrease exponentially over time.
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47. I-4 Damped Vibrations
• We therefore consider the amplitude as an
envelop: 𝐴𝑒𝑛𝑣 = 𝐴𝑒−𝛼𝑡 and the oscillatory
part as a carrier. We define:
𝑥 𝑡 = 𝐴𝑒𝑛𝑣cos(𝜔0𝑡 − 𝜙)
• The velocity can be obtained by differentiate
the position x(t) with respect to time
𝑣 𝑡 =
𝑑𝑥
𝑑𝑡
= −𝐴𝑒−𝛼𝑡 𝛼 cos 𝜔0𝑡 − 𝜙 − 𝜔0𝑠𝑖𝑛 𝜔0𝑡 − 𝜙
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Chapter 1 Applications 47

48. I-3 Damped Vibrations
• Example
• A 1 kg mass stretches a spring 49 cm. The system is immersed in a medium that imparts a
damping force equal to 4 times the instantaneous velocity of the mass. Find the equation of
motion if the mass is released from rest at a point 24 cm above equilibrium.
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Chapter 1 Applications 48

49. I-4 Damped Vibrations
Energy of an under critically damped
• Recall that the total energy stored in an undamped oscillator having angular
frequency  and amplitude A is
𝐸 =
1
2
𝑘𝐴2 =
1
2
𝑚𝜔2𝐴2
• But the amplitude of a damped oscillator diminishes over time. Therefore there
must be a loss of energy in the system over time.
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50. I-4 Damped Vibrations
Energy of an under critically damped
• The total energy is given By
𝐸 = 𝐸𝑝 + 𝐸𝑘 =
1
2
𝑘𝑥2
+
1
2
𝑚𝑣2
where x and v are the position and velocity of the mass.
• Substitution gives:
𝐸 =
1
2
𝑚𝐴2
𝜔2
𝑒−2𝛼𝑡
+
1
2
𝑚𝐴2
𝑒−2𝛼𝑡
𝛼2
𝑐𝑜𝑠2 𝜔0𝑡 − 𝜙 + 𝛼𝜔0𝑠𝑖𝑛2(𝜔0𝑡 − 𝜙)
• Take careful note of the quantities  and 0 which occur in this expression and they are different.
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Chapter 1 Applications 50

51. I-4 Damped Vibrations
• The first term in the expression of the total energy is a decaying function of the time,
while the second term oscillates and decays.
• The reason why the energy has an oscillating part is that energy loss is due to friction,
and the rate of energy loss is proportional to the velocity.
• In fact, at the instant when the velocity is zero, the rate of energy loss is also zero.
• The average total energy is just the first term:
𝐸𝑎𝑣𝑒 =
1
2
𝑚𝜔2𝐴2𝑒−2𝛼𝑡
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52. I-4 Damped Vibrations
• For most application it is sufficient to use only the average energy. Note in
particular that the average energy depends on the natural angular frequency 
and not on the damped angular frequency 0.
• An alternative way to write the average energy is:
𝐸𝑎𝑣𝑒 =
1
2
𝑚𝜔2𝐴𝑒𝑛𝑣
2
• This expression has the same form as for an undamped oscillator, but the
amplitude replace by the amplitude envelop.
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53. I-4 Damped Vibrations
Different quantities describing the damping
• The quantity v (in −𝑏𝑣) describes the strength of the damping force, while the
quantity 𝛼 =
𝑏
2𝑚
describes the acceleration due to damping.
• However, it is difficult to measure either b or  directly from a graph of the
motion.
• We shall give some alternative ways to describe the damping.
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Chapter 1 Applications 53

54. I-4 Damped Vibrations
Different quantities describing the damping
• The relaxation time ()
• For an under critically damped the amplitude envelop decays exponentially and the rate at
which it decays gives an indication of the damping.
• The relaxation time is defined as the time taken for the amplitude envelop to decay to
𝑒−1 = 0.368 ≈ 37% of its original value A0. (compare with radioactive decay 50%, half-
life)
• Obviously the larger is  the less strong is the damping.
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Chapter 1 Applications 54

55. I-4 Damped Vibrations
• From the graph of Aenv as a function of time
one can easily obtain the relaxation time as
show on the figure.
• Mathematically, if  is the relaxation time
then
𝐴𝑒𝑛𝑣 𝑡 + 𝜏 =
𝐴𝑒𝑛𝑣
𝑒
. Therefore
𝐴0𝑒−𝛼(𝑡+𝜏) = 𝐴0𝑒−𝛼𝑡𝑒−1 and so 𝜏 =
1
𝛼
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Chapter 1 Applications 55

56. I-4 Damped Vibrations
Different quantities describing the damping
• The logarithmic decrement ()
• The logarithmic decrement () is defined as the natural logarithm of the ratio of the
Aenv of the damping oscillator at any given time and the full (damped) period later:
𝛿 = 𝑙𝑛
𝐴𝑒𝑛𝑣(𝑡)
𝐴𝑒𝑛𝑣(𝑡 + 𝑇0)
• This provide a practical way to measure  in an experiment if the amplitude is
known as a function of time.
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57. I-4 Damped Vibrations
Different quantities describing the damping
• The logarithmic decrement ()
• The largest the logarithmic decrement the stronger is the damping.
• Mathematically:
𝛿 = 𝑙𝑛
𝐴0𝑒−𝛼𝑡
𝐴0𝑒−𝛼(𝑡+𝑇0)
= ln 𝑒𝛼𝑇0 = 𝛼𝑇0 =
𝑇0
𝜏
• If the ratio is small, then T0 <  and many oscillations can occur during one
relaxation time i.e. the damping is weak.
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58. I-4 Damped Vibrations
Different quantities describing the damping
• The quality factor (Q)
• For a damped oscillator the average energy decays exponentially ( 𝐸𝑎𝑣𝑒 =
1
2
𝑚𝜔2𝐴2𝑒−2𝛼𝑡).
• The rate at which it decays also gives an indication of the damping.
• Q is defined as the phase angle (in radian) required for the average energy to decay to
𝑒−1
= 0.368 ≈ 37% of its original value.
• Obviously the larger Q the less strong is the damping.
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Chapter 1 Applications 58

59. I-4 Damped Vibrations
Different quantities describing the damping
• The quality factor (Q)
• Mathematically, if t1 is the time (not phase angle) for the required decay, then it follows that
𝐸𝑎𝑣𝑒 𝑡 + 𝑡1 =
𝐸𝑎𝑣𝑒(𝑡)
𝑒
• Therefore:
1
2
𝑚𝜔2
𝐴2
𝑒−2𝛼(𝑡+𝑡1)
=
1
2
𝑚𝜔2
𝐴2
𝑒−2𝛼𝑡
𝑒−1
and 𝑡1 =
1
2𝛼
• During this time the increase in the phase angle is 𝑄 = 𝜔0𝑡1 =
𝜔0
2𝛼
. (show that 𝑄 =
𝜋
𝛿
)?
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60. I-4 Damped Vibrations
Different quantities describing the damping
• The fraction of energy lost per second
• The average energy at a time t of a damped oscillator is 𝐸𝑎𝑣𝑒(𝑡).
• After a one cycle or one oscillation, it is given by 𝐸𝑎𝑣𝑒(𝑡 + 𝑇0) and the ration of
the energy store initially to that lost in one cycle is:
𝐸𝑎𝑣𝑒(𝑡)
𝐸𝑎𝑣𝑒 𝑡 − 𝐸𝑎𝑣𝑒(𝑡 + 𝑇0)
=
1
1 − 𝑒−2𝜋/𝑄
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61. I-4 Damped Vibrations
Different quantities describing the damping
• The fraction of energy lost per second
• If Q is large (𝑄 ≫ 2𝜋), then we can approximate 𝑒
−
2𝜋
𝑄 = 1 −
2𝜋
𝑄
and the ratio becomes
1
1 − 1 −
2𝜋
𝑄
=
𝑄
2𝜋
• As long as Q is large:
𝑄 = 2𝜋 ×
𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒
𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
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