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In this exciting course on Oscillation, Waves, and Optics,
provide a clear and in-depth
comprehensive modules. Each module is carefully designed to
understanding of the fundamental concepts and principles of oscillation, waves, and optics.
into the fascinating
Today, in Module 1 of our course on Oscillation, Waves, and Optics, we will delve
damped and forced simple
topic of free Simple Harmonic Motion (SHM) and explore the concepts of
harmonic oscillators.

Aparticle will vibrate when there is a disturbance or displacement from the equilibrium
position, and
We refer to such a force
a force acts on the particle, bringing it back towards the equilibrium position.
as a restoring force. The simplest vibration in nature is when the restoring factor is directly proportional
linear
to its displacement and is called a simple harmonic motion. Aparticle executing SHM is called a
harmonic oscillator.
Thus, this restoring force F may be written

F=-sy (1)

Where s is a positive constant called the stiffness or spring constant or restoring force per unit
displacement, and the negative sign indicates that the force is acting backwards and away from the
direction of increasing displacement. Reciprocal of s is known as compliance.
Force MLT? = MT-2
Dimension of s is
distance L

Also, we know that F = mass Xacceleration = m m is the mass of the moving body.
Hence, by using Newton's second law of motion, the equation of motion (1.1) now becomes
²y
m=-sy
m+sx
dt2 = 0or,+
dt2 w'y = 0, (2)
where 2 S
m

MT-2
Dimension of is M
=T; therefore, = ) h¡s the dimension of 1/T that is the
\m/

frequency(v).
Expression 2 is the equation of motion of the linear harmonic oscillator.
We solve equn 2 by the method of trial solution. We consider y = et, by substituting, we get

dt2 =a'pt, (a' + w?)et = 0

Since the product of (a? + w') and et is zero, either of them has to be zero. e can not be zero as
it is the solution; it would be trivial, Hence (a + w) =0 or a = Fiw, where /' = -1.
The general solution is

y= Celat + Cye-iat
+ C(cos wt - isin wt)
=G,(cos wt + isin wt)
-C) sin wt
= (C, + Cz) cos wt + i(C
=P cos wt +Q sin wt

Where P = G + C, and Q = C
C;
and 8, such that P= A sin ß
two new arbitrary constants, A
simplicity, we replace P and Q with
For may be written as
and Q = A cos &. Then the solution (3)
sin wt = Asin(wt + 6)
cos 8
y= Asin&cos wt + A
Q = Asin 6, we may get
Also, by using P = Acos 8 and (4)
Ssin wt= Acos(wt- 8)
y=AcosS cos wt + A sin
equilibrium position.
displacement or amplitude of the body from its
Arepresents the
maximum
P
Now, A= p2 + Q2 and tan & Q

Putting 8 =0+in
2
eqn 4, we get
y= Acos(wt + 0)
i.e. 8 or 0 is called the epoch.
The initial phase of the particle,
then
of a single oscillation cycle,
IfT represents the duration
y(t + T) = y(t)
0)
Acos (w(t + T) + 0) = Acos(wt +
(5)
wT = 2n or, T = or, T = 27

and v==or,
21
w = 2v

A SIMPLE HARMONIC
VELOCITY, ACCELERATION AND ENERGY OF
OSCILLATOR
the velocity v is
From Egn. (3), we find that the magnitude of
/2

V=
dt
(Asin(wt +0) =Awcos(wt +8) =Aw(1 (6)

or, v = w(A? - y2)1/2

and a= ý = ç= -Aw²sin(wt + 0) = -w²y
to the displacement. This
The negative slgn indicates that the acceleration is in the opposite direction object towards the
is the characteristic behaviour of SHM, where the acceleration restores the
equlbrlum position.
 related to the stiffness or the restoring force of the system. Ahigher
The proportionalityconstant w? is acceleration for a given
restoring force, resulting in a higher
value of w' indicates a stronger
displacement.
energy, then from the law of conservation of energy, in the
IfT is the kinetic energy, Vis the potential
absence of any friction-type losses, we have
energy of the oscillator.
E =T + V= constant where E is the total

Also, Force F = -7v

or
_d=-sy
dy

or V=sy'+c
or V=mwA²sin²(wt +0) + c (7)

constant c we apply boundary

where c is an arbitrary constant. To estimate the value of the arbitrary
condition. If potential energy V=0 at x = 0, from egn 7, we obtain c = 0.
Hence, potential energy is V=;ma?A²sin? (wt + 6)
The kinetic energy of the oscillator is

T=;mý² =;mA'u² cos²(wt + 0)

Therefore the total energy is

E=T + V=;mAa² cos?(wt +0) +ma²A'sin?(wt + 8) =mA²w² (8)

Simple Harmonic Oscillations in an Electrical System

Oscillation happens in all kinds of physical systems, not only mechanical. Consider the electrical circuit
composed of a capacitor Cand an inductor L. We study the oscillation of the amount of charge across
the capacitor g(t). Figure x shows the schematics of LC circuit, where an inductance L is connected
across the plates of a capacitance C. The force equation of the mechanical and fluid examples now
becomes the voltage equation (balance of voltages) of the electrical circuit, but the form and solution
of the equations and the oscillatory behaviour of the systems are identical.

In the absence of resistance, the energy of the electrical system remains

constant and is exchanged
between the magnetic field energy stored in the inductance and the electric field energy
between the plates of the capacitor. At any instant, the voltage across the inductance is stored

dl
V= -L=-L dt2
 flowing and g is the charge on the capacitor, the negative sign showing that the
where / is the current voltage g/C across the capacitance so that
increase of current. This equals the
voltage opposes the
L-+=0

Where w² = 1/LC
The general solution is g = Asin (wt + 8)
throughout the cycle, as the
The energy stored in the magnetic field or inductive part of the circuit
with respect to time;
current increases from 0 to I, is formed by integrating the power at any instant
that is
d 1

The potential energy stored mechanically by the spring is now stored electrostatically by the
capacitance and equals Eç =cv² ==}Lu'g'. Usingthe solution of q, we get
2

1 1
Etot = EL + Eç LAw² cos² wt+;Lu'A' sin' wt =Lw?A?
The total energy is independent of time.

Complex number notation and phasor representation of simple

harmonic motion
The concept of a phasor is widely used in the fields of physics and engineering to denote a complex
number that symbolises a sinusoidal function. The amplitude (A), angular frequency (w), and initial
phase (0) of the function remain constant over time. It is related to a more general concept called
analytic representation,which decomposes a sinusoid into the product of a complex constant and a
factor depending on time and frequency. The complex constant, which depends on amplitude and
phase, is known as a phasor.
Areal-valued sinusoid with constant amplitude, frequency, and phase has the form:
A
cos(wt + 0)
With the inclusion of the imaginary component iA sin(wt +0), in accordance with Euler's formula we
can write

A
cos(wt + 0) + i4 sin(wt+ 0) = Aeltwt+9) = Aeie glat
whose real part is the original sinusoid. The advantage of utilising the complex
the fact that performing linear operations with other representation lies in
complex representations yields a complex
outcome, wherein the real component reflects the same linear operations conducted with the
real
components of the other complex sinusoids.

Phase Space Diagrams for an Oscillator

When discussing oscillation, one often must consider both
oscillator, especially when discussing potential energy (which the displacement and velocity of the
(which depends on velocity). Both the depends on position) and kinetic energy
displacement and velocity are functions of time, and there is a
 parametric graph of the velocity v(t)
y0° phase relationship between the two. Aphase-space plot is a time.
plotted as a function of the displacement x (t), with the changing variable being
projected onto its diameter, the
When the motion of a particle performing uniform circular motion is representation of SHM is
projection undergoes simple harmonic motion(SHM). The circular motion
frequency of the SHM, w.
the phasor, and the angular velocity of this circular motion is the
out simple harmonic motion.
phasor is a rotating arrow whose component on a vertical axis traces
A is
the length of the phasor
The reference circle is the circle traced out by the tip of the phasor, and
equal to the amplitude Aof the simple harmonic motion.
As the phasor rotates counterclockwise at a constant rotational speed
w, its vertical component varies
sinusoidally and therefore describe a simple harmonic motion.

positiou at instant !,
vertical component at ,

verticalcomponent at | = 0

2T
position at =0

Icycle

If the phasor completes one revolution in a periodT, then w =

And the frequency of the corresponding SHM isf=

Combining the previous two equations we get w = 2nf. w is often referred to as the angular
frequency and has the same unit (s-) as frequency (think: rad/sec).

Phasor: Energy of a simple harmonic oscillator

The rotational position of the tip of the phasor is called the phase
of the motion and is given by p(t) = wt + pi. Then, the vertical component of the phasor can be
written as

x(t) = Asin p(t) = Asin(wt + p) (simple harmonic motion)

Aisthe anmplitude of the phasor, and p, is the phase at t = 0s. This means there are 2 boundary
condítions: amplitude and initial phase
amplitude

phase at instant 1; bu) = l,t h,

Asin o,
2T
phasc at - 0: , ()

-A
 acceleration of the harmonic oscillator:
Now, we can obtain the velocity and
harmonic motion)
= wAcos(wt + o) (simple
dx
dt

=-o²Asin(wt + o:) (simple harmonic motion)

Comparing equations for x (t) and a(t), we can write

a, = -wx (simple harmonic motion)
Using Newton's 2nd law, X F, = ma,, another constraint!
ZE= -mw'x (simple harmonic motion)
oscillator as it moves from the equilibrium
The work done by the forces exerted on the harmonic
position toward the positivex direction is

W-|dx =- ma'xdx
This work causes a change in kinetic energy, given by
1
AK =-mw² xdx = -mw'
;ma'r -jmaz
For a closed system, AE = AK + AU = 0, which leads to
If we let U
(X%) = 0 (a free choice),then

E= K+U =;my² +;ma'*?

Using expressions for x(t) and v(t), we get
E =K+U =mu?A²
2 cos² (wt + p) +;mw²A²
2 sin?(wt + p)
at
lu
;mu?A? (simple harmonic motion)
e
Total energy determined by amplitude and frequency.
)E
rn

Damped Simple Harmonic Motion

)E In the case of ideal simple harmonic motion, wherein the total energy was conserved and the
yste displacement represents a sine wave pattern, seemingly indefinitely. In practical scenarios, a resistive
or viscous process inevitably results in dissipation of energy. For instance, the amplitude of a pendulum
leit
that is oscillating freely will decay over time due to the loss of energy. The existence of
impediment to
3) B
movement implies the operation of an additional force, which is taken as to be directly proportional
to the velocity. The force of friction is exerted in a direction that is opposite to the
arr direction of the
velocity (-r|). As a result, Newton's Second Law is modified
Case më = -Sx ri
where r is the constant of proportionality and has the
Wri mass of the particle and s is force dimensions of force per unit of velocity, m is
constant.
Now the equation becomes mi + rit sx = 0
As before, to solve the equation we adopt the method of trial solution, and solution of the form x =
Ceat can always be found. Since exponential function is dimensionless, C has the dimension of x,
and a has the dimension of time inverse T-1.
Taking Cas aconstant length we have i = aCet and i = a'Cet, then equn x may be written as
Ceat (ma' + ra + s) =0
so that either

Ceat = 0
Or

ma' + ra+s = 0
Solving the quadratic equation in a gives

2m -s/m
Note that r/2m and() ,and therefore, a, all have the dimensions of inverse time, T-', which we
expect from the form of et, The general solution for x is linear combination of the solutions involving
a, and a_.

The term inside square root : ( ( -4s/m) can be positive, zero or negative depending on the
relative magnitude of the two terms inside it. Each of these conditions gives one of the
three possible
solutions referred to earlier and each solution describes a particular kind of behaviour.
The conditions are:

(1)Bracket positive (>4s/m. Here the damping resistance term ( \2m/ dominates the stiffness
term sm, and heavy damping results in a dead beat system.
(2) Bracket zero (2m.= 4s /m. The balance between the two terms results in a
system.
critically damped
Neither (1) nor (2) gives oscillatory behaviour.

(3) Bracket negative(2m <4s /m. The system is lightly damped and gives
harmonic motion. oscillatory damped simple
Case 1: Heavy damping|> 4s/m)
Writing 2m =pand Zm.
) -s/m= q, we can replace
 mEm)sm-sme
*= Geil +qeszt =Gel + Czel

by
x=e-P(C,e + Ce -9t)
If now F = G + Gz and G= G- Gz, the displacement is given by
x=ep(F coshgt + Gsinh gt)
This represents non-oscillatory behaviour, but the actual displacement will depend upon the initial (or
boundary) conditions; that is, the value of xat time t= 0. Ifx = 0at t = 0then F= 0, and

x= Ge-rt/2m sinh
-sm
Figure xillustrates such behaviour when a heavily damped system is disturbed from equilibrium by a
sudden impulse (that is, given a velocity at t= 0). It will return to zero displacement quite slowly
without oscillating about its equilibrium position.

Case 2: Critical Damping( =s/m)

Using the notation of Case 1, we see that q = 0and that x=eP(C + Cz). This is, in fact, the limiting
case of the behaviour of Case Ias q changes from positive to negative. In this case the quadratic
equation in a has equal roots, which, in a differential equation solution, demands that C must be
written C = A+ Bt, where A is a constant length and Ba given velocity which depends on the
boundary conditions. It is easily verified that the value
x= (A+ Bt)e-rt/2m = (A+ Bt)e-pt

satisfies mi +r|t sx = 0when(\2m =4s/m.

Case 3: Damped Simple Harmonic Motion (<n)

Damped Simple HarmonicMotion conserved and the
In the case of ideal simple harmonic motion, wherein the total energy was
or
displacement represents a sine wave pattern, seemingly indefinitely. In practical scenarios, a resistive
pendulum
viscous process inevitably results in dissipation of energy. For instance, the amplitude of a
that is oscillating freely will decay over time due to the loss of energy. The existence of impediment to
movemernt implies the operation of an additional force, which is taken as to be directly proportional to
the velocity. The force of friction is exerted in a direction that is opposite to the direction of the velocity
(-r|). Asaresult, Newton's Second Law is modified
mi= -SX - r í

where r is the constant of proportionality and has the dimensions of force per unit of velocity, m is mass
of the particle, ands is the force constant.
Now the equation becomes mi + r| + sx = 0 (1)

As before, to solve the equation, we adopt the method of trial solution and a solution of the form x =
Cett can always be found. Since the exponential function is dimensionless, Chas the dimension of x,
and a has the dimension of time inverse T-1,

Taking Cas a constant length, we have i = aCe and ë = aCe,then egn 1 may be written as

Ceat(ma? + ra+ s) =0
so that either

Ceat = 0
or

ma' + ra +s =0
Solving the quadraticequation in a gives

a=-+
2m
J---t-² (2)
The solutions for displacement can be written as(y= and w =)

and x; ,and the general solution is of the form

*=x+x,= G e l e
+Gel (3)
 the right of (2) can be
Depending on the relative sizes of y and wn, the expression in brackets on
the nature of the motion in the three
positive, negative or zero. There willbe significant differences in
cases. From now on, therefore, we distinguish
dominates the stiffness
1. heavy damping (y/2 > wo), Here the dampingg resistance term r 2=4m 2
term s/m or natural resonant frequency Wo, and heavy damping results in a dead beat system.
2. critical damping (y/2 = wo), The balance between the two terms results in a critically damped
system.
3. weak damping (y/2 < wo), The system is lighty damped and gives oscillatory damped simple
harmonic motion.

Case 1. Heavy Damping (y/2 > wo)

Under the condition when y/2 > w, we find that the two values of a are now both real and negative.
Writing= pand -w= q, we get the

= X,+x =Gel-ptq)e + C,el-p-9)t =ep (C,eat + Gye-gt)

Now consider F = C + Cz and G= C - Cz; therefore, the displacement can be represented as
G
*=eet +e)+(et -e-)=e-Fcoshqt+Gsinhq)
This represents non-oscillatory behaviour, but the actual displacement will depend upon the initial (or
boundary) conditions; that is, the value of xat time t = 0.Ifx = 0at t = 0 then F= 0, and

x = ePGsinhgt = Gei' sinh (4)

Figure x illustrates such behaviour when a heavily damped system is disturbed from
equilibrium by a
sudden impulse (that is, given a velocity at t = 0). It will slowly return to zero
displacement without
oscillating about its equilibrium position.

n-)aat)

Figure 1: Behaviors of e-*G sinh xand

sinhx, cosh x functions
Case 2.Critical Damping (y/2= wo)
equation solution,
case the quadratic equation in a has equal roots, which, in a differential
In this given velocity
where C, is a constant length and Cz a
demands that C must be written C = C, + Ct, which is the product
might therefore look for asolution
which depends on the boundary conditions. We
easily verified that the value
of an exponential decay and a straight line. It is

x= (G +Ct)et
satisfies mä + ri + sx = 0.

Case 3. Damped Simple Harmonic Motion (y/2 < wo)

point of view, the most important
When y/2 < wo, the damping is light, and this gives from the present
oscillatory damped simple harmonic motion. The expression
w? is an
kind of behaviour,
as
imaginary quantity, the square root of a negative number, which can be rewritten

ly2 y2
= iw'

Form A

The general solution is

x= C1e

:ei((c, + Ca) cos aw't +i(C, -

=e C) sin w't] = e 'Acos w't + Bsin w't]
Form B

x=eGeo't +Cge-t']
A
If wenow choose C el and G= e-i
21

where Aand (and thus e) are constants which depend on the motion at t =0, we find after
substitution

2
:Ae sin(w't + )
The displacement, therefore, varies sinusoidally with time as in the case of simple harmonic
motion, but
now has a new frequency
 exponential term e .(y=r/m)
modified by the decaying
and its amplitude Ais
|1.5

+
X
y= exp(-0.01r) sin(0.1r)

=exp(-0.01r)

400 500

.5 sin(w't + )

sinusoidal function Ae i sin(w't + ).

Figure 2: Exponentially decaying behavior of

Evaluation of A and ¢

x= Aetsin(w't + )
and

|=(-)4e:sin(w't + ¢) + w'Ae cos(w't + )

=Ae -5sin(w't +¢) +t o' cos(w't + )
let x = X and = V at t = 0. Using the initial condition in the above set of equations we get
Xo = Asin()

,= a|-zain@) +o' coste)] =-Asin(9) +Ao'V+- sin (a)=-1~sin) + Aw' 1

=-x% + w'A-x

’A=
A- x =

Which gives sin($) == Xo/| x, +

Methods of Describing the Damping of an Oscillator

The energy of an oscillator is given by E = mw'A, that is energy is proportional to the amplitude. As
2 2
energy is proportional to e, energy decay will be proportional to (e) =(e zm) The larger the
value of the damping force r the more rapid the decay of the amplitude and energy. Thus we can use
the exponential foctor to express the rates at which the amplitude and energy are reduced.
Logarithmic Decrement

This measures the rate at which the amplitude dies away. Suppose we choose = n/2, then eqn x =
Ae i sin(w't+ ¢) becomes x = Ae i cos(w't). At time t = 0,x = A. If the period of oscillation is
r'(= 2n/w'), then one period later, the amplitude is given by A, = Ae ,so that A1 =e= e

Where &=r= loge is called the logarithmic decrement.

exp(-0 01x) cos(0. ir) A

U,= exp(-0.01)
Ae~r
Aear
Aele-ar'
Relaxation Time or Modulus of Decay
taken for the amplitude to decay
Another way of expressing the damping effect is by means of the time
to

e-= 0.368
of its original value A. This time is called the relaxation time or modulus of decay and the amplitude

A, = Aes = Ae-1
at atimet=2/y.

The Quality Factor or Q-value of a Damped Simple Harmonic Oscillator

This measures the rate at which the energy decays. Since the decay of the amplitude is represented by
A, = Ae
the decay of energy is proportional to

A,=Ae-27t
and may be written in terms of energy (E = 2;mw'A)as
E, = Ee-Yt = Ee-rt/m

where E is the energy value at t = 0. The time for the energy Eto decay to Ee-1 is given by t = 1/ys.
during which time the oscillator will have vibrated through w'/y rad.
We define the quality factor
w'm
Q=

as the number of radians through which the damped system oscillates as its energy decays to

E, = Ee-1
Ifr is small, then Q is very large and
r2
m

So that w' wo Thus, we write, to a very close approximation, Q = owhich is a constant

of the damped system.
 we can write
Sincenow equals
E, = Ee-r/m)t = Eel-wo/0)t

The fact that Qis a constant(=) implies that the ratio

energy stored in system
energy lost per cycle
is also a constant, for
Wom Vom
2Tt 2r r

is the number of cycles (or complete oscillations) through which the system moves in decaying to
E, = Ee-1

And if E, = Ee-rt/m, the energy lost per cycle is -AE =At = -E m

where At ==t,the period of oscillation.

Thus, the ratio

energy stored in system E v'm Vom

energy lost per cycle AE 2Tt

The Quality Factor or Q-value of a Damped Simple HarmonicOscillator

In the case of free SHM, the periodic time is 2n/wo. With damping this time
period is
-1/2
2u/w' =
(Assuming is small)
’T=T,(1+)
8w02)

Where T is the period in the absence of damping and To is the

period in the presence of damping.
Example 1: Damped SHM in an ElectricalCircuit
The force equation in the
mechanical oscillator is replaced by the voltage equation in the
circuit of inductance, resistance and electrical
capacitance
 IR

di
L
dt

L + 1R +!= 0
The sum of the voltages around the circuit is given from Kirchhoff's law as dt

We have, therefore
dl
L+IR+)

or,

Lä+Ró + =0
the charge
and by comparison with the solutions for x in the mechanical case we know immediately that

R R2
q= oexp

R2
which, for>gives oscillatory behaviour at afrequency

R2
w'=
412 LC

From the exponential decay term we see that R/L has the dimensions of inverse time T or w, so that
wL has the dimensions of R; that is, wl is measured in ohms. Similarly, since a = 1/LC; wgL = 1/
wC, so that 1/w,C is also measured in ohms.
Example 2: The heavily damped simple harmonic system

The heavily damped simple harmonic system is displaced a distance F from its equilibrium position and
released from rest. Show that in the expression for the displacement
x=eP(F cosh qt + Gsinh qt)
Where

2m
=p and J4m²
-W = q

that the ratio

G
F (r2 -4ms)1/2
Ans:

The system is released from rest, so we know its initial velocity is zero, i.e. dtdxlt=0 =0

x=eP(Fcosh qt +Gsinh qt) = e (et +e") + (et-e-9)|

(-p)e+ (-q-p)-e-=0
or, (q -p)+(-q-p)=0
or, F(q p-q-p)- G(q -p+q+p)= 0
or,Fp = Gq
or, = = .| r2 2m
1

s Vr2-4ms
V4m2 m

Example 3: The weakly damped simple harmonic system

The solution for damped simple harmonic motion is given by

x=e|Ceto'e +Cze-to']
Find the values of C, and Cz (y =r/m).
Given x= A
cos at t = 0,

and | -w'A sin at t= 0 only if r/m is very smallor /2.

Ans:

The initial displacement of the system is given by:

x=eCelo' +Cye-lo't] = Acos ¢at t= 0

’G + C, = Acos (1)

Now let the initialvelocity of the system to be:

|=(-)e#Ge a't + Cze-to't]+e(io)[Ceto't - Cqe-ta'] =-w'A sin ¢at t =0

--)G +G]+(io)[G - CG]=-'Asin ¢
-(-Acos¢ +(iw)[G, -G4] =-u'A sin ¢ (from eqn 1) (2)

if r/mis very small or n/2 egn 2 becomes

[C-C]= iA sin (3)
From eqn 1 and 3we get

CG=Acos +iA sin ¢) =Ae' and G, =Acos- iA sin ¢) =;Ae-i

Example 4: Numerical on damped simple harmonic oscillator
The frequency of a damped simple harmonic oscillator is given by

k 1 2

(a) If w'-w = 10-w show that Q =500 and that the logarithmic
decrement 8 = /500.
(b) If wo = 10 and m = 10-10 Kg show that the stiffness of the system is 100
N m-, and that the
resistive constant r is 2 x 10-7 N sm-1.

(c) If the maximum displacement at t= 0is 10- m, show

that the energy of the system is5 x 10-3
Jand the decay to e-' of this value takes 0.5 ms.

(d)Show that the energy loss in the first cycle is 2 x 10-5J.

Ans:

(a) w'-w = 10-w ’ w - -w~ = 10-w3=10-w ’ om = 500

2m

Also w' Wo, hence Q = w'm = 500

r
(1)
 Now &=r' =
2m w' 2m wo 500

(b) The stiffness of the system is given by:

k= w'm =1012x 10-10 = 100 Nm-1
10x10-10 =2x 10-7 N. sm-1
Now, Q=°0" ’ r= 500

(c) At t =0and maximum displacement, | = 0, energy is given by:

1 1 1
E=kr' +;m|² =kx'=;x 100 x10-* =5x 10-* U]
Time for energy to decay to e-1of initial value is given by:

m 10-10
0.5 ms
t=
2x 10-7=

(d) to calculate energy loss in the 1 cycle we use the formula for quality factor

0=Z1AE
Since energy loss in the 1 cycle is the same as energy loss per cycle, we have
5x 10-3
-AE = 2T =2I 500
2r x 10-5j

Summary of Important Results

Damped Simple Harmonic Motion

Equation of motion mi + r£ + sx = 0

Displacement x= Ae 2m sin(w't + ): a = S
m
r"

Logarithmic decrement 8: the logarithm of the ratio of two successive amplitudes one period t' apart.
rr'
8= An
loge Ant1 Zm

Relaxation Time: Time for amplitude to decay to Ae- from Age m, i.e., t=
Quality factor Q: Quality factor Q is the number of radians during which energy decreases to E =
Ege-1.
= 21
energy stored in system
energy lost per cycle