2. CONTENTS:
1.NEWTONโS LAW OF COOLING
2.RADIOACTIVE DECAY
3.L-R AND C-R CIRCUITS
4.FREE UNDAMPED AND DAMPED VIBRATIONS
5.FORCED OSCILLATIONS-RESONANCE PHENOMENON
6.SERIES LCR CIRCUIT
7.ANALOGY WITH MASS SPRING SYSTEM
8.LCR CIRCUIT WITH VOLTAGE SOURCE
9.COMPLEX IMPEDENCE
10. RESONANCE PHENOMENA
2
3. 3
Radioactive Decay
The D.E.
๐๐ข
๐๐ก
= โ๐๐ข, ๐>0
describes the decay phenomena where it is assumed that the
material ๐ข(๐ก) at any time ๐ก decays at a rate which is proportional to
the amount present. The solution is
๐ข(๐ก) = ๐๐โ๐๐ก
If initially at , ๐ก =0 ๐ข0 is the amount present then ๐ = ๐ข0
๐ข(๐ก) = ๐ข0๐โ๐๐ก
4. ๐๐ข
๐ข
= โ ๐๐๐ก (Separate the variables)
เถฑ
1
๐ข
๐๐ข = โ เถฑ๐๐๐ก (Integrate both sides)
ln ๐ข = โ๐๐ก + A (Apply integration formulas)
๐ln |๐ข| = ๐โ๐๐ก+๐ด (Raise both sides to exponential function of base ๐)
๐ข = ๐โ๐๐ก
๐๐ด
(Use inverse property ๐ln ๐
= ๐ and law of exponents ๐๐ฅ+๐ฆ
= ๐๐ฅ
๐๐ฆ
)
๐ข(๐ก) = ๐ถ๐โ๐๐ก
(Use absolute value definition ยฑ ๐๐ด
๐๐๐ก
and replace constant ยฑ ๐๐ด
with ๐ถ. )
4
5. Ex. A certain radioactive material is known to decay at rate
proportional to the amount present. If initially 500 mg of the
material is present and after 3 years 20% of the original mass was
decayed,find the expression for the mass at any time.
5
Sol. Let ๐ข(๐ก) denote the amount of material present at any time t. Then
according to the given condition
๐๐ข
๐๐ก
= โ๐๐ข
or
๐๐ข
๐ข
= โ๐. ๐๐ก
On integration log ๐ข = โ๐๐ก + ๐. . . . . . . . . . . . . (1)
6. Let ๐ข0 be the amount of the material at ๐ก = 0 so that when ๐ก = 0 ,๐ข = ๐ข0
๐๐๐๐ (1) log ๐ข0 = ๐
so from (1) log ๐ข = โ๐๐ก + log ๐ข0
log
๐ข
๐ข0
= โ๐๐ก
or
๐ข
๐ข0
= ๐โ๐๐ก
or ๐ข = ๐ข0๐โ๐๐ก
. . . . . . . . . . . . . . . (2)
After 3 years 20 % of the original mass decayed
Now mass ๐ข = ๐ข0 โ
20
100
๐ข0 =
4
5
๐ข0
๐๐๐๐ก๐๐๐ข๐๐
6
8. Ex. Radium decomposes at a rate proportional to the amount of radium
present. Suppose it is found that in 25 years approximately 1.1% of a certain
quantity of Radium has decomposed. Determine approximately how long will it
take for one half of the original amount of radium to decompose.
Sol. Let ๐ข(๐ก) denote the amount of material present at any time t.
Let ๐ข0 be the initial amount of radium at ๐ก = 0
By decay rule
๐๐ข
๐๐ก
= โ๐u
or
๐๐ข
๐ข
= โ๐. ๐๐ก
On integration log ๐ข = โ๐๐ก + ๐
or ๐ข = ๐๐โ๐๐ก. . . . . . . . . . . . . . . . (1)
8
9. ๐๐๐๐ก๐๐๐ข๐๐
using initial condition , at ๐ก = 0, ๐ข = ๐ข0 in (1)
๐ข0 = ๐ or ๐ข = ๐ข0๐โ๐๐ก...............(2)
when ๐ก = 25 , ๐ข = ๐ข0 โ
1.1
100
๐ข0
๐น๐๐๐ (2) 1 โ
1.1
100
๐ข0 = ๐ข0๐โ25๐
๐ = 0.000443
๐ข = ๐ข0๐โ0.000443๐ก. . . . . . . (3)
Now we have to find the time taken for half the radium to disintegrate
u =
๐ข0
2
๐ข0
2
= ๐ข0๐โ0.000443๐ก(๐๐๐๐ 3)
๐ก = 1565 years app.
9
11. History
โขIn the 17th century, Isaac Newton studied the nature of
cooling
โขIn his studies he found that if there is a less than 10 degree
difference between two objects the rate of heat loss is
proportional to the temperature difference
โขNewton applied this principle to estimate the temperature of
a red-hot iron ball by observing the time which it took to
cool from a red heat to a known temperature, and comparing
this with the time taken to cool through a known range at
ordinary temperatures.
โขAccording to this law, if the excess of the temperature of the
body above its surroundings is observed at equal intervals of
time, the observed values will form a geometrical
progression with a common ratio
โขHowever, Newtonโs law was inaccurate at high temperatures
โขPierre Dulong and Alexis Petit corrected Newtonโs law by
clarifying the effect of the temperature of the surroundings
11
12. โข Newton's Law of Cooling is used to model the temperature
change of an object of some temperature placed in an
environment of a different temperature. The law states that:
= the temperature of the object at time t
= the temperature of the surrounding environment (constant)
k = the constant of proportionality
โข This law says that the rate of change of temperature is
proportional to the difference between the temperature of the
object and that of the surrounding environment.
๐๐
๐๐ก
= โ๐(๐ โ ๐๐ด)
๐๐ด
๐
12
13. The Basic Concept
TO solve the differential equation steps are given below.
โข Separate the variables. Get all the T 's on one side and the t on
the other side. The constants can be on either side.
โข Integrate each side
โข Leave in the previous form or solve for T
Find antiderivative of each side
๐๐
(๐ โ ๐๐ด)
= โ๐๐๐ก
log( ๐ โ ๐๐ด) = โ๐๐ก + ๐0
๐ โ ๐๐ด = ๐โ๐๐ก+๐0
๐ = ๐๐ด + ๐โ๐๐ก+๐0 = ๐๐ด + ๐โ๐๐ก
๐๐0
T = ๐๐ด + ๐โ๐๐ก
๐
13
14. Ex. If a substance cools from 370k to 330k in 10minutes, when the temperature of the
surrounding air is 290k, find the temperature of the substance after 40 minutes.
Sol. Here ๐๐ด = 290 so that the solution is
๐ ๐ก = 290 + ๐๐โ๐๐ก
(T = ๐๐ด + ๐โ๐๐ก
๐ )
using condition at t=0 , ๐ = 370 (to find ๐)
370 =290+c ๐0
or ๐ = 80
Thus ๐ ๐ก = 290 + 80๐โ๐๐ก
using condition ๐ 10 = 330 (to find ๐)
Thus 330 =290+80 ๐โ๐ .10
so that ๐ =
๐๐๐2
10
= 0.069314718
The required solution is
๐ ๐ก = 290 + 80๐โ.06931๐ก
Putting ๐ก = 40 ๐๐๐๐ in the above solution
๐ 40 = 290 + 80๐โ.06931ร40 โ 295
14
15. Example 2: A body of temperature 80ยฐF is placed in a room of constant temperature
50ยฐF at time t = 0. At the end of 5 minutes the body has cooled to a temperature of
70ยฐF. (a) Find the temperature of the body at the end of 10 minutes. (b) When will
the temperature of the body be 60ยฐF?
โข Sol.
๐ฟ๐๐ก ๐ ๐๐ ๐กโ๐ ๐ก๐๐๐๐๐๐๐ก๐ข๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ฆ. ๐โ๐๐ ๐ ๐ก = 50 + ๐ ๐โ๐๐ก
๐ ๐๐๐๐ ๐๐ด = 50.
๐ด๐๐๐๐ฆ ๐กโ๐ ๐๐๐๐๐๐ก๐๐๐ ๐ 0 = 80, (๐ก๐ ๐๐๐๐ ๐);
80 = 50 + ๐ ๐0 ๐๐ ๐ = 30
๐ ๐ก = 50+ 30 ๐โ๐๐ก
o๐ ๐ข๐ ๐๐๐ ๐๐๐๐๐๐ก๐๐๐ ๐ 5 = 70 (๐ก๐ ๐๐๐ก๐๐๐๐๐๐ ๐);
70 = 50+ 30 ๐โ๐5
๐ ๐ ๐กโ๐๐ก ๐ =
1
5
log
3
2
๐ = .08109
๐โ๐ข๐ ๐กโ๐ ๐๐๐๐ข๐๐๐๐ ๐ ๐๐๐ข๐ก๐๐๐ ๐คโ๐๐โ ๐๐๐ฃ๐๐ ๐กโ๐ ๐ก๐๐๐๐๐๐๐ก๐ข๐๐ ๐๐
๐กโ๐ ๐๐๐๐ฆ ๐๐ก ๐๐๐ฆ ๐ก๐๐๐ ๐ก ๐๐
๐ ๐ก = 50+ 30 ๐โ.08109 ๐ก
(a). ๐ 10 = 50 + 30 ๐โ.08109(10) โ 63.330F
(b). 60 = 50 + 30 ๐โ.08109๐ก ๐ ๐๐๐ฃ๐๐๐ ๐ค๐ ๐๐๐ก ๐ก โ13.55mts
15
16. Ans. It takes about 8.88 more minutes For the object to cool to a temperature of 110ยฐ
110ยฐ
For the object to cool to a temperature of
110ยฐ
The Problem
Spencer and Vikalp are cranking out math problems at Safeway. Shankar is at home
making pizza. He calls Spencer and tells him that he is taking the pizza out from
the oven right now. Spencer and Vikalp need to get back home in time so that
they can enjoy the pizza at a warm temperature of 110ยฐF.
The pizza, heated to a temperature of 450ยฐF, is taken out of an oven and placed in a
75ยฐF room at time t=0 minutes. If the pizza cools from 450ยฐ to 370ยฐ in 1 minute,
how much longer will it take for its temperature to decrease to 110ยฐ?
16
17. A copper ball is heated to a
temperature 0f 100 0 C and it is
placed in water which is maintained
at a temperature of 400 C. If in 4
mins the temperature of the ball is
reduced to 600 ,find the time at
which the temperature of the ball is
500 C.
ANS.6.5 mins
17
18. Real Life Applications
โข To predict how long it takes for a hot cup of tea to cool down to a
certain temperature
โข To find the temperature of a soda placed in a refrigerator by a
certain amount of time.
โข In crime scenes, Newtonโs law of cooling can indicate the time
of death given the probable body temperature at time of
death and the current body temperature
โข
18
25. MASS โSPRING MECHANICAL SYSTEM
Airplanes, bridges, ships, machines, cars etc. are vibrating mechanical systems. The
simplest mechanical system is the mass-spring system which consists of a coil spring of
natural length L, suspended vertically from a fixed point support(such a ceiling or beam).A
constant mass โmโ attached to the lower end of the spring to a length(๐ฟ + ๐) and comes to
rest which is known as the static equilibrium position. Here ๐ > 0 is the static deflection
due to hanging the mass on the spring.
Now the mass is set in motion either by pushing or
pulling the mass from equilibrium position. Since
the motion takes place in the vertical direction,
we consider the downward direction as positive.
25
26. In order to determine the displacement ๐ฅ(t) of the mass from the static
equilibrium position, we use Newtonโs second law and Hookeโs law. The
mass ๐ is subjected to the following forces ;
1. A gravitational force ๐๐ acting downwards.
2. A spring restoring force โk(๐ฅ ๐ก + ๐) due to displacement of the spring
from rest position .
3. a frictional force of the medium , opposing the motion and of magnitude
โ ๐ แถ
๐ฅ(๐ก).
4. an external force F(๐ก)
The differential equation describing the motion of the mass is obtained by
Newtonโs second law as
26
27. ๐ แท
๐ฅ(๐ก)=๐๐ โ k(๐ฅ ๐ก + ๐) โ ๐ แถ
๐ฅ(๐ก)+๐น(๐ก)
Here k> 0 is known as spring constant , ๐ โฅ 0 damping constant , g is the
gravitational constant. Since the force on the mass exerted by the spring must
be equal and opposite to the gravitational force on the mass, we have
k๐ = ๐๐.Thus the D.E. modelling the motion of mass is
๐ แท
๐ฅ(๐ก)+๐ แถ
๐ฅ(๐ก)+ k ๐ฅ ๐ก = ๐น(๐ก)โฆโฆโฆ..(1)
which is a second order linear non-homogeneous equation with constant
coefficients. The displacement of the mass at any time ๐ก is ๐ฅ(๐ก) which is the
solution of the differential equation (1) .The three important cases of D.E. (1)
are referred to as free motion ,damped motion, and forced motion.
27
28. Free, Undamped Oscillations of a spring
In the absence of the external force ( ๐น ๐ก = 0) and neglecting the damping force(๐ = 0) D.E.
(1) reduce to ๐ แท
๐ฅ(๐ก)+ k๐ฅ=0โฆโฆโฆ..(2)
which is harmonic oscillator equation. Putting ๐2 =
๐
๐
, the equation (2) takes the form
แท
๐ฅ+ ๐2๐ฅ=0
whose general solution is given by ๐ฅ ๐ก = ๐1๐๐๐ ๐t+ ๐2๐ ๐๐ ๐t โฆโฆโฆ(3)
Introducing ๐1=A๐๐๐ ๐, ๐2=โA๐ ๐๐๐ equation (3) can be written as
๐ฅ ๐ก = A ๐๐๐ ๐ ๐๐๐ ๐t - A ๐ ๐๐๐ ๐ ๐๐ ๐t ,
i.e. ๐ฅ ๐ก =๐ด ๐๐๐ (๐t+๐)โฆโฆโฆโฆโฆ(4)
where ๐ด = ๐1
2 + ๐2
2 , tan๐=โ
๐2
๐1
. The constant A is called the amplitude of the motion
and gives the maximum displacement of the mass from its equilibrium position. Thus the free
, undamped motion of the mass is a simple harmonic motion , which is periodic.
28
29. The period of the motion is the time interval between two successive
maxima and is given by ๐ =
2๐
๐
= 2๐
๐
๐
the natural frequency of the
motion is the reciprocal of the period , which gives the number of
oscillations per second. Thus the natural frequency is the undamped
Frequency (i.e. frequency of the system without damping)
29
30. Free, Damped Motion of a Mass
Every system has some damping , otherwise the system continues to move forever.
Damping force opposes oscillations. Damping not only decreases the amplitude but
also alters the natural frequency of the system. With external force absent (๐น ๐ก = 0)
and damping present (๐ โ 0), the D.E.(1) takes the form
๐ แท
๐ฅ+๐ แถ
๐ฅ+ k๐ฅ=0 or แท
๐ฅ +
๐
๐
แถ
๐ฅ +
๐
๐
๐ฅ=0
Let 2๐ =
๐
๐
๐๐๐ ๐2
=
๐
๐
or แท
๐ฅ +2๐ แถ
๐ฅ +๐2
๐ฅ=0 โฆโฆโฆ..(5)
whose auxiliary equation is
๐2+2๐๐ + ๐2 = 0โฆโฆโฆโฆโฆ..(6)
The roots of eq(6) are ๐ =
โ2๐ยฑ 4๐2โ4๐2
2
= โ๐ ยฑ ๐2 โ ๐2 โฆโฆโฆ(7)
30
31. The motion of mass depends on the damping through the nature of the
discriminant ๐2 โ ๐2
Case 1; ๐2 โ ๐2 >0 i.e. ๐2 > 4๐๐ since ๐> ๐, the roots
๐1 = โ๐ + ๐2 โ ๐2 ๐๐๐ ๐2 = โ๐ โ ๐2 โ ๐2
are distinct ,real negative numbers. The general solution is
๐ฅ ๐ก = ๐1๐๐1๐ก + ๐2๐๐2๐ก โฆโฆโฆโฆโฆโฆโฆ.(8)
which tends to zero as ๐ก โ โ.Thus the damping is so great that no oscillations
can occur. The motion is said to be over critically damped (over damped).
Case 2; ๐2 โ ๐2=0 Here both the roots are equal , real negative number ๐. The
general solution is
๐ฅ ๐ก = (๐1+ ๐2๐ก)๐โ๐๐ก โฆโฆโฆโฆโฆโฆโฆ.(9)
This situation of motion is called critical damping and is not oscillatory .
31
32. Case 3; When ๐ < ๐ then ๐2 โ ๐2<0 so the roots of the auxiliary
equation (7) are complex and given by โ๐ ยฑ ๐2 โ ๐2๐.
The general solution is
๐ฅ(๐ก) = ๐โ๐๐ก ๐1cos( ๐2 โ ๐2)๐ก + ๐2sin( ๐2 โ ๐2)๐ก โฆโฆโฆโฆ.(10)
which can be written in the alternative form(as in eq 4)
๐ฅ(๐ก) = ๐ด๐โ๐๐ก
cos( ๐2 โ ๐2)๐ก + ๐ โฆโฆโฆโฆโฆ(11)
Here ๐ด= ๐1
2 + ๐2
2 and ๐ = ๐ก๐๐โ1
(โ
๐1
๐2
).
In this case the motion is said to be underdamped.
32
33. Forced Oscillation: Damped
In the presence of an external force F(t), also known as input or driving force, the
solutions of D.E. (1) are known as output or response of the system to the external
force. In this case, the oscillations are said to be forced oscillations, which are of
two types damped forced oscillations and undamped forced oscillations.
Case 1.The D.E in this case is ๐ แท
๐ฅ+c แถ
๐ฅ + ๐๐ฅ = ๐น(๐ก)
Suppose the external force ๐น(๐ก)=๐น1๐๐๐ ๐ฝ๐ก, (๐น1 > 0; ๐ฝ > 0)
Then D.E. takes the form ๐ แท
๐ฅ+c แถ
๐ฅ + ๐๐ฅ = ๐น1๐๐๐ ๐ฝ๐ก
Or แท
๐ฅ+2b แถ
๐ฅ+ ๐2๐ฅ= ๐ธ1๐๐๐ ๐ฝ๐กโฆโฆโฆโฆโฆโฆโฆโฆ.(12)
where 2b=
๐
๐
๐๐๐ ๐2
=
๐
๐
, ๐ธ1=
๐น1
๐
33
34. Assuming b < ๐ the complimentary function of (12) is
๐ฅ๐ = ๐ด๐โ๐๐ก cos( ๐2 โ ๐2)๐ก + ๐ โฆโฆโฆโฆโฆโฆโฆ(13) (as in eq 4)
which approaches to zero as ๐ก โ โ. The particular integral of (12) is
๐ฅ๐ =
1
๐ท2+2๐๐ท+๐2 ๐ธ1๐๐๐ ๐ฝ๐ก
or ๐ฅ๐=
๐ธ1
๐2โ๐ฝ2 2+4๐2๐ฝ2 ๐2
โ ๐ฝ2
๐๐๐ ๐ฝ๐ก + 2๐๐ฝ๐ ๐๐๐ฝ๐ก โฆโฆ..(14)
The general solution is
๐ฅ ๐ก = ๐ฅ๐+ ๐ฅ๐
34
35. Case 2. Undamped forced Oscillation: Resonance
In the undamped case ๐ = 0 and D.E. (1) takes the form
๐ แท
๐ฅ+๐๐ฅ = ๐น ๐ก = ๐น1๐๐๐ ๐ฝ๐ก
or แท
๐ฅ+ ๐2๐ฅ= ๐ธ1๐๐๐ ๐ฝ๐กโฆโฆโฆ(16) where ๐2=
๐
๐
,and ๐ธ1=
๐น1
๐
and ๐ , ๐ธ1,
๐ฝ are positive constant. The complimentary function of (16) is
๐ฅ๐ = ๐1๐๐๐ ๐๐ก+ ๐2๐ ๐๐๐๐ก โฆ โฆ โฆ โฆ โฆ โฆ โฆ (17)
Now we study the nature of solution of D.E. (16). When ๐ = ๐ฝ: In this
case the particular integral of D.E. (16), which is the forced solution, is
๐ฅ๐ =
1
๐ท2+๐2 ๐ธ1๐๐๐ ๐ฝ๐ก=
๐ธ1
2๐
๐ก ๐ ๐๐๐ฝ๐กโฆโฆโฆโฆ(18)
35
36. General solution :
The general solution in this case is
๐ฅ = ๐ฅ๐ + ๐ฅ๐ ๐๐ ๐ฅ = ๐1๐๐๐ ๐๐ก+ ๐2๐ ๐๐๐๐ก+
๐ธ1
2๐
๐ก ๐ ๐๐๐ฝ๐กโฆโฆโฆโฆ.(19)
The forced solution (the particular integral (18)) grows with time and
becomes larger and larger (because of the presence of โ๐กโ). Thus, in an
undamped system if ๐ = ๐ฝ i.e., the frequency ๐ฝ of external force matches
(equals) with the natural frequency ๐.
The phenomenon of unbounded oscillations occurs, which is known as
resonance. In resonance, for a bounded input the system responds with an
unbounded output. Thus resonance, the phenomenon of excitation of large
oscillation, is undesirable because the system may get destroyed due to these
unwanted large vibrations.
36
37. Example 1: An 8 lb weight is placed at one end of a spring suspended
from the ceiling. The weight is raised to 5 inches above the equilibrium
position and left free. Assuming the spring constant 12 lb/ft, find the
equation of motion, displacement function ๐(๐), amplitude, period,
frequency and maximum velocity.
37
39. Free ,Damped Problem
โข Example 2: A 2 lb weight suspended from one end of a spring stretches it
to 6 inches. A velocity of 5 ft/sec2 upwards is imparted to the weight at its
equilibrium position. Suppose a damping force ๐ท๐ acts on the weight.
Here 0<๐ท < ๐ and ๐ = แถ
๐ =velocity. Assuming the spring constant 4
lb/ft, (a) Determine the position and velocity of the weight at any time
(b) Express the displacement function ๐(๐) in the amplitude-phase form.
(c) Find the amplitude, period, frequency, maximum velocity.
39
44. RLC Circuit
An RLC-series circuit consists of a resistor, a conductor, a capacitor and an emf as
shown in the figure .
Using the Kirchhoffโs law, the sum of the voltage drops across the three elements
inductor, resistor and Capacitance equal to the external source ๐ธ. Thus, the RLC-circuit
is modeled by ๐ฟ
๐๐ผ
๐๐ก
+ ๐ ๐ผ +
1
๐ถ
๐=๐ธ(๐ก)โฆโฆโฆ. (1)
which contains two dependent variables ๐ and ๐ผ.
44
45. Since I =
๐Q
๐๐ก
. the above equation can be written as
๐ฟ
๐2Q
๐๐ก2 + ๐
๐๐
๐๐ก
+
1
๐ถ
๐ = ๐ธ(๐ก)โฆโฆโฆโฆโฆ (2)
which contains only one dependent variable ๐. Differentiating (1) w.r.t. โtโ, we obtain
๐ฟ
๐2I
๐๐ก2 + ๐
๐๐ผ
๐๐ก
+
1
๐ถ
๐ผ =
๐๐ธ
๐๐ก
โฆโฆโฆโฆโฆโฆโฆ (3)
which contains only one dependent variable ๐ผ. Thus, the charge ๐ and current ๐ผ at any time
in the RLC circuit are obtained as solutions of (2) and (3) which are both linear 2nd order
non homogeneous ordinary differential equations. The equation (3) is used more often,
since current I(๐ก) is more important than charge Q(๐ก), in most of the practical problems.
The RLC circuit reduces to an RL-circuit in absence of capacitor and to RC-circuit when no
inductor is present.
To find the solution we find the complimentary function and particular integral.
45
53. 53
PRACTICAL IMPOTTANCE
This analogy is strictly quantitative in the sense that to a given mechanical system
we can construct an electric circuit whose current will give the exact values of the
displacement in the mechanical system when suitable scale factors are introduced.
The practical importance of this analogy is almost obvious. The analogy may be
used for constructing an โ electrical modelโ of a given mechanical model ,
resulting in substantial savings of time and money because electric circuits are easy
to assemble , and electric quantities can be measured much more quickly and
accurately than mechanical ones.