1.NEWTON’S LAW OF COOLING
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
10. RESONANCE PHENOMENA
= −𝑘𝑢, 𝑘>0
describes the decay phenomena where it is assumed that the
material 𝑢(𝑡) at any time 𝑡 decays at a rate which is proportional to
the amount present. The solution is
𝑢(𝑡) = 𝑐𝑒−𝑘𝑡
If initially at , 𝑡 =0 𝑢0 is the amount present then 𝑐 = 𝑢0
𝑢(𝑡) = 𝑢0𝑒−𝑘𝑡
= − 𝑘𝑑𝑡 (Separate the variables)
𝑑𝑢 = − න𝑘𝑑𝑡 (Integrate both sides)
ln 𝑢 = −𝑘𝑡 + A (Apply integration formulas)
𝑒ln |𝑢| = 𝑒−𝑘𝑡+𝐴 (Raise both sides to exponential function of base 𝑒)
𝑢 = 𝑒−𝑘𝑡
(Use inverse property 𝑒ln 𝑘
= 𝑘 and law of exponents 𝑏𝑥+𝑦
𝑢(𝑡) = 𝐶𝑒−𝑘𝑡
(Use absolute value definition ± 𝑒𝐴
and replace constant ± 𝑒𝐴
with 𝐶. )
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.
Sol. Let 𝑢(𝑡) denote the amount of material present at any time t. Then
according to the given condition
= −𝑘. 𝑑𝑡
On integration log 𝑢 = −𝑘𝑡 + 𝑐. . . . . . . . . . . . . (1)
6. Let 𝑢0 be the amount of the material at 𝑡 = 0 so that when 𝑡 = 0 ,𝑢 = 𝑢0
𝑓𝑟𝑜𝑚 (1) log 𝑢0 = 𝑐
so from (1) log 𝑢 = −𝑘𝑡 + log 𝑢0
or 𝑢 = 𝑢0𝑒−𝑘𝑡
. . . . . . . . . . . . . . . (2)
After 3 years 20 % of the original mass decayed
Now mass 𝑢 = 𝑢0 −
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 𝑢(𝑡) denote the amount of material present at any time t.
Let 𝑢0 be the initial amount of radium at 𝑡 = 0
By decay rule
= −𝑘. 𝑑𝑡
On integration log 𝑢 = −𝑘𝑡 + 𝑐
or 𝑢 = 𝑐𝑒−𝑘𝑡. . . . . . . . . . . . . . . . (1)
using initial condition , at 𝑡 = 0, 𝑢 = 𝑢0 in (1)
𝑢0 = 𝑐 or 𝑢 = 𝑢0𝑒−𝑘𝑡...............(2)
when 𝑡 = 25 , 𝑢 = 𝑢0 −
𝐹𝑟𝑜𝑚 (2) 1 −
𝑢0 = 𝑢0𝑒−25𝑘
𝑘 = 0.000443
𝑢 = 𝑢0𝑒−0.000443𝑡. . . . . . . (3)
Now we have to find the time taken for half the radium to disintegrate
= 𝑢0𝑒−0.000443𝑡(𝑓𝑟𝑜𝑚 3)
𝑡 = 1565 years app.
•In the 17th century, Isaac Newton studied the nature of
•In 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
•Newton 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
•According 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
•However, Newton’s law was inaccurate at high temperatures
•Pierre Dulong and Alexis Petit corrected Newton’s law by
clarifying the effect of the temperature of the surroundings
12. • 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
• 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.
= −𝑘(𝑇 − 𝑇𝐴)
13. The Basic Concept
TO solve the differential equation steps are given below.
• 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.
• Integrate each side
• Leave in the previous form or solve for T
Find antiderivative of each side
(𝑇 − 𝑇𝐴)
log( 𝑇 − 𝑇𝐴) = −𝑘𝑡 + 𝑐0
𝑇 − 𝑇𝐴 = 𝑒−𝑘𝑡+𝑐0
𝑇 = 𝑇𝐴 + 𝑒−𝑘𝑡+𝑐0 = 𝑇𝐴 + 𝑒−𝑘𝑡
T = 𝑇𝐴 + 𝑒−𝑘𝑡
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 𝑇𝐴 = 290 so that the solution is
𝑇 𝑡 = 290 + 𝑐𝑒−𝑘𝑡
(T = 𝑇𝐴 + 𝑒−𝑘𝑡
using condition at t=0 , 𝑇 = 370 (to find 𝑐)
370 =290+c 𝑒0
or 𝑐 = 80
Thus 𝑇 𝑡 = 290 + 80𝑒−𝑘𝑡
using condition 𝑇 10 = 330 (to find 𝑘)
Thus 330 =290+80 𝑒−𝑘 .10
so that 𝑘 =
The required solution is
𝑇 𝑡 = 290 + 80𝑒−.06931𝑡
Putting 𝑡 = 40 𝑚𝑖𝑛𝑠 in the above solution
𝑇 40 = 290 + 80𝑒−.06931×40 ≈ 295
15. Example 2: A body of temperature 80°F is placed in a room of constant temperature
50°F at time t = 0. At the end of 5 minutes the body has cooled to a temperature of
70°F. (a) Find the temperature of the body at the end of 10 minutes. (b) When will
the temperature of the body be 60°F?
𝐿𝑒𝑡 𝑇 𝑏𝑒 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦. 𝑇ℎ𝑒𝑛 𝑇 𝑡 = 50 + 𝑐 𝑒−𝑘𝑡
𝑠𝑖𝑛𝑐𝑒 𝑇𝐴 = 50.
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑇 0 = 80, (𝑡𝑜 𝑓𝑖𝑛𝑑 𝑐);
80 = 50 + 𝑐 𝑒0 𝑜𝑟 𝑐 = 30
𝑇 𝑡 = 50+ 30 𝑒−𝑘𝑡
o𝑟 𝑢𝑠𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑇 5 = 70 (𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑘);
70 = 50+ 30 𝑒−𝑘5
𝑠𝑜 𝑡ℎ𝑎𝑡 𝑘 =
𝑘 = .08109
𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓
𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑎𝑡 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒 𝑡 𝑖𝑠
𝑇 𝑡 = 50+ 30 𝑒−.08109 𝑡
(a). 𝑇 10 = 50 + 30 𝑒−.08109(10) ≈ 63.330F
(b). 60 = 50 + 30 𝑒−.08109𝑡 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡 𝑡 ≈13.55mts
16. Ans. It takes about 8.88 more minutes For the object to cool to a temperature of 110°
For the object to cool to a temperature of
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 110°F.
The pizza, heated to a temperature of 450°F, is taken out of an oven and placed in a
75°F room at time t=0 minutes. If the pizza cools from 450° to 370° in 1 minute,
how much longer will it take for its temperature to decrease to 110°?
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
18. Real Life Applications
• To predict how long it takes for a hot cup of tea to cool down to a
• To find the temperature of a soda placed in a refrigerator by a
certain amount of time.
• In crime scenes, Newton’s law of cooling can indicate the time
of death given the probable body temperature at time of
death and the current body temperature
23. Example.Find the current at any time 𝑡 > 0 in a circuit having in series
a constant e.m.f. 40v, a resistor 10𝑜ℎ𝑚 , an inductor 0.2H given that
initial current is zero.
• 𝑆𝑂𝐿. 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑅𝐿 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠
+ 𝑅𝐼 = 𝐸 𝑡
𝐻𝑒𝑟𝑒 𝐿 = 0.2 , 𝑅 = 10, 𝐸 = 40
+ 10𝐼 = 40 𝑜𝑟
+ 50𝐼 = 200((first order linear D .E.))
𝐼𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝐼 𝑡 . 𝑒50𝑡 = (200. 𝑒50𝑡 𝑑𝑡 + 𝑐)
𝑢𝑠𝑖𝑛𝑔 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑤𝑒 𝑔𝑒𝑡 𝑐 = −4
𝐼 𝑡 = 4(1 − 𝑒−50𝑡) ans.
25. MASS –SPRING 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 ‘m’ attached to the lower end of the spring to a length(𝐿 + 𝑒) and comes to
rest which is known as the static equilibrium position. Here 𝑒 > 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.
26. In order to determine the displacement 𝑥(t) of the mass from the static
equilibrium position, we use Newton’s second law and Hooke’s law. The
mass 𝑚 is subjected to the following forces ;
1. A gravitational force 𝑚𝑔 acting downwards.
2. A spring restoring force −k(𝑥 𝑡 + 𝑒) due to displacement of the spring
from rest position .
3. a frictional force of the medium , opposing the motion and of magnitude
− 𝑐 ሶ
4. an external force F(𝑡)
The differential equation describing the motion of the mass is obtained by
Newton’s second law as
27. 𝑚 ሷ
𝑥(𝑡)=𝑚𝑔 − k(𝑥 𝑡 + 𝑒) − 𝑐 ሶ
Here k> 0 is known as spring constant , 𝑐 ≥ 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
k𝑒 = 𝑚𝑔.Thus the D.E. modelling the motion of mass is
𝑥(𝑡)+ k 𝑥 𝑡 = 𝐹(𝑡)………..(1)
which is a second order linear non-homogeneous equation with constant
coefficients. The displacement of the mass at any time 𝑡 is 𝑥(𝑡) 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.
28. Free, Undamped Oscillations of a spring
In the absence of the external force ( 𝐹 𝑡 = 0) and neglecting the damping force(𝑐 = 0) D.E.
(1) reduce to 𝑚 ሷ
which is harmonic oscillator equation. Putting 𝜔2 =
, the equation (2) takes the form
whose general solution is given by 𝑥 𝑡 = 𝑐1𝑐𝑜𝑠 𝜔t+ 𝑐2𝑠𝑖𝑛 𝜔t ………(3)
Introducing 𝑐1=A𝑐𝑜𝑠𝜙, 𝑐2=−A𝑠𝑖𝑛𝜙 equation (3) can be written as
𝑥 𝑡 = A 𝑐𝑜𝑠𝜙 𝑐𝑜𝑠 𝜔t - A 𝑠𝑖𝑛𝜙 𝑠𝑖𝑛 𝜔t ,
i.e. 𝑥 𝑡 =𝐴 𝑐𝑜𝑠(𝜔t+𝜙)……………(4)
where 𝐴 = 𝑐1
2 + 𝑐2
2 , tan𝜙=−
. 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.
29. The period of the motion is the time interval between two successive
maxima and is given by 𝑇 =
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)
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 (𝐹 𝑡 = 0)
and damping present (𝑐 ≠ 0), the D.E.(1) takes the form
𝑥+ k𝑥=0 or ሷ
Let 2𝑏 =
𝑥 +2𝑏 ሶ
whose auxiliary equation is
𝑟2+2𝑏𝑟 + 𝜔2 = 0……………..(6)
The roots of eq(6) are 𝑟 =
= −𝑏 ± 𝑏2 − 𝜔2 ………(7)
31. The motion of mass depends on the damping through the nature of the
discriminant 𝑏2 − 𝜔2
Case 1; 𝑏2 − 𝜔2 >0 i.e. 𝑐2 > 4𝑚𝑘 since 𝑏> 𝜔, the roots
𝑟1 = −𝑏 + 𝑏2 − 𝜔2 𝑎𝑛𝑑 𝑟2 = −𝑏 − 𝑏2 − 𝜔2
are distinct ,real negative numbers. The general solution is
𝑥 𝑡 = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑒𝑟2𝑡 ………………….(8)
which tends to zero as 𝑡 → ∞.Thus the damping is so great that no oscillations
can occur. The motion is said to be over critically damped (over damped).
Case 2; 𝑏2 − 𝜔2=0 Here both the roots are equal , real negative number 𝑏. The
general solution is
𝑥 𝑡 = (𝑐1+ 𝑐2𝑡)𝑒−𝑏𝑡 ………………….(9)
This situation of motion is called critical damping and is not oscillatory .
32. Case 3; When 𝑏 < 𝜔 then 𝑏2 − 𝜔2<0 so the roots of the auxiliary
equation (7) are complex and given by −𝑏 ± 𝜔2 − 𝑏2𝑖.
The general solution is
𝑥(𝑡) = 𝑒−𝑏𝑡 𝑐1cos( 𝜔2 − 𝑏2)𝑡 + 𝑐2sin( 𝜔2 − 𝑏2)𝑡 ………….(10)
which can be written in the alternative form(as in eq 4)
𝑥(𝑡) = 𝐴𝑒−𝑏𝑡
cos( 𝜔2 − 𝑏2)𝑡 + 𝜙 ……………(11)
Here 𝐴= 𝑐1
2 + 𝑐2
2 and 𝜙 = 𝑡𝑎𝑛−1
In this case the motion is said to be underdamped.
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 𝑚 ሷ
𝑥 + 𝑘𝑥 = 𝐹(𝑡)
Suppose the external force 𝐹(𝑡)=𝐹1𝑐𝑜𝑠𝛽𝑡, (𝐹1 > 0; 𝛽 > 0)
Then D.E. takes the form 𝑚 ሷ
𝑥 + 𝑘𝑥 = 𝐹1𝑐𝑜𝑠𝛽𝑡
𝑥+ 𝜔2𝑥= 𝐸1𝑐𝑜𝑠𝛽𝑡…………………….(12)
34. Assuming b < 𝜔 the complimentary function of (12) is
𝑥𝑐 = 𝐴𝑒−𝑏𝑡 cos( 𝜔2 − 𝑏2)𝑡 + 𝜙 …………………(13) (as in eq 4)
which approaches to zero as 𝑡 → ∞. The particular integral of (12) is
𝜔2−𝛽2 2+4𝑏2𝛽2 𝜔2
𝑐𝑜𝑠𝛽𝑡 + 2𝑏𝛽𝑠𝑖𝑛𝛽𝑡 ……..(14)
The general solution is
𝑥 𝑡 = 𝑥𝑐+ 𝑥𝑝
35. Case 2. Undamped forced Oscillation: Resonance
In the undamped case 𝑐 = 0 and D.E. (1) takes the form
𝑥+𝑘𝑥 = 𝐹 𝑡 = 𝐹1𝑐𝑜𝑠𝛽𝑡
𝑥+ 𝜔2𝑥= 𝐸1𝑐𝑜𝑠𝛽𝑡………(16) where 𝜔2=
and 𝜔 , 𝐸1,
𝛽 are positive constant. The complimentary function of (16) is
𝑥𝑐 = 𝑐1𝑐𝑜𝑠𝜔𝑡+ 𝑐2𝑠𝑖𝑛𝜔𝑡 … … … … … … … (17)
Now we study the nature of solution of D.E. (16). When 𝜔 = 𝛽: In this
case the particular integral of D.E. (16), which is the forced solution, is
36. General solution :
The general solution in this case is
𝑥 = 𝑥𝑐 + 𝑥𝑝 𝑜𝑟 𝑥 = 𝑐1𝑐𝑜𝑠𝜔𝑡+ 𝑐2𝑠𝑖𝑛𝜔𝑡+
The forced solution (the particular integral (18)) grows with time and
becomes larger and larger (because of the presence of ‘𝑡’). Thus, in an
undamped system if 𝜔 = 𝛽 i.e., the frequency 𝛽 of external force matches
(equals) with the natural frequency 𝜔.
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.
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 𝒙(𝒕), amplitude, period,
frequency and maximum velocity.
39. Free ,Damped Problem
• 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 𝜷𝝊 acts on the weight.
Here 0<𝜷 < 𝟏 and 𝝊 = ሶ
𝒙 =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 𝒙(𝒕) in the amplitude-phase form.
(c) Find the amplitude, period, frequency, maximum velocity.
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 Kirchhoff’s law, the sum of the voltage drops across the three elements
inductor, resistor and Capacitance equal to the external source 𝐸. Thus, the RLC-circuit
is modeled by 𝐿
+ 𝑅𝐼 +
which contains two dependent variables 𝑄 and 𝐼.
45. Since I =
. the above equation can be written as
𝑑𝑡2 + 𝑅
𝑄 = 𝐸(𝑡)…………… (2)
which contains only one dependent variable 𝑄. Differentiating (1) w.r.t. ‘t’, we obtain
𝑑𝑡2 + 𝑅
which contains only one dependent variable 𝐼. Thus, the charge 𝑄 and current 𝐼 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(𝑡) is more important than charge Q(𝑡), 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.
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 “ electrical model” 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.