This Unit is rely on introduction to Simple Harmonic Motion. the contents was prepared using the Curriculum of NTA level 4 at Mineral Resources Institute- Dodoma.

1. By
Peter Huruma Mammba
Department of General Studies
DODOMA POLYTECHNIC OF ENERGY AND EARTH RESOURCES MANAGEMENT (MADINI INSTITUTE)
–DODOMA
SIMPLE HARMONIC
MOTION

2. What is Simple Harmonic Motion?
Is a type of periodic motion or oscillation motion
where the restoring force is directly proportional to
the displacement and acts in the direction opposite to
that of displacement.

3. Periodic Motion
This is the motion which repeats itself after a regular interval
of time.
Examples are;
the motion of the earth around the sun whose period is 1 year,
The motion of the moon around the earth is also a periodic
motion whose time period is 27.3 days, etc.

4. Oscillatory or Vibratory Motion
Is when a body is in a periodic motion moves along the same
path to and fro about a definite point (equilibrium).
Examples are;
A pendulum swing back and forth, the vibration of guitar string,
a mass vibrating at the end of a spring etc.
N.B it may be noted that all oscillatory motion are periodic
motions but all periodic motions are not oscillatory.

5. Why S.H.M is important?
• The study and control of oscillation are two of the primary
goals of both physics and engineering.

6. S.H.M
• Consider a particle executing S.H.M about point
O with an amplitude a. The particle is said to
have completed 1 vibration (or oscillation) if
starting from t = 0 up to t = T.
• The vibration of the displacement x with t is
a sine relation given by;
o
𝑥 = 𝑎 𝑠𝑖𝑛𝜔𝑡

7. Sinusoidal motion of a particle executing
S.H.M
Time (s)
Displacement (x)
Period T
The displacement x is
defined as the distance
the body moves from
its equilibrium
Period T

8. Sine Function: Mathematically
x
y
2π
π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π
-1
1
y=sin(x)y=cos(x)

9. Sine function: employed for oscillations
x
y
π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π
-1
1
y=sin(x)
Time t (s)
Displacement y (m)
T/2 T 2T
-𝑎
𝑎
y= 𝑎 sin(ωt)

10. Sine function: employed for oscillations
Time t (s)
Displacement y (m)
T/2 T 2T
-𝑎
𝑎 y= a sin(ωt)
What do we need ? 1. Maximum displacement a
2. ωT = 2π
3. Initial condition

11. Amplitude of the motion (a)
Amplitude Is the greatest
displacement from the mean
equilibrium position.

12. Time Period of the Motion (T)
Time period is the time
taken to complete 1
vibration.

13. Frequency of Oscillation (f)
Frequency Is the number of oscillations that are
completed each second.
That is;
SI unit of f is hertz
𝟏𝑯𝒆𝒓𝒕𝒛 = 𝟏𝑯𝒛 = 𝟏𝒐𝒔𝒄𝒊𝒍𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓 𝒔𝒆𝒄𝒐𝒏𝒅 = 𝟏𝒔−𝟏
𝑓 = 1
𝑇

14. Angular frequency of S.H.M (𝝎)
 Angular Frequency is a scalar measure of rotation rate.
• Angular frequency ω (also referred to by the
terms angular speed, radial frequency, circular
frequency, orbital frequency, radian frequency).

15. What is 𝝎 measured in?
𝝎 is measure in radians per second
One revolution is equal to 2π radians, hence;
𝜔 =
2𝜋
𝑇
= 2𝜋𝑓
In SI units, 𝝎 is normally presented in radians per second, even
when it does not express a rotational value.

16. Phase
Phase is the position of a point in time (an instant) on
a waveform cycle.
• Phase difference is the difference, expressed in degrees or
time, between two waves having the same frequency and
referenced to the same point in time.

17. In phase and Out of phase
Two oscillators that have
the same frequency and no
phase difference are said to
be in phase.
Two oscillators that have the
same frequency and different
phases have a phase difference,
and the oscillators are said to
be out of phase with each
other.

18. In phase vis Out of Phase

19. Analysis of one-dimensional S.H.M in terms of
Uniform Circular Motion
• It is possible to analyze S.H.M in terms of a uniform circular
motion.
• Suppose the particle is moving with a uniform circular speed
𝝎 along a circumference of a circle of radius a and center O.

20. Analysis…
• M moves to and fro along the straight line 𝐴𝐴′
; 𝑂 being
the equilibrium position. i.e. the motion of M is the
S.H.M.
• To test the statement above, let us find the displacement
of M (=y).

21. Analysis…
• At this instant, the expression of
displacement of M is as shown below;
sin 𝜃 =
𝑂𝑀
𝑂𝑃
𝑂𝑀 = 𝑂𝑃 sin 𝜃
• But OM = y; OP= a; 𝜃 = 𝜔𝑡
∴ 𝑦 = 𝑎sin 𝑤𝑡

22. General Initial Conditions for one
dimensional S.H.M
• Suppose the initial time (𝑡 = 0) is when the object is at P,
with an initial angular position 𝜙. At time t it is at s,
having turned through an additional angle wt.
• From the basic trigonometry, an expression for the
displacement x at time t in terms of 𝒂, wt and 𝜙 is; 𝜙 is sometime
called Phase
constant of
the motion.
∴ 𝑥 𝑡 = 𝑎 sin
2𝜋𝑡
𝑇
+ 𝜙 = 𝑎 sin 𝜔𝑡 + 𝜙

23. Velocity in S.H.M
• If the displacement 𝑥 at time t is given by the function 𝑥(𝑡)
then the gradient of the displacement-time graph at the time t
is given by 𝑑𝑥(𝑡)
𝑑𝑡 and this is also the instantaneous
velocity 𝑣 𝑥(𝑡) of the object at that time.
∴ 𝑣 𝑥 𝑡 =
𝑑𝑥
𝑑𝑡
(𝑡) …..(i)

24. Velocity…
• Since, 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , substitute this equation into
equation (i);
𝑣 𝑥 𝑡 = 𝑎
𝑑
𝑑𝑡
sin 𝜔𝑡 + 𝜙
∴ 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙

25. The speed of the particle executing S.H.M
as a function displacement.
• From the displacement equation, 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , we have;
𝑥2
𝑎2 = 𝑠𝑖𝑛2
𝜔𝑡 + 𝜙 ……...(i)
• Also, from velocity equation, 𝒗 𝒙 𝒕 = 𝒂𝝎 𝒄𝒐𝒔 𝝎𝒕 + 𝝓 ,we
have;
𝑣2
𝑎2 𝜔2 = 𝑐𝑜𝑠2
𝜔𝑡 + 𝜙 ……(ii)

26. The speed of the particle…
• Substitute equation (i) and (ii) into trigonometric
identity; 𝑠𝑖𝑛2
𝜃 + 𝑐𝑜𝑠2
𝜃 = 1; then we have;
𝑥2
𝑎2
+
𝑣2
𝑎2 𝜔2
= 1
∴ 𝑣 = 𝜔 𝑎2 − 𝑥2

27. Acceleration in S.H.M
• Note that; the gradient of the velocity-time graph at time t is
given by 𝑑𝑣 𝑥(𝑡)
𝑑𝑡 and is also the instantaneous acceleration
𝐴 𝑥(𝑡) of the object at time t.
∴ 𝐴 𝑥 𝑡 =
𝑑𝑣 𝑥
𝑑𝑡
𝑡 =
𝑑2 𝑥
𝑑𝑡2 (𝑡) …..(ii)

28. Acceleration…
• Since, 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙 , substitute this equation into equation (ii);
𝐴 𝑥 𝑡 = 𝑎𝜔
𝑑
𝑑𝑡
cos 𝜔𝑡 + 𝜙
𝐴 𝑥 𝑡 = −𝑎𝜔2 sin 𝜔𝑡 + 𝜙
but 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙
∴ 𝐴 𝑥 𝑡 = −𝜔2
𝑥

29. In our derivation of equation of 𝒗 𝒙 𝒕 and 𝑨 𝒙 𝒕 we have
arbitrary chosen to express the displacement by the sine
function.
If we could use cosine function representation, we could have
produce the following equivalent expression;
𝑥 𝑡 = a cos(𝜔𝑡 + 𝜙)
𝑣 𝑥 𝑡 = −𝑎𝜔 sin(𝜔𝑡 + 𝜙)
𝐴 𝑥 𝑡 = −𝑎𝜔2 cos 𝜔𝑡 + 𝜙 = −𝜔2 𝑥

30. Graph Representation of S.H.M
• Graph representation of the displacement,
velocity and acceleration for an object in one
dimensional S.H.M as represented by equations
𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙
and 𝐴 𝑥 𝑡 = −𝜔2 𝑥 respectively.
∴For the case 𝜙 = 0

31. Important Properties of a particle
executing S.H.M
1. The displacement, velocity and
acceleration all vary sinusoidal with
time but are not in phase.

32. Important Properties…
2. The acceleration of the
particle is proportional to the
displacement but in the opposite
direction.

33. Important Properties…
3. The Frequency and period of the motion are independent of
the amplitude of the motion.
4. The particle moves to and fro about the mean (equilibrium)
position in a straight line

34. Example 1
• A simple harmonic oscillation is represented by;
𝑥 = 0.34 𝑐𝑜𝑠(3000𝑡 + 0.74)
Where x and t are in mm and sec respectively. Determine;
i. The amplitude
ii. The frequency and the angular frequency
iii. The time period.

35. Solution
i. Amplitude;
∴ 𝑎 = 0.34 𝑚𝑚
ii. A. Angular frequency;
∴ 𝜔 = 3000 𝐻𝑧
ii. B. Frequency;
𝑓 =
𝜔
2𝜋
=
3000
2𝜋
∴ 𝑓 =
1500
𝜋
Hz
iii. Time period;
𝑇 =
1
𝑓
∴ 𝑇 =
𝜋
1500
sec

36. Example 2
•A particle executes S.H.M of amplitude 25 cm and
time period 3 s. What is the minimum time
required for the particle to move between two
points 12.5 cm on either side of the mean
position?

37. Solution
• 𝑦 = 𝑎 sin
2𝜋
𝑇
𝑡
Here 𝑦 = 12.5 𝑐𝑚; 𝑎 = 25 𝑐𝑚; 𝑇 = 3 𝑠
12.5 = 25 sin
2𝜋
3
𝑡
∴ 𝒕 = 𝟏𝟒. 𝟑𝟐 𝒔

38. Example 3
• The amplitude of a particle executing S.H.M with a
frequency of 60 Hz is 0.01 m. Determine the maximum value
of he acceleration of the particle.

39. Solution
• The maximum acceleration of a particle executing S.H.M is
given by;
𝐴 𝑚𝑎𝑥 = 𝜔2
𝑎
𝐴 𝑚𝑎𝑥 = (2𝜋𝑓)2
𝑎
𝐴 𝑚𝑎𝑥 = (2𝜋 𝑥 60)2
𝑥 0.01
∴ 𝑨 𝒎𝒂𝒙 = 𝟏𝟒𝟐𝟏. 𝟐𝟐𝟑 𝒎 𝒔−𝟐

40. Example 4
•A particle executing S.H.M has a maximum
displacement of 4 cm and its acceleration at a
distance of 1 cm from its mean position is 3
𝒄𝒎 𝒔−𝟐
. What will be its velocity when it is at a
distance of 2 cm from its mean position?

41. Solution
• 𝐴 𝑥 𝑡 = −𝜔2
𝑥
𝜔 =
𝐴
𝑥
=
3
1
𝝎 = 𝟏. 𝟕𝟑𝟐 𝒓𝒂𝒅 𝒔−𝟏
• 𝑣 = 𝜔 𝑎2 − 𝑥2
𝑣 = 1.732 𝑥 42 − 22
∴ 𝒗 = 𝟔 𝒎 𝒔−𝟏

42. Forces acting in S.H.M
• From Newton’s second law of motion,
𝑭 = 𝒎𝑨
The acceleration expressed in terms of displacement of the particle
executing S.H.M is;
𝑨 𝒙 𝒕 = −𝝎 𝟐
𝒙
Substitute the acceleration equation into the equation force then we have;
∴ 𝑭 = −𝒎𝝎 𝟐
𝒙

43. Forces acting in S.H.M
• In S.H.M the magnitude of the force acting at any time is linearly
proportional to the distance from the equilibrium at that time and
the direction of the force is opposite to that of the displacement.
Thus, the force causing S.H.M always acts in a direction that
tends to reduce the displacement and for this reason it is usually
called restoring force.

44. Hooke’s Law
Law: The force exerted on a body by an external source
is directly proportional to the mean displacement the
object has displaced from its mean position.
𝐹 𝛼 − 𝑥
𝐹 = −𝑘𝑥

45. Force Law for S.H.M
• For a spring constant here being;
𝑘 = 𝑚𝜔2
The angular frequency 𝜔 of the S.H.M of the block is related
to 𝑘 and m of the block which yield;
𝜔 =
𝑘
𝑚

46. • The period of the S.H.M of the block is also related to 𝑘 and
m of the block which is;
𝑇 = 2𝜋
𝑚
𝑘

47. Example
A block whose mass m is 680 g is fastened to a spring whose spring
constant k is 65 N/m. The block is pulled a distance x = 11 cm from its
equilibrium position at x = 0 on a frictionless surface and released from
rest at t = 0.
(a) What are the angular frequency, the frequency, and the period of
the resulting motion?
(b) What is the maximum acceleration of the oscillating block.

48. Solution
(a) The angular frequency is;
𝜔 =
𝑘
𝑚
𝜔 =
65 N/m
0.68 𝐾𝑔
∴ 𝝎 = 𝟗. 𝟕𝟖 𝒓𝒂𝒅/𝒔
(a) The period is;
𝑇 = 2𝜋
𝑚
𝑘
𝑇 = 2𝜋
0.68 𝐾𝑔
65 N/m
∴ 𝑻 = 𝟎. 𝟔𝟒𝟐𝟕𝒔

49. Solution…
(a) The frequency is;
𝑓 =
𝜔
2𝜋
=
1
𝑇
𝑓 =
1
0.6427𝑠
𝒇 = 𝟏. 𝟓𝟓𝟔 𝑯𝒛
(b) Maximum acceleration;
𝑎 𝑚𝑎𝑥 = 𝜔2
𝑥 𝑚𝑎𝑥
𝑎 𝑚𝑎𝑥 = 9.78 𝑟𝑎𝑑/𝑠 2
𝑥 0.11 𝑚
𝒂 𝒎𝒂𝒙 = 𝟏𝟎. 𝟓𝟐𝟏 𝒎𝒔−𝟐

50. Springs in Parallel
• The 𝐹𝑡𝑜𝑡 puling the mass m
at distance x is distributed
among the three springs as
shown on the equation
below;
𝐹𝑡𝑜𝑡 = 𝐹1 + 𝐹2 + 𝐹3
i.e. displacement of each spring is
equal to x
𝐹𝑡𝑜𝑡 = 𝑘1 𝑥 + 𝑘2 𝑥 + 𝑘3 𝑥

51. Spring in parallel…
• An equivalent spring constant
𝑘 𝐸 for springs connected in
parallel is given by;
𝑘 𝐸 = 𝑘1 + 𝑘2 + 𝑘3
Hooke’s law for the combination
of springs is given by;
𝐹𝑡𝑜𝑡 = 𝑘 𝐸 𝑥
• The springs in parallel
will execute a S.H.M
whose period is given by;
𝑇 = 2𝜋
𝑚
𝑘 𝐸

52. Example
Three springs with force constants 𝑘1= 10.0 N/m, 𝑘2=
12.5 N/m, and 𝑘3= 15.0 N/m are connected in parallel to
a mass of 0.500 kg. The mass is then pulled to the right
and released. Find the period of the motion.

53. Solution
• The period of the motion is;
𝑇 = 2𝜋
𝑚
𝑘 𝐸
𝑇 = 2𝜋
𝑚
𝑘1 + 𝑘2 + 𝑘3
𝑇 = 2𝜋
0.5
10 + 12.5 + 15
∴ 𝑻 = 𝟎. 𝟕𝟐𝟓𝟓 𝒔

54. Springs in Series
• The total displacement x of the
spring connected in series is
the sum of the displacements
of each spring that is;
𝒙 = 𝒙 𝟏 + 𝒙 𝟐 + 𝒙 𝟑
i.e. Force in each spring is
transmitted equally;
𝐹𝑡𝑜𝑡 = 𝐹1 = 𝐹2 = 𝐹3

55. Spring Series….
• Since, 𝒙 = 𝒙 𝟏 + 𝒙 𝟐 + 𝒙 𝟑
• Then, 𝑥 = 𝐹𝑡𝑜𝑡
1
𝑘1
+
1
𝑘2
+
1
𝑘3
• Therefore; equivalent spring
constant for spring connected
in series is;
1
𝑘 𝐸
=
1
𝑘1
+
1
𝑘2
+
1
𝑘3
• The springs in series will execute
a S.H.M whose period is given
by;
𝑇 = 2𝜋
𝑚
𝑘1 𝑘2 𝑘3
𝑘2 𝑘3 + 𝑘1 𝑘3 + 𝑘1 𝑘2

56. Example
Three springs with force constants 𝑘1= 10.0 N/m, 𝑘2=
12.5 N/m, and 𝑘3= 15.0 N/m are connected in series to a
mass of 0.500 kg. The mass is then pulled to the right
and released. Find the period of the motion.

57. Solution
• The period of the motion is
𝑇 = 2𝜋
𝑚
𝑘1 𝑘2 𝑘3
𝑘2 𝑘3 + 𝑘1 𝑘3 + 𝑘1 𝑘2
𝑇 = 2𝜋
0.5
10𝑥12.5𝑥15
12.5𝑥15 + 10𝑥15 + 10𝑥12.5
∴ 𝑇 = 2.207 𝑠

58. Conservation of Energy
The vibrating spring system can also be
described in terms of the law of conservation
of energy.
 When the spring is stretched to its maximum
displacement a, work is done on the spring, and
hence the spring contains potential energy.
The kinetic energy is equal to zero at this point
because v = 0 at the maximum displacement.
The total energy of the system is thus;
𝐸𝑡𝑜𝑡 = 𝑃𝐸 + 𝐾𝐸

59. Kinetic Energy (KE)
From KE equation, 𝐾𝐸 = 1
2 𝑚𝑣2
……….(i)
Then, substitute equation; 𝑣 = 𝑎𝜔 cos(𝜔𝑡 + 𝜙) into equation (i),
then we have;
𝐾𝐸 = 1
2 𝑎2
𝑚𝜔2
𝑐𝑜𝑠2
𝜔𝑡 + 𝜙
But 𝐹 = −𝑘𝑥 = 𝑚𝑎 = −𝑚𝜔2
𝑥; 𝝎 𝟐
= 𝒌
𝒎
∴ 𝐾𝐸 = 1
2 𝑎2
𝑘 𝑐𝑜𝑠2
𝜔𝑡 + 𝜙

60. Potential Energy (PE)
• PE of the system is given by the amount of work required to move system
from position 0 𝑡𝑜 𝑥 under the position of applied force (𝑘𝑥).
• Work done of the particle executing S.H.M; 𝑊 = −𝐹𝑑𝑥
𝑊 = 𝑘𝑥𝑑𝑥
0
𝑤
𝑊 = 𝑘
0
𝑥
𝑥 𝑑𝑥
𝑊 = 1
2 𝑘𝑥2

61. PE…
• From equation; 𝑥 = 𝑎 sin 𝜔𝑡 + 𝜙 substitute into equation
𝑊 = 1
2 𝑘𝑥2
𝑃𝐸 = 𝑊 = 1
2 𝑘𝑎2
𝑠𝑖𝑛2
(𝜔𝑡 + 𝜙)
∴ 𝑃𝐸 = 1
2 𝑘𝑎2
𝑠𝑖𝑛2
(𝜔𝑡 + 𝜙)

62. Total Mechanical Energy
• Total mechanical energy of the system;
𝐸𝑡𝑜𝑡 = 𝑃𝐸 + 𝐾𝐸
• But,
𝐾𝐸 = 1
2 𝑎2
𝑘 𝑐𝑜𝑠2
𝜔𝑡 + 𝜙
𝑃𝐸 = 1
2 𝑘𝑎2
𝑠𝑖𝑛2
(𝜔𝑡 + 𝜙)
Then;
𝐸𝑡𝑜𝑡 = 1
2 𝑘𝑎2
𝑠𝑖𝑛2
𝜔𝑡 + 𝜙 + 𝑐𝑜𝑠2
𝜔𝑡 + 𝜙

63. Total Mechanical Energy of the System…
From trigonometric identity;
𝑠𝑖𝑛2
𝜃 + 𝑐𝑜𝑠2
𝜃 = 1
Then;
𝐸𝑡𝑜𝑡 = 1
2 𝑘𝑎2
𝑠𝑖𝑛2
𝜔𝑡 + 𝜙 + 𝑐𝑜𝑠2
𝜔𝑡 + 𝜙
𝐸𝑡𝑜𝑡 = 1
2 𝑘𝑎2
= 1

64. Total Mechanical Energy of the System…
• Thus 𝐸𝑡𝑜𝑡 of the oscillator remains constant
as 𝑥 is regained after every half cycle.
• If no energy is dissipated then all the PE
becomes KE and vice versa.
• Figure on the right side shows the variation
of KE and PE of harmonic oscillator with
time where phase φ is set to zero for
simplicity.

65. Example
• Conservation of energy applied to a spring. A horizontal spring has a
spring constant of 29.4 N/m. A mass of 300 g is attached to the spring
and displaced 5.00 cm. The mass is then released. Find
(a) the total energy of the system,
(b)the maximum velocity of the system, and
(c) the potential energy and kinetic energy for x = 2.00 cm.

66. Solution…
(a) The total Energy of the system(𝑬 𝒕𝒐𝒕) is;
𝐸𝑡𝑜𝑡 = 1
2 𝑘𝑎2
𝐸𝑡𝑜𝑡 = 1
2 𝑥 29.4 𝑥 (0.05)2
∴ 𝑬 𝒕𝒐𝒕 = 𝟑. 𝟔𝟕𝟓 𝒙 𝟏𝟎−𝟐
𝑱

67. Solution…
(b). The maximum velocity occurs when x = 0 and the potential energy
is zero;
𝐸𝑡𝑜𝑡 = 𝐾𝐸 𝑚𝑎𝑥 = 1
2 𝑚𝑣 𝑚𝑎𝑥
2
𝑣 𝑚𝑎𝑥=
2𝐸𝑡𝑜𝑡
𝑚
1
2
𝑣 𝑚𝑎𝑥=
2 𝑥 3.675 𝑥 10−2
0.3
1
2
∴ 𝑣 𝑚𝑎𝑥 = 0.495 𝑚𝑠−1

68. Solution…
(c)The potential energy at 2.00 cm
is;
𝑃𝐸 = 1
2 𝑘𝑥2
=1
2 𝑥 29.4 𝑥 0.02 2
∴ 𝑃𝐸 𝑎𝑡 2.0𝑐𝑚 = 5.88 x 10−3
𝐽
• The kinetic energy at 2.00 cm
is;
K𝐸 = 𝐸𝑡𝑜𝑡 − 𝑃𝐸 𝑎𝑡 2.00𝑐𝑚
K𝐸 = 3.675 𝑥 10−2 𝐽 − 5.88 𝑥 10−3 𝐽
∴ 𝐾𝐸 𝑎𝑡 2.0𝑐𝑚 = 0.03087 𝐽