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PHYSICS 025 CHAPTER 9 1 CHAPTER 9: Simple Harmonic Motion (6 Hours)
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Page 1: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

1

CHAPTER 9: Simple Harmonic Motion

(6 Hours)

Page 2: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

2

At the end of this chapter,students should be able to: a) Explain SHM as periodic

motion without loss of energy.

b) Introduce and use SHM formulae:

Learning Outcome:9.1 Simple harmonic motion (1 hour)

xdt

xda 2

2

2

REMARKS :

Examples : simple pendulum, frictionless horizontal and vertical spring oscillations.

Page 3: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

3

9.1 Simple harmonic motion

9.1.1 Simple harmonic motion (SHM) is defined as the periodic motion without loss of energy in

which the acceleration of a body is directly proportional to its displacement from the equilibrium position (fixed point) and is directed towards the equilibrium position but in opposite direction of the displacement.

OR mathematically,

body theofon accelerati : a

xa 2

where

Oposition, mequilibriu thefromnt displaceme : xfrequency)ular locity(angangular ve : ω

(9.1)2

2

dt

xd

Page 4: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

4

The angular frequency, always constant thus

The negative sign in the equation 9.1 indicates that the

direction of the acceleration, a is always opposite to the

direction of the displacement, x. The equilibrium position is a position at which the body would

come to rest if it were to lose all of its energy. When the body in SHM is at the point of equilibrium. displacement x = 0 acceleration a = 0 resultant force on the body F = 0

xa

Page 5: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

5

o The displacement x is measured from the point of equilibrium. The maximum distance from the point of equilibrium is known as the amplitude of the simple harmonic motion.

o The period T of a simple harmonic is the time taken for a complete oscillation.

o The frequency f of the motion is the number of complete oscillations per second.

The unit of frequency is the hertz (Hz).

1,frequency f

period

Page 6: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

6

Equation 9.1 is the hallmark of the linear SHM. Examples of linear SHM system are simple pendulum,

horizontal and vertical spring oscillations as shown in Figures 9.1a, 9.1b and 9.1c.

m

a

O xx

sF

Figure 9.1a

Page 7: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

7

Figure 9.1b

x

x a

sF

O

m

m

O xx

a

sF

Figure 9.1c

Page 8: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

ExerciseA particle in simple harmonic motion makes 20 complete oscillations in 5.0 s. what isa. the frequency

b. the periodof the motion

8

Page 9: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

Solution:a. frequency, f =

b. period, T =

9

204.0

5.0Hz Hz

1 10.25

4.0s s

f

Page 10: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

10

At the end of this chapter, students should be able to: a) Write and use SHM displacement equation

b) Derive and apply equations for :

i) velocity,

ii) acceleration,

iii) kinetic energy,

iv) potential energy,

Learning Outcome:9.2 Kinematics of SHM (2 hours)

tAx sin

22

2

1xmU

222

2

1xAmK

xdt

xd

dt

dva 2

2

2

22 xAdt

dxv

Page 11: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

11

9.2 Kinematics of SHM9.2.1 Displacement, x Uniform circular motion can be translated into linear SHM and

obtained a sinusoidal curve for displacement, x against angular

displacement, graph as shown in Figure 9.6.

Figure 9.6

S

T

MN

O

x

)rad(02

2

3 2

A

A1x

11

A

P

Simulation 9.3

Page 12: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

12

At time, t = 0 the object is at point M (Figure 9.6) and after time t it moves to point N , therefore the expression for displacement, x1 is given by

In general the equation of displacement as a function of time in SHM is given by

The S.I. unit of displacement is metre (m).Phase It is the time-varying quantity . Its unit is radian.

111 sinAx where and t tAx sin1

tAx sin

angular frequency

amplitude

displacement from equilibrium position

time

Initial phase angle (phase constant)

phase

(9.4)

t

Page 13: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

13

Initial phase angle (phase constant), It is indicate the starting point in SHM where the time, t = 0 s. If =0 , the equation (9.4) can be written as

where the starting point of SHM is at the equilibrium position, O.

For examples:

a. At t = 0 s, x = +A

Equation :

tAx sin

AA O

tAx sin 0sinAA

rad 2

2sin

tAx OR tAx cos

Page 14: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

14

b. At t = 0 s, x = A

Equation :

c. At t = 0 s, x = 0, but v = vmax

Equation :

tAx sin 0sinAA

rad 2

3

2

3sin

tAx

AA Orad

2

OR

OR

2sin

tAx

OR tAx cos

AA O

maxv

tAx sin OR tAx sin

Page 15: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

15

From the definition of instantaneous velocity,

Eq. (9.5) is an equation of velocity as a function of time in SHM.

The maximum velocity, vmax occurs when cos(t+)=1 hence

9.2.2 Velocity, v

dt

dxv and tAx sin

)sin( tAdt

dv

)sin( tdt

dAv

)cos( tAv (9.5)

Av max (9.6)

Page 16: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

16

The S.I. unit of velocity in SHM is m s1. If = 0 , equation (9.5) becomes

Relationship between velocity, v and displacement, x From the eq. (9.5) :

From the eq. (9.4) :

From the trigonometry identical,

tAv cos

A

xt sin

1cossin 22

tAx sin

)cos( tAv (1)

(2)

and t

tt 2sin1cos (3)

Page 17: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

17

By substituting equations (3) and (2) into equation (1), thus

2

1

A

xAv

2

222

A

xAAv

22 xAv (9.7)

Page 18: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

18

From the definition of instantaneous acceleration,

Eq. (9.8) is an equation of acceleration as a function of time in SHM.

The maximum acceleration, amax occurs when sin(t+)=1

hence

9.2.3 Acceleration, a

dt

dva and tAv cos

)cos( tAdt

da

)cos( tdt

dAa

)sin(2 tAa (9.8)

2max Aa (9.9)

Page 19: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

19

The S.I. unit of acceleration in SHM is m s2. If = 0 , equation (9.8) becomes

Relationship between acceleration, a and displacement, x From the eq. (9.8) :

From the eq. (9.4) :

By substituting eq. (2) into eq. (1), therefore

tAa sin2

tAx sin

)sin(2 tAa (1)

xa 2

(2)

Page 20: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

20

Caution : Some of the reference books use other general equation for

displacement in SHM such as

The equation of velocity in term of time, t becomes

And the equation of acceleration in term of time, t becomes

)sin( tAdt

dxv

)cos(2 tAdt

dva

tAx cos (9.10)

(9.11)

(9.12)

Page 21: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

21

Potential energy, U Consider the oscillation of a spring as a SHM hence the

potential energy for the spring is given by

The potential energy in term of time, t is given by

9.2.4 Energy in SHM

2

2

1kxU and

2mk

22

2

1xmU (9.13)

22

2

1xmU tAx sinand

tAmU 222 sin2

1(9.14)

Page 22: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

22

Kinetic energy, K The kinetic energy of the object in SHM is given by

The kinetic energy in term of time, t is given by

2

2

1mvK and

22 xAv

222

2

1xAmK (9.15)

2

2

1mvK tAv cosand

tAmK 222 cos2

1(9.16)

Page 23: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

23

Total energy, E The total energy of a body in SHM is the sum of its kinetic

energy, K and its potential energy, U .

From the principle of conservation of energy, this total energy is always constant in a closed system hence

The equation of total energy in SHM is given by

UKE

22222

2

1

2

1xmxAmE

22

2

1AmE

OR 2

2

1kAE

constant UKE

(9.17)

(9.18)

Simulation 9.4

Page 24: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

24

An object executes SHM whose displacement x varies with time t according to the relation

where x is in centimetres and t is in seconds.

Determine

a. the amplitude, frequency, period and phase constant of the

motion,

b. the velocity and acceleration of the object at any time, t ,c. the displacement, velocity and acceleration of the object at

t = 2.00 s,

d. the maximum speed and maximum acceleration of the object.

Example 3 :

22sin00.5

tx

Page 25: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

25

Solution :

a. By comparing

thus

i.

ii.

iii. The period of the motion is

iv. The phase constant is

tAx sinwith

22sin00.5

tx

1s rad 2 cm 00.5A

and f 2 22 f

Hz 00.1f

T

100.1

Tf

1

s 00.1T

rad 2

Page 26: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

26

Solution :

b. i. Differentiating x respect to time, thus

22sin00.5

tdt

d

dt

dxv

seconds.in is and s cmin is where 1 tv

22cos200.5

tv

22cos0.10

tv

Page 27: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

27

Solution :

b. ii. Differentiating v respect to time, thus

22cos0.10

tdt

d

dt

dva

seconds.in is and s cmin is where 2 ta

22sin20.10

ta

22sin0.20 2 ta

Page 28: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

28

Solution :

c. For t = 2.00 s

i. The displacement of the object is

ii. The velocity of the object is

OR

200.22sin00.5

x

cm 00.5x

200.22cos0.10

v1s cm 00.0 v22 xAv

22 00.500.52 v1s cm 00.0 v

Page 29: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

29

Solution :

c. For t = 2.00 s

iii. The acceleration of the object is

OR

222 s cm 197s cm 0.20 a

xa 2 00.52 2 a

200.22sin0.20 2 a

222 s cm 197s cm 0.20 a

Page 30: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

30

Solution :

d. i. The maximum speed of the object is given by

ii. The maximum acceleration of the object is

Av max

1max s cm 0.10 v

200.5max v

Aa 2max

22max s cm 0.20 a

00.52 2max a

Page 31: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

31

The length of a simple pendulum is 75.0 cm and it is released at an angle 8 to the vertical. Calculate

a. the frequency of the oscillation,

b. the pendulum’s bob speed when it passes through the lowest

point of the swing.

(Given g = 9.81 m s2)

Solution :

Example 4 :

A

8

m

L

AO

A

Page 32: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

32

Solution :

a. The frequency of the simple pendulum oscillation is

b. At the lowest point, the velocity of the pendulum’s bob is

maximum hence

8 m; 75.0 L

Tf

1

g

LT 2

Hz 576.0f

and

L

gf

2

1

75.0

81.9

2

1

f

fLv 28sinmax

8sinLA andAv max

576.028sin75.0max v1

max s m 378.0 v

Page 33: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

33

A body hanging from one end of a vertical spring performs vertical SHM. The distance between two points, at which the speed of the body is zero is 7.5 cm. If the time taken for the body to move between the two points is 0.17 s, Determine

a. the amplitude of the motion,

b. the frequency of the motion,

c. the maximum acceleration of body in the motion.

Solution :

a. The amplitude is

b. The period of the motion is

Example 5 :

A

A

Om

s 17.0tcm .57

m10 75.32

105.7 22

A

17.022 tTs 34.0T

Page 34: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

34

Solution :

b. Therefore the frequency of the motion is

c. From the equation of the maximum acceleration in SHM, hence

34.0

11

Tf

Hz 94.2f

2max 2 fAa

f 2and2

max Aa

22max 94.221075.3 a

2max s m 8.12 a

Page 35: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

35

An object of mass 450 g oscillates from a vertically hanging light spring once every 0.55 s. The oscillation of the mass-spring is started by being compressed 10 cm from the equilibrium position and released.

a. Write down the equation giving the object’s displacement as a

function of time.

b. How long will the object take to get to the equilibrium position

for the first time?

c. Calculate

i. the maximum speed of the object,

ii. the maximum acceleration of the object.

Example 6 :

Page 36: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

36

Solution :

a. The amplitude of the motion is

The angular frequency of the oscillation is

and the initial phase angle is given by

Therefore the equation of the displacement as a function of time is

s 55.0 kg; 450.0 Tm

cm 10

0

m

cm 10

0t55.0

22 T

1s rad 4.11

cm 10A

tAx sin 0sinAA

rad 2

tAx sin

24.11sin10

tx

seconds.in is and cmin is where tx

OR tx 4.11cos10

Page 37: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

37

Solution :

b. At the equilibrium position, x = 0

s 55.0 kg; 450.0 Tm

24.11sin10

tx

s 138.0t

24.11sin100

t

0sin2

4.11 1

t

24.11 t

4

55.0

4

TtOR

s 138.0t

Page 38: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

38

Solution :

c. i. The maximum speed of the object is

ii. The maximum acceleration of the object is

s 55.0 kg; 450.0 Tm

Av max

4.111.0max v1

max s m 14.1 v

2max Aa

2max 4.111.0a

2max s m 0.13 a

Page 39: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

39

An object of mass 50.0 g is connected to a spring with a force constant of 35.0 N m-1 oscillates on a horizontal frictionless surface with an amplitude of 4.00 cm. Determine

a. the total energy of the system,

b. the speed of the object when the position is 1.00 cm,

c. the kinetic and potential energy when the position is 3.00 cm.

Solution :

a. By applying the equation of the total energy in SHM, thus

Example 7 :

m 1000.4;m N 0.35 kg; 100.50 213 Akm

2

2

1kAE

221000.40.352

1 E

J 1080.2 2E

Page 40: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

40

Solution :

b. The speed of the object when x = 1.00 102 m

22 xAv

22 xAm

kv

m 1000.4;m N 0.35 kg; 100.50 213 Akm

andm

k

22223

1000.11000.4100.50

0.35v

1s m 03.1 v

Page 41: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

41

Solution :

c. The kinetic energy of the object when x = 3.00 102 m is

and the potential energy of the object when x = 3.00 102 m is

222

2

1xAmK

22

2

1xAkK

m 1000.4;m N 0.35 kg; 100.50 213 Akm

and2mk

2222 1000.31000.40.352

1 K

J 1023.1 2K

2

2

1kxU

J 1058.1 2U

221000.30.352

1 U

Page 42: Physics Chapter 9-Simple Harmonic Motion

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42

Exercise 9.1 :1. A mass which hangs from the end of a vertical helical spring is

in SHM of amplitude 2.0 cm. If three complete oscillations take 4.0 s, determine the acceleration of the mass

a. at the equilibrium position,

b. when the displacement is maximum.

ANS. : U think ; 44.4 cm s2

2. A body of mass 2.0 kg moves in simple harmonic motion. The

displacement x from the equilibrium position at time t is given by

where x is in metres and t is in seconds. Determine

a. the amplitude, period and phase angle of the SHM.

b. the maximum acceleration of the motion.

c. the kinetic energy of the body at time t = 5 s.

ANS. : 6.0 m, 1.0 s, ; 24.02 m s2; 355 J

62sin0.6

tx

rad 3

Page 43: Physics Chapter 9-Simple Harmonic Motion

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43

3. A horizontal plate is vibrating vertically with SHM at a frequency of 20 Hz. What is the amplitude of vibration so that the fine sand on the plate always remain in contact with it?

ANS. : 6.21104 m4. An object of mass 2.1 kg is executing simple harmonic motion,

attached to a spring with spring constant k = 280 N m1. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m s1. Calculatea. the amplitude of the motion.b. the maximum velocity attained by the object.

ANS. : 5.17102 m; 0.597 m s1

5. A simple harmonic oscillator has a total energy of E.a. Determine the kinetic energy and potential energy when the displacement is one half the amplitude.b. For what value of the displacement does the kinetic energy equal to the potential energy?

ANS. : AEE2

2 ;

4

1 ,

4

3

Page 44: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

44

At the end of this chapter, students should be able to: Sketch, interpret and distinguish the following graphs:

displacement - time velocity - time acceleration - time energy - displacement

Learning Outcome:

9.3 Graphs of SHM (2 hours)

Page 45: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

45

9.3 Graphs of SHM

9.3.1 Graph of displacement-time (x-t) From the general equation of displacement as a function of time

in SHM,

If = 0 , thus The displacement-time graph is shown in Figure 9.7.

tAx sin

tAx sin

Amplitude

Figure 9.7

x

t0

4

T T

A

A

2

T

4

3T

Period

Simulation 9.5

Page 46: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

46

For examples:

a. At t = 0 s, x = +A

Equation:

Graph of x against t:

2sin

tAx OR tAx cos

x

t0

4

T T

A

A

2

T

4

3T

Page 47: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

47

b. At t = 0 s, x = A Equation:

Graph of x against t:x

t0

4

T T

A

A

2

T

4

3T

2

3sin

tAx OR

2sin

tAx

OR tAx cos

Page 48: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

48

c. At t = 0 s, x = 0, but v = vmax

Equation:

Graph of x against t:

tAx sin OR tAx sin

x

t0

4

T T

A

A

2

T

4

3T

Page 49: Physics Chapter 9-Simple Harmonic Motion

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49

How to sketch the x against t graph when 0Sketch the x against t graph for the following expression:

From the expression, the amplitude, the angular frequency,

Sketch the x against t graph for equation

23sincm 5

πtx

cm 5A

T

2s rad 3 1 s

3

2T

tx 3sin5(cm)x

)s(t0

5

5

3

2

3

1

4

T

Page 50: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

50

Because of

Sketch the new graph.

4rad

2

Tt

4

Thence shift the y-axis to the left by

(cm)x

)s(t0

5

5

3

2

3

1

If = negative value

shift the y-axis to the left

If = positive value

shift the y-axis to the right

RULES

Page 51: Physics Chapter 9-Simple Harmonic Motion

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51

From the general equation of velocity as a function of time in SHM,

If = 0 , thus The velocity-time graph is shown in Figure 9.8.

9.3.2 Graph of velocity-time (v-t)

tAv cos

tAv cos

v

t0

4

T T

A

A

2

T

4

3T

Figure 9.8

Page 52: Physics Chapter 9-Simple Harmonic Motion

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52

From the relationship between velocity and displacement,

thus the graph of velocity against displacement (v-x) is shown in Figure 9.9.

22 xAv

v

x0

A

A

AA

Figure 9.9

Page 53: Physics Chapter 9-Simple Harmonic Motion

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53

From the general equation of acceleration as a function of time in SHM,

If = 0 , thus The acceleration-time graph is shown in Figure 9.10.

9.3.3 Graph of acceleration-time (a-t)

tAa sin2

tAa sin2

Figure 9.10

a

t0

4

T T

2A

2

T

4

3T

2A

Page 54: Physics Chapter 9-Simple Harmonic Motion

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54

From the relationship between acceleration and displacement,

thus the graph of acceleration against displacement (a-x) is shown in Figure 9.11.

The gradient of the a-x graph represents

xa 2

Figure 9.11

a

x0

2A

2A

AA

2,gradient m

Page 55: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

55

E

x

From the equations of kinetic, potential and total energies as a term of displacement

thus the graph of energy against displacement (a-x) is shown in Figure 9.12.

9.3.4 Graph of energy-displacement (E-x)

22

2

1xmU 222

2

1xAmK 22

2

1AmE and;

constant2

1 22 AmE

22

2

1xmU

222

2

1xAmK Figure 9.12

Page 56: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

56

Energy

t

22

2

1AmE

tAmU 222 sin2

1

tAmK 222 cos2

1

The graph of Energy against time (E-t) is shown in Figure 9.13.

Figure 9.13 Simulation 9.6

Page 57: Physics Chapter 9-Simple Harmonic Motion

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57

The displacement of an oscillating object as a function of time is shown in Figure 9.14.

From the graph above, determine for these oscillations

a. the amplitude, the period and the frequency,

b. the angular frequency,

c. the equation of displacement as a function of time,

d. the equation of velocity and acceleration as a function of time.

Example 8 :

)cm(x

s)(t0

0.15

0.15

8.0

Figure 9.14

Page 58: Physics Chapter 9-Simple Harmonic Motion

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58

Solution :

a. From the graph,

Amplitude,

Period,

Frequency,

b. The angular frequency of the oscillation is given by

c. From the graph, when t = 0, x = 0 thus

By applying the general equation of displacement in SHM

m 15.0As 8.0T

8.0

11

Tf

Hz 25.1f

8.0

22 T

1s rad 5.2 0

tAx sin tx 5.2sin15.0seconds.in is and metresin is where tx

Page 59: Physics Chapter 9-Simple Harmonic Motion

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59

Solution :

d. i. The equation of velocity as a function of time is

ii. and the equation of acceleration as a function of time is

tdt

d

dt

dxv 5.2sin15.0

seconds.in is and s min is where 1 tv

tv 5.2cos5.215.0tv 5.2cos375.0

tdt

d

dt

dva 5.2cos375.0

seconds.in is and s min is where 2 ta

ta 5.2sin5.2375.0ta 5.2sin938.0 2

Page 60: Physics Chapter 9-Simple Harmonic Motion

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60

Figure 9.15 shows the relationship between the acceleration a and

its displacement x from a fixed point for a body of mass 2.50 kg at which executes SHM. Determine

a. the amplitude,

b. the period,

c. the maximum speed of the body,

d. the total energy of the body.

Example 9 :

Figure 9.15

)s m( 2a

)cm(x0

80.0

80.0

00.400.4

Page 61: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

61

Solution :

a. The amplitude of the motion is

b. From the graph, the maximum acceleration is

By using the equation of maximum acceleration, thus

OR The gradient of the a-x graph is

m 1000.4 2A

2max Aa

2max s m 80.0 a

T

2

2

max

2

TAa

s 40.1T

and

2

2 21000.480.0

T

2

212

12

1000.40

80.00gradient

xx

yy

22

20

T

s 40.1T

kg 50.2m

Page 62: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

62

Solution :

c. By applying the equation of the maximum speed, thus

d. The total energy of the body is given by

Av max T

2

TAv

2max

1max s m 180.0 v

and

40.1

21000.4 2

max

v

22

2

1AmE

222

1000.440.1

250.2

2

1

E

J 1003.4 2E

kg 50.2m

Page 63: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

63

Figure 9.16 shows the displacement of an oscillating object of mass 1.30 kg varying with time. The energy of the oscillating object consists the kinetic and potential energies. Calculate

a. the angular frequency of the oscillation,

b. the sum of this two energy.

Example 10 :

Figure 9.16

)m(x

)s(t0

2.0

2.0

4 52 31

Page 64: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

64

Solution :

From the graph,

Amplitude,

Period,

a. The angular frequency is given by

b. The sum of the kinetic and potential energies is

m 2.0A

4

22 T

s 4T

1s rad 2

22

2

1AmE

22

2.02

30.12

1

E

J 1042.6 2E

kg 30.1m

Page 65: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

65

1x

2x

Considering two SHM with the following equations,

is defined as

For examples,

a.

9.3.5 Phase difference,

1111 sin tAx

1122 tt

2222 sin tAx

12 phasephase

x

t

A

0

A

T2

T

2sin2

tAx

tAx cos2 OR

tAx sin1

Page 66: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

66

1x

2x

Thus the phase difference is given by

If > 0 , hence

b.

rad 2

Δ

ttΔ

2

x2 leads the x1 by phase difference ½ rad and

constant with time.

A

0

A

T2

T

x

t

2sin2

tAx

tAx cos2 OR

tAx sin1

Page 67: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

67

1x

2x

Thus the phase difference is given by

If < 0 , hence

c.

rad 2

Δ

ttΔ

2

x2 lags behind the x1 by phase difference

½ rad and constant with time.

A

0

A

T2

T

x

t tAx sin2

tAx sin2 OR

tAx sin1

Page 68: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

68

1x

2x

Thus the phase difference is given by

If = ± , hence

d.

If = 0 , hence

rad Δ ttΔ

x2 is antiphase with the x1 and constant with

time.

A

0

A

T2

T

x

t

tAx sin2 tAx sin1

0Δ ttΔ

The phase difference is

x2 is in phase with the x1 and constant with

time.

Simulation 9.7

Simulation 9.8

Page 69: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

69

Figure 9.17 shows the variation of displacement, x with time, t for an object in SHM.

a. Determine the amplitude, period and frequency of the motion.

b. Another SHM leads the SHM above by phase difference of

0.5 radian where the amplitude and period of both SHM are

the same. On the same axes, sketch the displacement, x against

time, t graph for both SHM.

Example 11 :

Figure 9.17

)cm(x

)s(t0

4

4

0.1 0.2 0.3

Page 70: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

70

Solution :

a. From the graph,

Amplitude,

Period,

The frequency is given by

b. Equation for 1st SHM (from the graph):

cm 4A

0.2

11

Tf

s 0.2T

Hz 5.0f

tAx sin1

ftAx 2sin1

2sin41

tx

Page 71: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

71

1x

2x

Solution :

b. The 2nd SHM leads the 1st SHM by the phase difference of 0.5 radian thus

Equation for 2nd SHM :

2

ttΔ

rad

tx sin42

rad 2

Δ

22

tt

)cm(x

)s(t0

4

4

0.1 0.2 0.3

Page 72: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

72A

maxa

max

AO

maxv

maxa

maxv

maxa

max

max

t x v a K USummary :

0

4

T

2

T

4

3T

T

A

0

A

0

A

0

A

0

A

0

2A

0

2A

0

2A

0

0

0

22

2

1 mA

22

2

1 mA

2

2

1kA

2

2

1kA

2

2

1kA

0

0

22 xAv xa 2

2

2

1mvK

2

2

1kxU

Page 73: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

Learning outcome9.4 Period of simple harmonic motion

Derive and use expression for period of SHM, T for simple

pendulum and single spring

Simple pendulum Simple

73

g

LT 2

k

mT 2

Page 74: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

74

Amplitude (A) is defined as the maximum magnitude of the displacement

from the equilibrium position. Its unit is metre (m).Period (T) is defined as the time taken for one cycle. Its unit is second (s). Equation :

Frequency (f) is defined as the number of cycles in one second. Its unit is hertz (Hz) :

1 Hz = 1 cycle s1 = 1 s1

Equation :

9.4.1 Terminology in SHM

2

f

fT

1

f 2 OR

Page 75: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

75

A. Simple pendulum oscillation Figure 9.2 shows the oscillation of the simple pendulum of

length, L.

9.4.2 System of SHM

mx

L

PO

T

cosmgsinmg Figure 9.2

gm

Page 76: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

76

A pendulum bob is pulled slightly to point P. The string makes an angle, to the vertical and the arc length,

x as shown in Figure 9.2. The forces act on the bob are mg, weight and T, the tension in

the string. Resolve the weight into

the tangential component : mg sin the radial component : mg cos

The resultant force in the radial direction provides the centripetal force which enables the bob to move along the arc and is given by

The restoring force, Fs contributed by the tangential

component of the weight pulls the bob back to equilibrium position. Thus

r

mvmgT

2

cos

sins mgF

Page 77: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

77

The negative sign shows that the restoring force, Fs is

always against the direction of increasing x. For small angle, ;

sin in radian arc length, x of the bob becomes straight line (shown in

Figure 9.3) then

L

x

L

xsin

thus

L

xmgFs

Figure 9.3

Page 78: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

78

By applying Newton’s second law of motion,

Thus

By comparing

Thus

sFmaF

L

mgxma

xL

ga

xa Simple pendulum executes linear SHM

L

g2

with xa 2xL

ga

andT

2

Page 79: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

79

Therefore

The conditions for the simple pendulum executes SHM are the angle, has to be small (less than 10). the string has to be inelastic and light. only the gravitational force and tension in the string acting

on the simple pendulum.

g

LT 2

pendulum simple theof period : Twhere

onaccelerati nalgravitatio : gstring theoflength : L

(9.2)

Simulation 9.1

Page 80: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

80

B. Spring-mass oscillation

Vertical spring oscillation

mm

1x

xO O

Figure 9.4a Figure 9.4b Figure 9.4c

F

gm

a

gm

1F

Page 81: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

81

Figure 9.4a shows a free light spring with spring constant, k hung vertically.

An object of mass, m is tied to the lower end of the spring as shown in Figure 9.4b. When the object achieves an equilibrium

condition, the spring is stretched by an amount x1 . Thus

The object is then pulled downwards to a distance, x and released as shown in Figure 9.4c. Hence

then

0F 0WF 01 Wkx

1kxW

maF maWF1 and xxkF 11

makxxxk 11

xm

ka

xa Vertical spring oscillation executes linear SHM

Page 82: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

82

Horizontal spring oscillation Figure 9.5 shows a spring is

initially stretched with a

displacement, x = A and then released.

According to Hooke’s law,

The mass accelerates toward

equilibrium position, x = 0 by

the restoring force, Fs hence

m

m

m

m

m

0s F

sF

sF

sF

0s F

0x Ax Ax

a

a

0t

4

Tt

2

Tt

4

T3t

Tt

a

kxF s

maF s

kxma

xm

ka

executes linear SHM

Then

Figure 9.5

xa

Page 83: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

83

By comparing

Thus

Therefore

The conditions for the spring-mass system executes SHM are The elastic limit of the spring is not exceeded when the

spring is being pulled. The spring is light and obeys Hooke’s law. No air resistance and surface friction.

with xa 2xm

ka

m

k2 and

T

2

k

mT 2 noscillatio spring theof period : T

where

object theof mass : mconstant) (forceconstant spring : k

Simulation 9.2

(9.3)

Page 84: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

84

A certain simple pendulum has a period on the Earth surface’s of 1.60 s. Determine the period of the simple pendulum on the surface of Mars where its gravitational acceleration is 3.71 m s2.

(Given the gravitational acceleration on the Earth’s surface is

g = 9.81 m s2)

Solution :

The period of simple pendulum on the Earth’s surface is

But its period on the surface of Mars is given by

Example 1 :

2M

2EE s m 71.3 ;s m 81.9 s; 60.1 ggT

EE 2

g

lT (1)

MM 2

g

lT (2)

Page 85: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

85

Solution :

By dividing eqs. (1) and (2), thus

2M

2EE s m 71.3 ;s m 81.9 s; 60.1 ggT

M

E

M

E

2

2

gl

gl

T

T

E

M

M

E

g

g

T

T

81.9

71.360.1

M

T

s 60.2M T

Page 86: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

86

A mass m at the end of a spring vibrates with a frequency of 0.88 Hz. When an additional mass of 1.25 kg is added to the mass

m, the frequency is 0.48 Hz. Calculate the value of m.

Solution :

The frequency of the spring is given by

After the additional mass is added to the m, the frequency of the spring becomes

Example 2 :

kg 25.1 Hz; 48.0 Hz; 88.0 21 Δmff

11

1

Tf and

k

mT 21

m

kf

2

11 (1)

Δmm

kf

2

12 (2)

Page 87: Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

87

Solution :

By dividing eqs. (1) and (2), thus

Δmmkmk

f

f

21

21

2

1

m

Δmm

f

f

2

1

m

m 25.1

48.0

88.0

kg 529.0m

kg 25.1 Hz; 48.0 Hz; 88.0 21 Δmff

Page 88: Physics Chapter 9-Simple Harmonic Motion

PHYSICS CHAPTER 9

88

THE END…Next Chapter…

CHAPTER 10 :Mechanical waves


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