PHYSICS CHAPTER 9

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CHAPTER 9: CHAPTER 9: Simple Harmonic MotionSimple Harmonic Motion

(5 Hours)(5 Hours)

PHYSICS CHAPTER 9

SIMPLE HARMONIC MOTIONSIMPLE HARMONIC MOTION 9.1 Simple Harmonic Motion (SHM)9.1 Simple Harmonic Motion (SHM) 9.2 Kinematics of Simple Harmonic Motion9.2 Kinematics of Simple Harmonic Motion 9.3 Graphs of Simple Harmonic Motion9.3 Graphs of Simple Harmonic Motion 9.4 Period of Simple Harmonic Motion9.4 Period of Simple Harmonic Motion

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PHYSICS CHAPTER 9

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain simple harmonic motion (SHM) as periodic simple harmonic motion (SHM) as periodic

motion without loss of energy.motion without loss of energy. Examples of linear SHM system are simple Examples of linear SHM system are simple

pendulum, frictionless horizontal and vertical spring pendulum, frictionless horizontal and vertical spring oscillationsoscillations

Introduce Introduce andand use use SHM according to formulae: SHM according to formulae:

Learning Outcome:

9.1 Simple harmonic motion (1 hour)

xdt

xda 22

2

PHYSICS CHAPTER 9

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9.1 Simple harmonic motion9.1.1 Simple harmonic motion (SHM) is defined as a periodic motion without loss of energy in a periodic motion without loss of energy in

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

dtxd

PHYSICS CHAPTER 9

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The angular frequencyangular frequency, always constantconstant thus

The negative signnegative sign in the equation 9.1 indicates that the direction of the acceleration, direction of the acceleration, aa is always opposite to the is always opposite to the direction of the displacement, direction of the displacement, xx.

The equilibrium position is a positionposition at which the bodybody would come to restcome to rest if it were to lose all of its energylose all of its energy.

Equation 9.1 is the hallmarkhallmark 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.

xa

m

a

O xx

sF

Figure 9.1aFigure 9.1a

PHYSICS CHAPTER 9

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Figure 9.1bFigure 9.1b

x

x asF

O

m

m

O xx

a

sF

Figure 9.1cFigure 9.1c

PHYSICS CHAPTER 9

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

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

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

1 Hz = 1 cycle s1 = 1 s1

Equation :

9.1.2 Terminology in SHM

2

f

fT 1

f 2 OR

PHYSICS CHAPTER 9Equilibrium Position-- a point where the acceleration of the body undergoing oscillation is

zero. -- At this point, the force exerted on the body is also zero.Restoring Force-- the force which causes simple harmonic motion to occcur. This force is

proportional to the displacement from equilibrium & always directed towards equilibrium.

xkFs

Equation for SHM -- Consider a system that consists of a block of mass, m attached to the end

of a spring with the block free to move on a horizontal, frictionless surface.

PHYSICS CHAPTER 9-- when the block is displaced to one side of its equilibrium position &

released, it moves back & forth repeatedly about a maximum values of displacement x.

-- the spring exerts a force that tends to restore the spring to its equilibrium position.

xkFs

-- Maximum value of x is called amplitude, A-- It can be negative (–) or positive (+).

-- Given by Hooke’s law :

PHYSICS CHAPTER 9

-- Fs is known as restoring force.-- Applying Newton’s 2nd Law to the motion of the block :

netF ma

)alueconstant v:(mkx

mka

-- the acceleration, a is proportional to the displacement of the block & its direction is opposite the direction of the displacement.

-- denote ratio k/m with symbol ω2 :

xa 2

-- any system that satify this equation is said to exhibit Simple Harmonic Motion ( SHM )

[ equation for SHM ]

xa

amxk

-- from above equation, we find that:

xkFs

PHYSICS CHAPTER 9

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Write Write andand use use equation of displacement for SHM,equation of displacement for SHM,

Derive and applyDerive and apply equations for : equations for : velocity, velocity,

acceleration, acceleration,

kinetic energy,kinetic energy,

potential energy, potential energy,

Learning Outcome:9.2 Kinematics of SHM (2 hours)

tAx sin

22

21 xmU

222

21 xAmK

xdt

xddtdva 2

2

2

22 xAdtdxv

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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.Figure 9.66

S

T

MN

O

x

)rad(02

23 2

A

A1x

11

A

P

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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 displacementequation of displacement as a function of time in SHM is given by

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

111 sinAx where and t tAx sin1

tAx sin

angular angular frequencyfrequency

amplitudeamplitude

displacement from displacement from equilibrium positionequilibrium position

timetime

Initial phase angle Initial phase angle (phase constant)(phase constant)

phasephase

(9.4)

t

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Initial phase angle (phase constant),Initial phase angle (phase constant), It is indicate the starting point in SHMstarting point in SHM where the time, tt = = 0 s0 s. If ==00 , the equation (9.4) can be written as

where the starting pointstarting point of SHM is at the equilibrium equilibrium positionposition, 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

PHYSICS CHAPTER 9

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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

23sin tAx

AA Orad

2

OR

OR

2sin tAx

OR tAx cos

tAx sin OR tAx sinAA O

maxv maxv maxv

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From the definition of instantaneous velocity,

Eq. (9.5) is an equation of velocity as a function of time in SHM. The maximum velocity, maximum velocity, vvmaxmax occurs when coscos((t+t+))==11 hence

9.2.2 Velocity, v

dtdxv and tAx sin

)sin( tAdtdv

)sin( tdtdAv

)cos( tAv (9.5)

Av max (9.6)

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The S.I. unit of velocity in SHM is m sm s11. If = = 00 , equation (9.5) becomes

Relationship between velocity,Relationship between velocity, vv and displacement, and displacement, xx From the eq. (9.5) :

From the eq. (9.4) :

From the trigonometry identical,

tAv cos

Axt sin

1cossin 22

tAx sin

)cos( tAv (1)

(2)

and t

tt 2sin1cos (3)

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

2

1

AxAv

2

222

AxAAv

22 xAv (9.7)

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From the definition of instantaneous acceleration,

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

The maximum acceleration, maximum acceleration, aamaxmax occurs when sin(sin(t+t+))==11 hence

9.2.3 Acceleration, a

dtdva and tAv cos

)cos( tAdtda

)cos( tdtdAa

)sin(2 tAa (9.8)

2max Aa (9.9)

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The S.I. unit of acceleration in SHM is m sm s22. If = = 00 , equation (9.8) becomes

Relationship between acceleration,Relationship between acceleration, aa and displacement, and displacement, xx 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)

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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( tAdtdxv

)cos(2 tAdtdva

tAx cos (9.10)

(9.11)

(9.12)

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Potential energy, Potential energy, UU 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

21 kxU and 2mk

22

21 xmU (9.13)

22

21 xmU tAx sinand

tAmU 222 sin21

(9.14)

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

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

2

21 mvK and 22 xAv

222

21 xAmK (9.15)

2

21 mvK tAv cosand

tAmK 222 cos21

(9.16)

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

energy,energy, KK and its potential energy, and its potential energy, UU .

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

The equation of total energy in SHM is given by

UKE

22222

21

21 xmxAmE

22

21 AmE

OR 2

21 kAE

constant UKE

(9.17)

(9.18)

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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 9.1 :

22sin00.5 tx

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Solution :Solution :a. By comparing

thusi. ii.

iii. The period of the motion is

iv. The phase constant is

tAx sinwith

22sin00.5 tx

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Solution :Solution :b. i. Differentiating x respect to time, thus

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Solution :Solution :b. ii. Differentiating v respect to time, thus

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Solution :Solution :c. For t = 2.00 s i. The displacement of the object is

ii. The velocity of the object is

OR

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Solution :Solution :c. For t = 2.00 s iii. The acceleration of the object is

OR

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

ii. The maximum acceleration of the object is

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The length of a simple pendulum is 75.0 cm and it is released at an angle 8 to the vertical. Frequency of the oscillation is 0.576 Hz. Calculate the pendulum’s bob speed when it passes through the lowest point of the swing.

(Given g = 9.81 m s2)Solution :Solution :

Example 9.2 :

A

8

m

L

AO

A

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Solution :Solution : At the lowest pointlowest point, the velocityvelocity of the pendulum’s bob is maximummaximum hence

8 m; 75.0 L

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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, Determinea. the amplitude of the motion,b. the frequency of the motion,c. the maximum acceleration of body in the motion.Solution :Solution :

a. The amplitude is

b. The period of the motion is

Example 9.3 :

A

A

Om

s 17.0tcm .57

m10 75.32105.7 2

2

A

17.022 tTs 34.0T

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Solution :Solution :b. Therefore the frequency of the motion is

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

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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 9.4 :

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Solution :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

0t

cm 10A

tAx sin

OR

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Solution :Solution : b. At the equilibrium position, xx = 0 = 0

s 55.0 kg; 450.0 Tm

OR

PHYSICS CHAPTER 9

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Solution :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

PHYSICS CHAPTER 9

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 and ω is 26.46 rads-1 . Determinea. 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 :Solution :

a. By applying the equation of the total energy in SHM, thusm 1000.4;m N 0.35 kg; 100.50 213 Akm

Example 9.5 :

PHYSICS CHAPTER 9 b) The speed of the object when xx = 1.00 ×10 = 1.00 ×10–2–2 m m

c) The kinetic energy of the object

and the potential energy of the object

PHYSICS CHAPTER 9

An object of mass 3.0 kg executes linear SHM on a smooth horizontal surface at frequency 10 Hz & with amplitude 5.0 cm. Neglect all resistance forces. Determine :(a) total energy of the system(b) The potential & kinetic energy when the displacement of the object is 3.0 cm.

Solution:Given : m = 3.0 kg

A = 5 cm = 0.05 mf = 10 Hz → knowing : ω = 2πf

Example 9.6 :

PHYSICS CHAPTER 9)(b

To calculate Kinetic energy :

b)

PHYSICS CHAPTER 9

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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. :ANS. : U think ; 44.4 cm sU think ; 44.4 cm s22

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. Determinea. 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. :ANS. : 6.0 m, 1.0 s, 6.0 m, 1.0 s, ; 24.0; 24.02 2 m sm s22; 355 J; 355 J

62sin0.6 tx

rad 3

PHYSICS CHAPTER 9

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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. :ANS. : 6.21106.211044 m m4. 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. :ANS. : AEE22 ;

41 ,

43

PHYSICS CHAPTER 9

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

displacement - timedisplacement - time velocity - timevelocity - time acceleration - timeacceleration - time energy - displacementenergy - displacement

Learning Outcome:

9.3 Graphs of SHM (1 hour)

PHYSICS CHAPTER 9

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9.3 Graphs of SHM9.3.1 Graph of displacement-time (x-t) From the general equation of displacement as a function of time

in SHM,

If = = 00 , thus The displacement-time graph is shown in Figure 9.7.

tAx sin tAx sin

Amplitude

Figure 9.7Figure 9.7

x

t04T T

A

A

2T

43T

Period

PHYSICS CHAPTER 9

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For examples:

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

Graph of x against t:

2sin tAx OR tAx cos

x

t04T T

A

A

2T

43T

PHYSICS CHAPTER 9

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

Graph of x against t:x

t04T T

A

A

2T

43T

23sin tAx OR

2sin tAx

OR tAx cos

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

Equation:

Graph of x against t: tAx sin OR tAx sin

x

t04T T

A

A

2T

43T

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How to sketch the How to sketch the xx against against tt graph when graph when 0 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

22sincm 2 πtx

cm 2A

T 2s rad 2 1 s 1T

tx 2sin2(cm)x

)s(t0

2

2

15.0

4T

PHYSICS CHAPTER 9

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Because of

Sketch the new graph.

4rad

2Tt

4T

hence shift the y-axis to the right by

If = negative valueshift the y-axis to the left

If = positive valueshift the y-axis to the right

RULES

0

2

2

15.0

(cm)x

)s(t

PHYSICS CHAPTER 9

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

If = = 00 , 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

t04T T

A

A

2T

43T

Figure 9.8Figure 9.8

PHYSICS CHAPTER 9

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From the relationship between velocity and displacement,

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

22 xAv

v

x0

A

A

AA

Figure 9.9Figure 9.9

PHYSICS CHAPTER 9

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

If = = 00 , 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 9.10Figure 9.10

a

t04T T

2A

2T

43T

2A

PHYSICS CHAPTER 9

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From the relationship between acceleration and displacement,

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

The gradient of the gradient of the a-xa-x graph graph represents

xa 2

Figure 9.11Figure 9.11

a

x0

2A

2A

AA

2,gradient m

PHYSICS CHAPTER 9

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E

x

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

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

9.3.4 Graph of energy-displacement (E-x)

22

21 xmU 222

21 xAmK 22

21 AmE and;

constant21 22 AmE

22

21 xmU

222

21 xAmK Figure 9.12Figure 9.12

PHYSICS CHAPTER 9

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Energy

t

22

21 AmE

tAmU 222 sin21

tAmK 222 cos21

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

Figure 9.13Figure 9.13

PHYSICS CHAPTER 9

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

From the graph above, determine for these oscillationsa. 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 9.7 :

)cm(x

s)(t0

0.15

0.15

8.0

Figure 9.14Figure 9.14

PHYSICS CHAPTER 9

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Solution :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

0

tAx sin

PHYSICS CHAPTER 9

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Solution :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

PHYSICS CHAPTER 9

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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. Determinea. the amplitude,b. the period,c. the maximum speed of the body,d. the total energy of the body.

Example 9.8 :

Figure 9.15Figure 9.15

)s m( 2a

)cm(x0

80.0

80.0

00.400.4

PHYSICS CHAPTER 9

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Solution :Solution :a. The amplitude of the motion isb. 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 2A2

max s m 80.0 a

kg 50.2m

PHYSICS CHAPTER 9

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

d. The total energy of the body is given by

kg 50.2m

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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 9.9 :

Figure 9.16Figure 9.16

)m(x

)s(t0

2.0

2.0

4 52 31

PHYSICS CHAPTER 9

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Solution :Solution :From the graph,

Amplitude, Period,

a. The angular frequency is given by

b. The sum of the kinetic and potential energies is

kg 30.1m

PHYSICS CHAPTER 9

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)ms( -2a

)s(t0

2

2

8.0 0.14.0 6.02.0

Exercise 9.2 :

1. The graph shows the SHM acceleration-time graph of a 0.5 kg mass attached to a spring on a smooth horizontal surface. By using the graph determine(a) the spring constant(b) the amplitude of oscillation(c) the equation of displacement x varies with time, t.ANS: 30.8 NmANS: 30.8 Nm-1-1, 0.032 m, , 0.032 m, tx 5.2cos032.0

PHYSICS CHAPTER 9

68A

maxa

max

AO

maxv

maxa

maxv

maxa

max

max

t x v a K USummary :Summary :

0

4T

2T

43T

T

A

0

A

0

A

0

A

0

A

0

2A

0

2A

0

2A

0

0

0

22

21 mA

22

21 mA

2

21 kA

2

21 kA

2

21 kA

0

0

22 xAv xa 2

2

21 mvK

2

21 kxU

PHYSICS CHAPTER 9

69

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and use Derive and use expression for period of oscillation, expression for period of oscillation, TT for for

simple pendulum and single spring.simple pendulum and single spring.

(i) simple pendulum:(i) simple pendulum:

(ii) single spring:(ii) single spring:

Learning Outcome:9.4 Period of simple harmonic motion (1 hour)

glT 2

kmT 2

PHYSICS CHAPTER 9

70

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

length, L.

mx

L

PO

T

cosmgsinmg Figure 9.2Figure 9.2

gm

PHYSICS CHAPTER 9

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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, weightweight and T, the tension in tension in

the stringthe string. Resolve the weight into

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

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

The restoring forcerestoring force, Fs contributed by the tangentialtangential component of the weight pullspulls the bob back to equilibrium position. Thus

rmvmgT

2

cos

sins mgF

PHYSICS CHAPTER 9

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The negative signnegative sign shows that the restoring forcerestoring force, Fs is

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

sinsin in radianin radian arc lengtharc length, x of the bob becomes straight line straight line (shown in

Figure 9.3) then

L

x

Lxsin

thus

LxmgFs

Figure 9.3Figure 9.3

PHYSICS CHAPTER 9

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By applying Newton’s second law of motion,

Thus

By comparing

Thus

sFmaF

Lmgxma

xLga

xa Simple pendulum executes linear SHM

Lg

2

with xa 2xLga

andT 2

PHYSICS CHAPTER 9

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Therefore

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

on the simple pendulum.

gLT 2

pendulum simple theof period : Twhere

onaccelerati nalgravitatio : gstring theoflength : L

(9.2)

Simulation 9.1

PHYSICS CHAPTER 9

75

B.B. Spring-mass oscillationSpring-mass oscillationVertical spring oscillation

mm

1x

xO O

Figure 9.4aFigure 9.4a Figure 9.4bFigure 9.4b Figure 9.4cFigure 9.4c

F

gm

a

gm

1F

PHYSICS CHAPTER 9

76

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

xmka

xa Vertical spring oscillation executes linear SHM

PHYSICS CHAPTER 9

77

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

4Tt

2Tt

4T3t

Tt

a

kxF s

maF skxma

xmka

executes linear SHM

Then

Figure 9.5Figure 9.5xa

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78

By comparing

Thus

Therefore

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

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

with xa 2xmka

mk

2 andT 2

kmT 2 noscillatio spring theof period : T

where

object theof mass : mconstant) (forceconstant spring : k

(9.3)

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79

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

But its period on the surface of Mars is given by

Example 9.10 :

2M

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

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80

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

2M

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

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81

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 :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 9.11 :

kg 25.1 Hz; 48.0 Hz; 88.0 21 Δmff

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82

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

kg 25.1 Hz; 48.0 Hz; 88.0 21 Δmff

PHYSICS CHAPTER 9

1. 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. Calculate

a. the amplitude of the motion.b. the maximum velocity attained by the object.ANS. :ANS. : 5.17x105.17x1022 m; 0.597 m s m; 0.597 m s11

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

a) the period of the oscillation,b) the pendulum’s bob speed and acceleration when it passes

through the lowest point of the swing. (Given g = 9.81 m s2)ANS.: 1.74s; 0.378msANS.: 1.74s; 0.378ms-1-1

83

Exercise 9.3 :

PHYSICS CHAPTER 9

84

3. The acceleration of free fall on the Moon is 1/6 the acceleration of free fall on the earth. If the period of a simple pendulum on the earth is 1.0 second, what would its period be on the Moon.ANS: 2.45 sANS: 2.45 s

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85

THE END…Next Chapter…

CHAPTER 10 :Mechanical waves