Download - Note chapter9 0708-edit-1
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PHYSICS 025 CHAPTER 9
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CHAPTER 9: Simple Harmonic Motion
(6 Hours)
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PHYSICS 025 CHAPTER 9
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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.
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PHYSICS 025 CHAPTER 9
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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
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PHYSICS 025 CHAPTER 9
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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
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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
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PHYSICS 025 CHAPTER 9
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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
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PHYSICS 025 CHAPTER 9
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Figure 9.1b
x
x a
sF
O
m
m
O xx
a
sF
Figure 9.1c
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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
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PHYSICS 025 CHAPTER 9
Solution:a. frequency, f =
b. period, T =
9
204.0
5.0Hz Hz
1 10.25
4.0s s
f
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PHYSICS 025 CHAPTER 9
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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
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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 9.6
S
T
MN
O
x
)rad(02
2
3 2
A
A1x
11
A
P
Simulation 9.3
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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 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
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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
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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
2
3sin
tAx
AA Orad
2
OR
OR
2sin
tAx
OR tAx cos
AA O
maxv
tAx sin OR tAx sin
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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, 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)
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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)
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By substituting equations (3) and (2) into equation (1), thus
2
1
A
xAv
2
222
A
xAAv
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, 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)
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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)
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PHYSICS 025 CHAPTER 9
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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( tAdt
dxv
)cos(2 tAdt
dva
tAx cos (9.10)
(9.11)
(9.12)
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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)
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PHYSICS 025 CHAPTER 9
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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)
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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
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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 3 :
22sin00.5
tx
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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
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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
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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
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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
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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
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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
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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. 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
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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
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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, 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
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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
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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 6 :
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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
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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
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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
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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
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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
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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
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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. : 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
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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. : 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
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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)
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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
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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
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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
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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
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PHYSICS 025 CHAPTER 9
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
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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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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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
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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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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PHYSICS 025 CHAPTER 9
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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)
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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
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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)
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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
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PHYSICS CHAPTER 9
88
THE END…Next Chapter…
CHAPTER 10 :Mechanical waves