1 Simple harmonic motion Vibration / Oscillation to-and-fro repeating movement.
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Transcript of 1 Simple harmonic motion Vibration / Oscillation to-and-fro repeating movement.
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Simple harmonic motion
Vibration / Oscillation to-and-fro repeating
movement
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Simple harmonic motion S.H.M. A special kind of oscillation
X YO
X, Y : extreme points O: centre of oscillation / equilibrium position A: amplitude
A
A special relation between A special relation between the displacement and the displacement and acceleration of the particleacceleration of the particle
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Exploring the acceleration and displacement of S.H.M.
- 1 .5
- 1
- 0 .5
0
0 .5
1
1 .5
0 4 5 9 0 1 3 5 1 8 0 2 2 5 2 7 0 3 1 5 3 6 0 4 0 5
time
displacement
acceleration
aa– – xx graph graph
- 0 .3
- 0 .2
- 0 .1
0
0 .1
0 .2
0 .3
- 0 .5 - 0 .4 - 0 .3 - 0 .2 - 0 .1 0 0 .1 0 .2 0 .3 0 .4 0 .5
displacement
acceleration
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Exploring the acceleration and displacement of S.H.M.
a – x graph a
x
a a ∝∝ x x a a ∝∝ -x -x
Definition of Simple harmonic motion (S.H.M.)Definition of Simple harmonic motion (S.H.M.) An oscillation is said to be an S.H.M if An oscillation is said to be an S.H.M if (1)(1) the the magnitude magnitude of acceleration is directly proportional to of acceleration is directly proportional to
distance from a fixed point (centre of oscillation), and distance from a fixed point (centre of oscillation), and
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-3 -2 -1 0 1 2 3-3 -2 -1 0 1 2 3x
aa(-) (-) aa(-) (-) aa(-) (-) aa(+) (+) aa(+) (+) aa(+) (+) aa = 3 = 3kk aa = 2 = 2kk aa = = kk aa = 0 = 0 aa = = kk aa = 2 = 2kk aa = 3 = 3kk
Definition of Simple harmonic motion (S.H.M.)Definition of Simple harmonic motion (S.H.M.) An oscillation is said to be an S.H.M if An oscillation is said to be an S.H.M if (1)(1) the magnitude of acceleration is directly proportional to the magnitude of acceleration is directly proportional to
distance from a fixed point, and distance from a fixed point, and (2)(2) the acceleration is always the acceleration is always directed towards that pointdirected towards that point..
Note: For S.H.M., direction of accelerationacceleration and displacementdisplacement is always opposite to each other.
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Equations of S.H.M.
P
O
r
aa = = rr22 r2 sin
r2 cos
Y’ O’ P’ P’ X’
x
x
For the projection P’ • Moves from X’ to O’ to Y’ and returns through O’ to X’ as P completes each revolution.Displacement• Displacement from O • x = r cos = r cos tAccelerationAcceleration of P’ = component of acceleration of P along the x-axis • a = -r2 cos (-ve means directed towards O)∴ aa = - = -22 xxThe motion of P’ is simple harmonic.
= t
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Equations of S.H.M.
P
O
r
Y’ O’ P’ P’ X’
x
x
PeriodThe period of oscillation of P’ = time for P to make one revolution T = 2 / angular speed ∴ T = 2T = 2//
VelocityVelocity of P’ = component of velocity of P along the x-axisvv = - = -rr sin sin = - = -rr sin sin tt
r
rsin
rcos
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Equations of S.H.M.
P
O
r
Y’ O’ P’ P’ X’
x
x
Motion of P’ Amplitude of oscillation
= Radius of circle ⇒⇒ AA = = rr
Displacement x:xx = A cos = A cos
Velocity v:vv = - = -AA sin sin
Acceleration a:aa = - = -22AA cos cos
r
rsin
rcos
A
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Relation between the amplitude of oscillation A and x, , and v:
2222
22222
22
1
cos,sin
xAv
Axv
A
x
A
v
A
x
A
v
Note: Maximum speed = A at x = 0 (at centre of oscillation / equilibrium positio
n).
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Example 1Example 1A particle moving with S.H.M. has velocities of 4 cm sA particle moving with S.H.M. has velocities of 4 cm s-1-1 and and 3 cm s3 cm s-1-1 at distances of 3 cm and 4 cm respectively from its at distances of 3 cm and 4 cm respectively from its equilibrium. Findequilibrium. Find(a)(a) the amplitude of the oscillationthe amplitude of the oscillation Solution:
By v2 = 2(A2 – x2)when x = 3 cm, v = 4 cm s-1,
x = 4 cm, v = 3 cm s-1. 42 = 2(A2 – 32) --- (1) 32 = 2(A2 – 42) --- (2)(1)/(2):16/9 = (A2 – 9) / (A2 – 16)9A2 – 81 = 16 A2 - 256 A2 = 25A = 5 cm∴ amplitude = 5 cm
4 ms-1
3 ms-1O
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(b)(b) the period,the period,(c)(c) the velocity of the particle as it passes through the velocity of the particle as it passes through the equilibrium position.the equilibrium position.
(b) Put A = 5 cm into (1) 42 = 2(52 – 32) 2 = 1⇒ = 1 rad s-1
T = 2/ = 2 s(c) at equilibrium position, x = 0
By v2 = 2(A2 – x2)v2 = 12(52 – 02)v = 5 cm s-1
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Isochronous oscillations
Definition: period of oscillation is independent of its amplitude.
Examples: Masses on springs and simple pendulums
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Phase difference of x-t, v-t and a-t graphs
0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T
A A 2A
x v a
t t t
tAx cos tAa cos2tAv sin
2A
x
y
v
aA A
Vectors x, v and a rotate with the same angular velocity . Their projections on the y-axis give the above x-t, v-t and a-t graphs.
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Phase difference of x-t, v-t and a-t graphs
2A
x
y
v
a
A A
Note:1 a leads v by 90o or T/4.
(v lags a by 90o or T/4)2 v leads x by 90o or T/4.
(x lags v by 90o or T/4)3 a leads x by 180oor T/2.
(a and x are out of phase or antiphase)
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Energy of S.H.M.(Energy and displacement)
From equation of S.H.M. v2 = 2(A2 – x2), ∴ K.E. = ½ mv2 = ½ m2(A2 –
x2)
x
K.E.
0 A-ANote:1. K.E. is maximum when x = 0 (equilibrium position) 2. K.E. is minimum at extreme points (speed = 0)
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Potential energy P.E. = ½ kx2 ∵ 2 = k/m ∴ P.E. = ½ m2x2
P.E. is maximum at extreme points.
(Spring is most stretched.) P.E. is minimum when x = 0.
(Spring is not stretched)
x
P.E.
0 A-A
x = Ax = -A
Centre of oscillation
ix
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Total energy = K.E. + P.E.
x
Total energy
0 A-A
Energy
K.E.
P.E.
½ m2A2
2222222
2
1
2
1
2
1AmxmxAm (constant)
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Energy and time
tAvandtAx sincos
tAmtAmmv 22222 sin2
1sin
2
1
2
1
tAmtAmxm 2222222 cos2
1cos
2
1
2
1
22
2
1Am
From equation of S.H.M.
K.E. =
P.E. =
Total energy = K.E. + P.E.
(constant)
ttAmtAmtAm 2222222222 cossin2
1cos
2
1sin
2
1
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Total energy
Energy
P.E. = ½ m2A2cos2t
0 T/4 T/2 3T/4 T Time
½ m2A2
K.E. = ½ m2A2sin2t
t = 0t = T/2
Centre of oscillation
ix
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Examples of S.H.M.Mass on spring – horizontal oscillationHooke’s law: F = kx where k is the force constant and x is the extension By Newton’s second law
T = -makx = -maa = -(k/m)x
which is in the form of a = -2x Hence, the motion of the mass i
s simple harmonic, and 2 = k/m Period of oscillation
Natural length (l)
Extension (x)
T
Centre of oscillation
ix
k
m
22
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Mass on spring – vertical oscillation At equilibrium,
T’ = mgke = mg
Displaced from equilibrium,T – mg = -mak(e + x) – mg = -ma
Natural length (l)
TT
Extension at equilibrium
Displacement from
equilibrium
x
e
mgmg
T’T’
mgmg
Centre of oscillation
In equilibrium
Spring unstretched
Displaced from
equilibrium which is in the form of a = -2x Hence, the motion of the mass i
s simple harmonic
and 2 = k/m. Period of oscillation
mamgxk
mgk )(
xm
ka
k
m
22
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Effective mass of spring
Not only the mass oscillates when it is released, but also the spring itself.
The period of oscillation is affected by the mass of the spring.
Hence, the equation k
mT 2
k
mmT s
2should be rewritten as
where ms is the effective mass of the spring.
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Measurement of effective mass of spring To find the effective mass, we can do an experiment by
using different masses m and measure the corresponding periods T.
Use the results to plot a graph of T2 against m which is a straight line but it does not pass through the origin.
x
xx
x
x
T2
m
Line of best fit
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k
mmT s
22 4sm
km
kT
222 44
∴ or
k
mmT s
2
∴
x
xx
x
x
T2
m
k
24slope =
smk
24y-intercept =
∴ effective mass ms
24
ky-intercept
In theory, effective mass of a spring is about ⅓ of the mass of
string.
Usually, we would neglect the effective mass for simplicity.
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Combined Springs OscillationCase 1: Springs in parallel
Let x be the common extension of the spring.
the springs are in parallel, ∴ upward force FF = FF11 + FF22
F = k1x + k2x = (k1 + k2)x Note: k1 + k2 is the equivalent force
constant of the system. When the mass is set into vibration,
the oscillation is simple harmonic. Period of oscillation
FF
F2 = k2x
where k = k1 + k2
k
mT 2
F1 = k1x
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Case 2: Springs in series Let x1 and x2 be the extensions of the first
and the second spring respectively. The total extension x = x1 + x2 ∵ the springs are in series, ∴ upward force F = F1 = F2
F = k1x1 = k2x2
FF
FF11 = = k k11xx11
F2 = k2 x2
22
11 k
Fxand
k
Fx
∵ x = x1 + x2
21 k
F
k
Fx
21
11
kk
xF
kxF
21
111
kkkor where
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Case 2: Springs in series
FF
FF11 = = k k11xx11
F2 = k2 x2
Note: the equivalent force constant of the
system is k where 21
111
kkk
k
mT 2
When the mass is set into vibration, the oscillation is simple harmonic.
Period of oscillation
21
111
kkk where
.
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Case 3: The mass is connected by two springs on both sides
T1
Equilibrium position
T2
Under compression
Under extension
Force constant k1
Force constant k2
x
Suppose the springs are initially unstretched.
When the mass is displaced to the right by x, 1st spring is extended but 2nd spring is compressed.Resultant force on the bob F = T1 + T2
∴ F = k1x + k2x = (k1 + k2)xNote: k1 + k2 is the equivalent force constant of the system.
The oscillation is simple harmonic. Period of oscillation
k
mT 2 where k = k1 + k2.
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Simple pendulum Resolve tangentially (perpendicular to the string)
mg sin = -ma where a is the acceleration along the arcIf is small (i.e. <10o), sin ≈ and x ≈ l ,mg sin = -ma becomesmg = -maa = -g(x/l) = -(g/l)xwhich is in the form of a = -2x
Hence, the motion of the bob is simple harmonic and 2 = g/l
Period of oscillation T
OP
A
mgmg cos
mg sin
l
x
T
g
l
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A simple pendulum has a period of 2 s and an A simple pendulum has a period of 2 s and an amplitude of swing 5 cm. amplitude of swing 5 cm. Calculate the maximum magnitudes of Calculate the maximum magnitudes of (a)(a) velocity, and velocity, and (b)(b) acceleration of the bob.acceleration of the bob.
Solution:
(a) maximum magnitude of velocity
= A = (5) = 5 cm s-1 (b) maximum magnitude of velocity
= 2A = 5cm s-2
1srad2
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By
T