Post on 03-Jan-2016
Simple Harmonic Motion
Periodic Motion:
any motion which repeats itself at regular, equal intervals of time.
Simple Harmonic Motion
It is actually anything but simple!
Simple harmonic motion is a special case of periodic motion
Characteristics of simple harmonic motion:1. the displacement is a sinusoidal function of time, it ranges from
zero to a maximum displacement (amplitude),2. Velocity is maximum when displacement is zero,3. Acceleration is always directed toward the equilibrium point, and is
proportional to the displacement but in the opposite direction,4. The period does not depend on the amplitude.
Equilibrium position
Positive amplitude
Negative amplitude
A
O
B
1. It is periodic oscillatory motion about a central equilibrium point,
Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from
zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point, and is
proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.
Dis
pla
cem
en
t
Time
A A
B B
O
Equilibrium position
Positive amplitude
Negative amplitude
A
O
B
Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from
zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,(equilibrium
point)4. Acceleration is always directed toward the equilibrium point, and
is proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.
Dis
pla
cem
en
t
Time
A A
B B
O
Velo
city
Time
A
O
B
O remember velocity is the gradient of displacement
Equilibrium position
Positive amplitude
Negative amplitude
A
O
B
Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from
zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point,
and is proportional to the displacement but in the opposite direction,
5. The period does not depend on the amplitude.
Dis
pla
cem
en
t
Time
A A
B B
O
Velo
city
Time
A
O
B
OA
ccele
rati
on
Time
A
O
Bremember acceleration is the gradient of velocity
Equilibrium position
Positive amplitude
Negative amplitude
A
O
B
Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from
zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point, and
is proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.
Dis
pla
cem
en
t
Equilibrium position
Positive amplitude
Negative amplitude
A
O
B
Velo
city
Time
A
O
B
OA
ccele
rati
on
Time
A
O
B
Simple Harmonic Motion• Equilibrium: the position at which no net force acts on
the particle.• Displacement: The distance of the particle from its
equilibrium position. Usually denoted as x(t) with x=0 as the equilibrium position.
• Amplitude: the maximum value of the displacement with out regard to sign. Denoted as xmax or A.
What is the equation of this graph?
Hint
HIVO, HOVIS
A cos graph takes 2 radians to complete a cycle
Dis
pla
cem
en
t
Time
A A
B B
O
Relation between Linear SHM and Circular Motion
B
B A
A
O
Dis
pla
cem
en
t
Time
A A
B B
O
Velo
city
Time
A
O
B
O
Acc
ele
rati
on
Time
A
O
B
Relation between Linear SHM and Circular Motion
B
B A
A
O
Dis
pla
cem
ent
Time
A A
B B
O
At A, t = 0
the displacement is maximum positive x = xo
B
B
xo
O A
A
xo
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At O, t =
the displacement is zero x = 0
B
B
xo
O A
A
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At B, t =
the displacement is maximum negative x = -xo
B
B
xo
O A
A
-xo
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At O, t =
the displacement is zero x = 0
B
B
xo
O A
A
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At A, t = T
the displacement is maximum positive x = xo
B
B
xo
O A
A
xo
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At A, t=0
the displacement is maximum positive x = xo
θ = 0°
B
B
xo
O A
A
xo
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At O, t =
the displacement is zero x = 0
θ = 90°
B
B
xo
O A
A
Dis
pla
cem
ent
Time
A A
B B
O
At B, t =
the displacement is maximum negative x = -xo
θ = π 180°
B
B
xo
O A
A
-xo
T = periodic time
= the time to complete 1 cycle
Dis
pla
cem
ent
Time
A A
B B
O
At O, t =
the displacement is zero x = 0
θ = 270°
B
B
xo
O A
A
Dis
pla
cem
ent
Time
A A
B B
O
At A, t = T
the displacement is maximum positive x = xo
θ = 2π 360°
A
A
xo
O B
B
xo
T = periodic time
= the time to complete 1 cycle
M2 reminder
w =
Units
change∈angletime taken
=θt
rads per sec
Angular velocity =
Dis
pla
cem
ent
Time
A A
B B
O
At a random point M , the displacement is x
A
A
xo
O B
B
x
x
M
θ
since ω = θ = ω t
cos θ = cos (ω t) = Using trigonometry
x = xo cos (ω t)
and θ = θ°
T = periodic time
= the time to complete 1 cycle
xx o
Dis
pla
cem
en
t
Time
B
O
B A A
If timing begins when x = 0 at the centre of the SHM
Dis
pla
cem
ent
Time
A A
B B
O
If however timing begins when x = 0 at the centre of the SHM then the displacement equation is :
A
A
xo
O B
B
x
x
M
θ
x = xo sin (θ)
T = periodic time
= the time to complete 1 cycle
ω = θ = ω t
So x = xo sin (ω t)
Displacement Formulae
If x = x0 when t = 0 then use x = xo cos (ω t)
This is usually written as x = A cos (ω t)
where A is the amplitude of the motion
If x = 0 when t = 0 then use x = xo sin (ω t)
This is usually written as x = A sin (ω t)
where A is the amplitude of the motion
Memory aid Sin graphs start at the origin(centre)Cos graphs start at the top(end)
Dis
pla
cem
ent
Time
A A
B B
O
What is the displacement of a ball from the end undergoing SHM between points A and B starting at A where the distance between them is 1 m and the angle is rads
B
A
Amp
O B
A
x
x
M
( 𝜋7 )
x = A cos (ω t) when t = 0 A =
x = 0.5 cos = 0.45 m so it is 0.05m from the end
0.45
0.5 i.e the Amplitude
0.45
Periodic Time
ω = So t =
T =
So the time to complete one cycle is T =
To complete one cycle t = T and q = 2p
=
Tom is 2 pies over weight
What is the displacement of a ball from the centre undergoing SHM between points A and B starting at the A, after 2secs where the distance between them is 1 m and the periodic time is 6
Dis
pla
cem
ent
Time
A A
B
O
Amp
x = A cos (ω t) when t = 0 A =
x = 0.5 cos 2ω
To find ω use the fact that ω =
To complete one whole cycle then q = 2p and t = 6 So ω = = =
x = 0.5 cos 2ω = 0.5 cos 2 = -0.25 B O Ax
M
-0.25
2
0.5 i.e the Amplitude
Memory aid
Tom is 2 pies overweight T=
-0.25
What is the displacement from the of a ball going through SHM between points A and B starting at the centre, after 3secs where the distance between them is 0.8m and the periodic time is 8sx = A sin (ω t) when t = 0 A =
x = 0.4 sin 3ω
To find ω use the fact that ω =
To complete one whole cycle then q = 2p and t = 8 So ω = = =
x = 0.4 sin 3ω = 0.4 sin 3 = B O Ax
M
Dis
pla
cem
ent
A
B B
O
xo0.4
√23
0.4 i.e the Amplitude
Memory aid
Tom is 2 pies overweight T=
Calculus Memory Aid
s
v
a
Some Vehicles AccelerateThanks to Chloe Barnes
Starving Vultures Attack
Differentiate Integrate
Velocity Equation.
since v =
v = Velo
city
Time
A
O
B
O
x = A cos (ω t)
Characteristics of simple harmonic motion:Velocity is maximum when displacement is zero,
Dis
pla
cem
en
t
-ω A sin (ω t)
dxdt
since x = A cos (ω t)
Acceleration Equation.
a =
since a =
v = -ω A sin (ω t)
a =
Velo
city
Time
A
O
B
O
Acc
ele
rati
on
Time
A
O
B
Characteristics of simple harmonic motion:Acceleration is always directed toward the
equilibrium point, and is proportional to the displacement but in the opposite direction,
d vdt
-ω2 A cos (ω t)
-ω2 x
Equation for velocity where v depends on x rather than t
v = -ω Asin (ω t) v depends on t
Acceleration = rate of change of velocity =
dv
dt
Velocity = rate of change of displacement =
So acceleration =
dvv
dx
Chain Rule for acceleration in terms of displacement
=
dv dx
dx dt
Equation for velocity where v depends on x rather than t
v = -ω A sin (ω t) v depends on t
2 2.. dv
acceleration x x v xdx
∫ vdv=∫−ω2 xdx−ω2 x2
2+c
)
0 = c =
v2=ω2 Amp2−ω2 x2
Separate the variables
=
Equation for velocity where v depends on x rather than t
v2=ω2 Amp2−ω2 x2
v2=ω2(Amp¿¿2−x¿¿2)¿¿
The max velocity occurs at the centre where x = 0
Vmax = w Amp
x = A cos (ωt) if x = A when t = 0
v = -ω A sin (ωt) if x = A when t = 0
a = -ω2 x
T = Tom is 2 pies over weight
)
x = A sin (ωt) if x = 0 when t = 0
Vmax = ω A
Differentiate the displacement equation
Summary
Stop and look at Course notes
Horizontal elastic string
Consider a GENERAL point P where the extension is x .
Resolving 0 – T = m but
– = m so =
Comparing with the S.H.M eqn = –w2x w2 =
T+ve x
xl
xl
xl
x
m
l
m
l
T = for an elastic string
Ex.1 String l = 0.8m l = 20N m = 0.5kg
Bob pulled so that extension = 0.2m.
a) Find time for which string taught
b) Time for 1 oscillation
c) Velocity when ext. = 0.1m
T+ve x
x
l
Horizontal elastic string
Resolving 0 – T = mbut T = for an elastic string
– = 0 so =
Comparing with the S.H.M eqn = –w2x
T+ve x
x
l
xl
20 x
0.8
20 x50x
0.8 0.5
Hence w = 50
As the string goes slack at the natural length it then travels with constant velocity before the string goes taught again.
l = 0.8m l = 20N m = 0.5kg
w2 = 50
a) 2π2π
T= =50
This is the periodic time when the string is tight
b) Time for which initially taught is a T so t =2 50
As the initial extension is 0.2 m then
Amp = 0.2m
Velocity when string goes slack v = w Amp as x = 0
vslack = 500.2 = 1.414m/s
So the distance travelled when string is slack is 4 0.8
So time for which slack
So total time for 1 oscillation = + = 3.15secs
distance 4 0.8
speed 1.414
2
50
4 0.8
1.414
c) Vel. when ext = 0.1
v2 = w2(Amp2 – x2)
v2 = 50(0.22 – 0.12) = 1.5ms–1 v = 1.225ms–1
Two Connected Horizontal Elastic Strings
A and B are two fixed points on a smooth surface with AB = 2m. A string of length 1.6m and modulus 20N is stretched between A and B. A mass of 3kg is attached to a point way along the string from A and pulled sideways 9cm towards A. Show that the motion is S.H.M and find the max acceleration and max velocity.
AB
N1 N2E
0.8 0.2 0.8
T1 xT2
0.2
As x is increasing towards A make +vex is measured from the equilibrium position
1
x 20(0.2-x)T = =
0.8ll
T1 =
+ve direction T1 – T2 = 3a
2
x 20(0.2+x)T = =
0.8ll
- = 3
A B
N1 N2E
0.8 0.2 0.8
T1 xT2
0.2
T1 =
- = 3
-50x = 3
simplifying
- x = which is SHM
ω2=503
A B
N1 N2E
0.8 0.2 0.8
T1 xT2
0.2
T1 =
= - x ω2=503
250 0.091.5ms
3-- ´
=
Max velocity = wAmp = √ 503
0.09 = 0.37ms-1
A B