Transcript simple harmonic motion

```By
PROF. G. C. ONYEDIM
Geophysics Research Group
Department of Physics
Topics to Cover in Section 3
• Introduction
• Simple Harmonic Motion (SHM)
• Force Law in SHM
• Energy in SHM
• Angular SHM
• The Simple Pendulum
• The Physical Pendulum
• SHM and Uniform Circular Motion
• Damped SHM
• Forced Oscillations and
1. Introduction
The present topic deals with motions
that repeat themselves
Practical examples include:
• Child on a swing
• The piston in an engine
• The Pendulum in a clock
• The diaphragm of a loudspeaker
• Atoms in molecules or in solid lattice
• Boat on sea wave
• Radio waves, microwaves and visible light are
oscillating magnetic and electric field vectors
In nature, some of such motions will
gradually die out as a result of frictional
forces.
Therefore, to sustain such motion, there
must be some impressed force.
Next we shall define the basic parameters
that are used to describe SHM
2. Simple Harmonic Motion (SHM)
Consider a body of mass m that moves back
and forth along the x-axis. It’s motion is
analogous to that of a mass moving in a
circle but viewed from the side.
The motion can be shown to sweep out a
sine wave and is therefore described as
HARMONIC
Repeats itself at regular
intervals.
Mass in Linear SHM along x-axis
Parameters of SHM
(a) Frequency: Number of complete oscillation per
second and represented by the f or ν.
Unit of frequency is cycles/sec
Hertz
Hz
(b) Period: Time required to complete one
oscillation and represented by T.
Unit of T is sec. T is also related to frequency by
T = 1/f
(1)
(c) Amplitude: The maximum displacement in the
motion and represented by
xm
SHM as a periodic sinusoid
Analogy of Oscillating Spring and
Sine Wave
Equations of SHM
In SHM, the restoring force F is directly
proportional to the displacement x and is
governed by
Hooke's Law.
That is F = - kx
(2)
where k = spring constant and x is
displacement
Thus, Newton’s equation of motion for the
system becomes:
F = ma = md2x/dt2 = -kx
The solution for the equation is:
(ωt + φ) is called the phase
ω is the angular frequency (radians per sec)
φ is the phase constant which depends on
how the motion started.
Assume φ = 0 at time t = 0.
After one period T from any time t, the mass
returns to the original position. That is:
X = Xm Cos(ωt) = xm Cosω(t + T)
ωt + 2π = ω(t + T) = ωt + ωT
Therefore ω = 2π/T = 2πf
(4)
• When the displacement of a particle at any
time “t” can be given by this formula:
x(t )  xm cos(t   )
Its motion is called simple harmonic motion
Simple Harmonic Motion
x(t )  xm cos(t   )
• The factors xm, ω and  are all constants
• xm is the amplitude of the oscillation
• ω is angular frequency of the oscillation
• The time varying quantity (ωt + ) is called
the phase of the motion and  is called the
phase constant or phase angle
Simple Harmonic Motion
Let   0 then xt   xm cos t
SHM repeats itself after T time
 xt  T   xt 
•and cosine function repeats itself when its argument (the phase) has increased by
the value 2π
Simple Harmonic Motion
 (t  T )  t  2
2

 2f
T
 is called angular frequency
Simple Harmonic Motion
Keeping ω & Φ constant
varying xm …
• We can see that the curves look identical except
that one is ‘taller’ than the other
• These two curves have a different maximum
displacement – or amplitude xm
Simple Harmonic Motion
• Keeping Φ & xm constant
• Varying ω =2π/T …
• let T´ = T/2
Simple Harmonic Motion
• Keeping xm & ω constant
• varying phase angle 
•The 2nd curve has ‘slid over’ by a constant amount relative to the 1st curve…
The Velocity of SHM
• To get the velocity of the particle
• differentiate the displacement function with respect to
time
x(t )  xm cos(t   )
d
v(t )  xm cos(t   )
dt
The Velocity of SHM
v(t )   xm sin( t   )

or vt    xm  xt 
2
2
for all times
2
2

The Velocity of SHM
• velocity is a sine function
• It is T/4 period (or π/2) out of phase with the
displacement
The Acceleration of SHM
• lets differentiate once again to get the acceleration
function:
d v(t ) d
a(t ) 
  xm sin( t   )
dt
dt
a(t )   xm cos(t   )
2
The Acceleration of SHM
• We can combine our original equation for the
displacement function and our equation for the
acceleration function to get:
a(t )   xm cos(t   )
2
a(t )   x(t )
2
The Acceleration of SHM
a(t )   x(t )
2
• This relationship of the acceleration being
proportional but opposite in sign to the
displacement is the hallmark of SHM
• Specifically, the constant of proportionality is the
square of the angular frequency
• The acceleration
function is onehalf period (or π
phase with the
displacement
and the
maximum
magnitude of the
acceleration is
ω2xm
• When the
displacement is at a
maximum, the
acceleration is also
at a maximum (but
opposite in sign)
• And when the
displacement and
acceleration are at a
maximum, the
velocity is zero
• Similarly,
when the
displacement
is zero, the
velocity is at a
maximum
SHM – CIRCULAR MOTION
ANALOGY
Velocity in SHM
X = Xm Cos(ωt + φ)
We differentiate to obtain Velocity as follows:
V = dx/dt = - Xm ωSin(ωt + φ)
i.e.
V = - ωXm Sin(ωt + φ)
(5)
Acceleration
The acceleration a is calculated by taking the
derivative of velocity v or the second
derivative of displacement x as follows:
a = dv/dt = d/dt[- Xm ωSin(ωt + φ)]
i.e.
i.e.
a = - ω2 Xm Cos(ωt + φ)
a = - ω2 X(t)
(6)
(7)
Thus, (i) Acceleration is proportional to X(t)
and (ii)
,,
is opposite displacement X(t)
3. FORCE IN SHM
Using the equation for acceleration shown
above, Newton’s 2nd Law becomes
F = ma = -m[ω2 X(t)] = -mω2 [X(t)]
(8)
Let mass m be attached to spring of spring
constant k and made to perform SHM
between points +X and –X along the x-axis.
Hooke’s Law is F = - kx
(9)
Compare Eqns.8 and 9 results in k = mω2 (10)
ω = (k/m) ½ = Angular Frequency
(11)
Ponderable: Car springs
When a family of four people with a total mass of 200 kg
step into their 1200-kg car, the car springs compress 3.0
cm
1) What is the spring constant for the car springs,
assuming that they act as a single spring?
2) How far will the car lower if loaded with 300 kg?
3) What are the period and frequency of the car after
hitting a bump? Assume the shock absorbers are
poor, so the car really oscillates up and down.
Thus, the Period T for the oscillation becomes
T = 2π/ω = 2π (m/k)½
(12)
Large k = Stiff Spring
Short Period OR
Low Frequency
Large mass (m) Long Period/High Frequency
Spring provides ‘Springiness’
Mass provides
‘Inertia’
4.
ENERGY IN SHM
As body eg mass attached to spring oscillates
in SHM, energy is transformed from P.E. of
spring to K.E. of mass in such a way that Total
Energy at any time (t) is CONSTANT.
P.E. due to spring is
U(t) = ½ k[X(t)]2 = ½k Xm2 Cos2(ωt + φ) (13)
Kinetic Energy of oscillating mass m is
K(t) = ½mv2 = ½m(-ωXm)2 Sin2(ωt + φ)
(14)
Using Eqn. 11 i.e. ω = (k/m) ½
Then : K(t) = ½ k Xm2 Sin2(ωt + φ)
(15)
The Total Energy E(t) of the oscillator becomes:
E(t) = U(t)+K(t)
= ½kXm2 Cos2(ωt+φ) + ½ kXm2 Sin2(ωt+φ)
= ½kXm2 [Cos2(ωt+φ) + Sin2(ωt+φ)]
= ½kXm2
(16)
(16) Shows that the Total Energy is CONSTANT
and independent of time since it is a function
of the Maximum Displacement Xm
Plot of SHM Energy as a Function of Time
Plot of Energy as a Function of Displacement
The oscillator needs the two complementary
elements:
• Spring to store Potential Energy and
• Mass to store Kinetic Energy
5.
ANGULAR SHM
Another form of SHM is
Angular SHM such as
that
triggered
by
twisting and releasing
wire that suspends a
circular disk as shown
in the diagram.
The device is referred to as a Tortional Pendulum
Twisting causes a Restoring Torque τ on wire
which is proportional to the twist angle θ
i.e.
τ = -κθ
(17)
where κ (kappa) is a constant that depends on
parameters of the spring (Length, radius and
type of material)
Compare (17) with Hooke’s Law F= -kx which
caused SHM of period T = 2π (m/k)½
Therefore, Period for Angular SHM becomes
T = 2π (I/κ)½
(18)
Here, I is known as the moment of Inertia
which depends on the mass (m) of the
oscillating element and a characteristic
distance (r) of the mass distribution from a
reference point.
6. SHM of a Simple Pendulum
Diagram shows mass
m suspended on
string of length L.
If system is displaced
by angle θ, it
the
central
equilibrium axis.
Inertia comes from mass m
Springiness comes from Force of gravity
Restoring Force is F = -mgSinθ
For small θ, Sinθ
θ
(19)
Thus, F = -mgθ = -mg(x/L) = -(mg/L)x = kx (20)
i.e.
k = mg/L
(21)
Thus, Period T = 2π (m/k)½ = 2π (L/g)½
(22)
7. SHM of Physical Pendulum
Here, displacement is angular due torestoring Torque as
shown in diagram
Torque = τ = Fh = -mgSinθh
= -mgθh
(24)
τ = -(mgh)θ
(25)
Therefore k = mgh
(26)
Resulting in the equation for
the period of the Physical
Pendulum as
T = 2π (I/k)½ = 2π (I/mgh)½ (27)
I = Rotational Moment of Inertia (MI)
h = Distance between CG and point of support
7. SHM of Physical Pendulum
Here, displacement is angular due to restoring
Torque as shown in diagram
Torque = τ = Fh = -mgSinθh
(23)
τ = -mgθh
(24)
τ = -(mgh)θ
(25)
Therefore k = mgh
(26)
Resulting in the equation for the period of the
Physical Pendulum as
T = 2π (I/k)½ = 2π (I/mgh)½
(27)
I = Rotational Moment of Inertia (MI)
h = Distance between CG and point of support
The Simple Pendulum
 An application of Simple Harmonic Motion
 A mass m at the end of a massless
rod of length L
L
 There is a restoring force
which acts to restore the
mass to =0

m
F   mg sin 
T
• Compare to the spring F=-kx
• The pendulum does not
display SHM
mgsin
mg
 But for very small  (rad), we can make the
 simple pendulum approximation
sin     F  mg This is SHM
since s  r  L Arc length
s
mg Looks like spring force
F  mg  
s
L
L
mg
Like the spring
( Fs  kx)  k 
constant
L
 Now, consider the angular frequency of the
spring
k
mg / L
g




m
m
L
1 g Simple
f 
pendulum
2 L
Simple
pendulum
angular
frequency
frequency
• With this , the same equations expressing the
displacement x, v, and a for the spring can be used
for the simple pendulum, as long as  is small
• For  large, the SHM equations (in terms of sin and
cos) are no longer valid  more complicated
functions are needed (which we will not consider)
• A pendulum does not have to be a point-particle
The Physical Pendulum
 A rigid body can also be a pendulum
 The simple pendulum has a moment of inertia
 Rewrite  in terms of I
g
mg
mgL



2
L
mL
mL
mgL
1 mgL

or f 
I
2
I
I  mL
2
L
cg
m
mg
 L is the distance from the rotation axis to the
center of gravity
Damped Harmonic Motion
 Simple harmonic motion in which the amplitude is
steadily decreased due to the action of some nonconservative force(s), i.e. friction or air resistance
(F=- bv, where b is the damping coefficient)
 Classifications of damped harmonic motion:
1. Underdamped – oscillation, but amplitude
decreases with each cycle (shocks)
2. Critically damped – no oscillation, with
smallest amount of damping
3. Overdamped – no oscillation, but more
damping than needed for critical
• Apply Newton’s 2nd Law
 Fx  kx  bv  max
2
dx
d x
 kx  b
m
2
dt
dt
d 2x
b dx
k


kx  0
2
dt
m dt
m
• Another 2nd-order ordinary differential
equation, but with a 1st-order term
b
 2m
t
x  A0e cos(t  0 )
• The solution is
• Where   02  (b /2m)2
0  k /m
• Type of damping determined by comparing
0
and
b/2m

Envelope of
damped
motion
A=A0e-bt/2m
SHM
underdamped
Overdamped
Critically
damped
Underdamped
SHM
Forced Harmonic Motion
 Unlike damped harmonic motion, the amplitude
may increase with time
 Consider a swing (or a pendulum) and apply a
force that increases with time; the amplitude will
increase with time
Forced HM
SHM
 Consider the spring-mass system, it has a
frequency
1
f  2 k / m  f 0
 We call this the natural frequency f0 of the
system. All systems (car, bridge, pendulum, etc.)
have an f0
 We can apply a time-dependent external driving
force with frequency fd (fd f0) to the spring-mass
system
F (t )  Fext cos( 2 f d t )
 This is forced harmonic motion, the amplitude
increases
 But if fd=f0, the amplitude can increase
dramatically – this is a condition called resonance
Reference
• Halliday; Resnick; Walker. Fundamental of
Physics 8th edition extended
```