#### Transcript Slide 1

```4.1c Further Mechanics
SHM & Oscillations
Breithaupt pages 34 to 49
January 4th 2012
AQA A2 Specification
Lessons
1 to 3
4 to 6
7 to 9
Topics
Simple harmonic motion
Characteristic features of simple harmonic motion.
Condition for shm: a = − (2πf )2 x
X = A cos 2πf t and v = ± 2πf √(A2 − x 2)
Graphical representations linking x, v, a and t .
Velocity as gradient of displacement-time graph.
Maximum speed = 2πfA.
Maximum acceleration = (2πf )2 A.
Simple harmonic systems
Study of mass-spring system. T = 2π√(m / k)
Study of simple pendulum. T = 2π√(l / g)
Variation of Ek, Ep and total energy with displacement, and with time.
Forced vibrations and resonance
Qualitative treatment of free and forced vibrations.
Resonance and the effects of damping on the sharpness of resonance.
Phase difference between driver and driven displacements.
Examples of these effects in mechanical systems and stationary wave
situations
Oscillations (definitions)
An oscillation is a repeated
Time
Time
Time
Time
Time
====1=
05T
3T
7T
period
T
period
// 8
48
2
4
The fixed point, known as
the equilibrium position, is
where the oscillating object
returns to once the
oscillation stops.
The time period, T of an
oscillation is the time taken
for an object to perform one
complete oscillation.
equilibrium position
The amplitude, A of an
oscillation is equal to the
maximum value of the
displacement, x
Frequency, f in hertz is equal to
the number of complete
oscillations per second.
also: f = 1 / T
x = -+
0AA
Angular frequency, ω in
radians per second is given by:
ω = 2π f
equilibrium position
or
ω = 2π / T
Simple Harmonic Motion
Many oscillating systems undergo a pattern of oscillation
that is, or approximately the same as, that known as
Simple Harmonic Motion (SHM).
Examples
(including some that are approximate):
A mass hanging from the end of a spring
Molecular oscillations
A pendulum or swing
A ruler oscillating on the end of a bench
The oscillation of a guitar or violin string
Tides
Breathing
The pattern of SHM motion
The pattern of SHM is the same as the side view of an
object moving at a constant speed around a circular path
The amplitude of the
oscillation is equal to
+A
-A
The object moves quickest as
it passes through the central
equilibrium position.
The time period of the
oscillation is equal to the
time taken for the object to
complete the circular path.
Conditions required for SHM
When an object is performing SHM:
1. Its acceleration is proportional to its displacement
from the equilibrium position.
2. Its acceleration is directed towards the
equilibrium position.
Mathematically the above can be written: a = - k x
where k is a constant and the minus sign indicates that the
acceleration, a and displacement, x are in opposite
directions
Acceleration variation of SHM
a
The constant k is equal to
(2πf )2 or ω2 or (2π/T)2
+ ω2 x
Therefore:
a = - (2πf )2 x
x
-A
+A
(given on data sheet)
or
a=-
ω2
a=-
(2π/T)2
- ω2 x
x
or
x
gradient = - ω2 or - (2πf )2
Question
A body oscillating with SHM has a period of 1.5s and
amplitude of 5cm. Calculate its frequency and maximum
acceleration.
(a) f = 1 / T
= 1 / 1.5s
frequency = 0.67 Hz
(b) a = - (2πf )2 x
maximum acceleration is when x = A (the amplitude)
a = - (2πf )2 A
= - (2π x 0.6667Hz)2 x 0.015m
maximum acceleration = 0.88 ms-2
Displacement equations of SHM
The displacement, x of the oscillating object
varies with time, t according to the equations:
x = A cos (2π f t)
given on data sheet
or
x = A cos (ω t)
or
x = A cos ((2π / T) t)
Note: At time, t = 0, x = +A
Question
A body oscillating with SHM has a frequency of 50Hz and
amplitude of 4.0mm. Calculate its displacement and
acceleration 2.0ms after it reaches its maximum
displacement.
Maximum displacement = amplitude, A = 4.0mm
(a) x = A cos (2π f t)
= 4.0mm x cos (2π 50 x 2.0ms)
= 4.0mm x cos (2π 50 x 0.0020s)
= 0.6283
displacement = 0.628 mm
(b) a = - (2πf )2 x
= - (2π x 50)2 x 0.6283mm
= - (100π)2 x 0.0006283m
acceleration = - 62 ms-2
Velocity equations of SHM
The velocity, v of an object oscillating with SHM
varies with displacement, x according to the
equations:
v = ± 2πf √(A2 − x 2)
given on data sheet
or
v = ± ω√(A2 − x 2)
or
v = ± (2π / T) √(A2 − x 2)
Question
A body oscillating with SHM has a period of 4.0ms and amplitude of
30μm. Calculate (a) its maximum speed and (b) its speed when its
displacement is 15μm.
f=1/T
= 1 / 4.0ms
= 1 / 0.004s
= 250Hz
(a) v = ± 2πf √(A2 − x 2)
maximum speed occurs when x = 0
vmax = ± 2πf √(A2) = ± 2πf A
= ± 2π x 250 x 30μm
= ± 2π x 250 x 0.000 030m
maximum speed = 0.0471 ms-1
(b) v = ± 2πf √(A2 − x 2)
= ± 2π x 250 x √((0.000 030m)2
− (0.000 015m)2)
= ± 500π x √((0.000 030m)2
− (0.000 015m)2)
= ± 500π x √((9 x 10 -10)
− (2.25 x 10 -10))
= ± 500π x √(6.75 x 10 -10)
= ± 500π x 2.598 x 10 -5
speed = 0.0408 ms-1
Variation of x, v and a with time
The acceleration, ax depends on the resultant force, Fspring on the mass.
Note that the acceleration ax is always in the opposite direction to the
displacement, X.
SHM time graphs
+A
x
x
v
a
time
v
T/4
T/2
3T/4
T
5T/4
3T/2
a
-A
x = A cos (2π f t)
v = - 2π f A sin (2π f t)
a = - (2π f )2 A cos (2π f t)
Note:
vmax = ± 2π f A
amax = ± (2π f )2 A
The velocity curve is the gradient of the displacement curve
and the acceleration curve is the gradient of the velocity curve.
SHM summary table
Displacement
Velocity
Acceleration
+A
A0?
-B(2π
? f )2 A
C0
?
± 2π f A
D?
0
E-?A
0
2
+
E (2π
? f) A
SHM graph question
The graph below shows how the acceleration, a of an object
undergoing SHM varies with time. Using the same time axis show
how the displacement, x and velocity, v vary in time.
a
x
a
v
v
time
T/4
T/2
3T/4
T
5T/4
3T/2
x
acceleration is proportional to
minus displacement
displacement
The spring-mass system
If a mass, m is hung from a
spring of spring constant, k
and set into oscillation the
time period, T of the
oscillations is given by:
T = 2π√(m / k)
Mass on spring - Fendt
AS Reminder: Spring Constant, k:
This is the force in newtons required to cause a change
of length of one metre.
k = F / ΔL
unit of k = Nm-1
Question
A spring extends by 6.0 cm when a mass of 4.0 kg is hung from it near
the Earth’s surface (g = 9.8ms-2). If the mass is set into vertical
oscillation state or calculate the period (a) near the Earth’s surface and
(b) on the surface of the Moon where g = 1.7 ms-2.
The spring constant, k is given by:
k = F / ΔL
= (4.0 x 9.8) / 0.060m
= 653.3 Nm-1
(a) T = 2π√(m / k)
= 2π √(4.0 / 653.3)
= 2π √(0.006122)
time period = 0.49s
(b) g does not appear in the time period equation of the spring.
Therefore the period on the Moon is the same = 0.49s
The simple pendulum
A simple pendulum consists of:
• a point mass
• undergoing small oscillations (less than 10°)
• suspended from a fixed support
• by a massless, inextendable thread of length, L
• within a gravitational field of strength, g
The time period, T is given by:
T = 2π√(L / g)
Simple Pendulum - Fendt
Question
Calculate: (a) the period of a pendulum of length 20cm on
the Earth’s surface (g = 9.81ms-2) and (b) the pendulum
length required to give a period of 1.00s on the surface of
the Moon where g = 1.67ms-2.
(a) T = 2π√(L / g)
= 2π √(0.20m / 9.81ms-2)
= 2π √(0.204)
time period = 0.90s
(b) T = 2π√(L / g)
becomes: T2 = 4π2 (L / g)
becomes: L = T2g / 4π2
= ((1.00)2 x 1.67) / 4π2
length = 0.0423m (4.23cm)
Free oscillation
A freely oscillating object
oscillates with a constant
amplitude.
The total of the potential and
kinetic energy of the object
will remain constant.
EP + EK = a constant
This occurs when there are no
frictional forces acting on the
object such as air resistance.
Energy variation in free oscillation
Position 5
1 – Maximum
2
3
4
Equilibriumdisplacement
position
Potential
E
=Total
0EP +energy,
energy,
EK
EEPT= Total energy, ET
T =
P
Kinetic
E
0 energy,
energy,
EK E
=T0
K = Total
1
2
5
3
4
equilibrium position
position
EP
EK
1
ET
0
3
0
ET
5
ET
0
Question
A simple pendulum consists
of a mass of 50g attached to
the end of a thread of length
60cm. Calculate: (a) the
period of the pendulum
(g = 9.81ms-2) and (b) the
maximum height reached by
the mass if the mass’s
maximum speed is 1.2 ms-1.
(b) The maximum speed
occurs at the equilibrium
position, when the mass is at
its lowest position so that its
potential energy. EP = 0
(a) T = 2π√(L / g)
= 2π √(0.60m / 9.81ms-2)
= 2π √(0.06116)
time period = 1.56s
The maximum height occurs
when EP = EK
kinetic energy, EK = ½ m v2
= ½ x (0.050kg) x (1.2ms-1)2
= 0.036 J
so m x g x h = 0.036 J
h = 0.036 / (0.050 x 9.81)
maximum height = 0.074m
Energy variation with an oscillating spring
The strain potential
energy is given by:
EP = ½ k x2
Therefore the maximum
potential energy of an
oscillating spring
system
= ½ k A2
= Total energy of the
system, ET
But: ET = EP + EK
½ k A2 = ½ k x2 + EK
And so the kinetic
energy is given by:
EK = ½ k A2 - ½ k x2
EK = ½ k (A2 - x2)
Energy versus displacement graphs
energy
ET
EK
The kinetic energy curve is an
inverted parabola, given by:
EK = ½ k (A2 - x2)
EP
-A
The potential energy curve is
parabolic, given by:
EP = ½ k x2
0
+A
displacement, x
The total energy ‘curve’ is a
horizontal line such that:
ET = EP + EK
Energy versus time graphs
Displacement varies with time
according to:
x = A cos (2π f t)
Therefore the potential energy
curve is cosine squared,
given by:
EP = ½ k A2 cos2 (2π f t)
energy
½ kA2 ET
EP
EK
0
T/4
T/2
3T/4
T
time, t
½ k A2 = total energy, ET.
and so: EP = ET cos2 (2π f t)
Kinetic energy is given by:
EK = ET - EP
EK = ET - ET cos2 (2π f t)
EK = ET (1 - cos2 (2π f t))
EK = ET sin2 (2π f t)
EK = ½ k A2 sin2 (2π f t)
Damping
Damping occurs when frictional
forces cause the amplitude of an
oscillation to decrease.
The amplitude falls to zero with the
oscillating object finishing in its
equilibrium position.
The total of the potential and kinetic
energy also decreases.
The energy of the object is said to
be dissipated as it is converted to
thermal energy in the object and its
surroundings.
Types of damping
1. Light Damping
In this case the amplitude gradually
decreases with time.
The period of each oscillation will
remain the same.
The amplitude, A at time, t will be
given by: A = A0 exp (- C t)
where A0 = the initial amplitude and
C = a constant depending on the
system (eg air resistance)
displacement
A0
critical damping
heavy damping
time
light damping
2. Critical Damping
In this case the system
returns to equilibrium,
without overshooting, in the
shortest possible time after it
has been displaced from
equilibrium.
3. Heavy Damping
In this case the system
returns to equilibrium more
slowly than the critical
damping case.
Forced oscillations
All undamped systems of bodies have a frequency with which
they oscillate if they are displaced from their equilibrium position.
This frequency is called the natural frequency, f0.
Forced oscillation occurs when a system is made to oscillate by a
periodic force. The system will oscillate with the applied
frequency, fA of the periodic force.
The amplitude of the driven system will depend on:
1. The damping of the system.
2. The difference between the applied and natural frequencies.
Resonance
The maximum amplitude occurs when the
applied frequency, fA is equal to the natural
frequency, f0 of the driven system.
This is called resonance and the natural
frequency is sometimes called the resonant
frequency of the system.
Resonance curves
amplitude of driven
system, A
very light damping
light damping
more damping
driving
force
amplitude
f0
applied force frequency, fA
Notes on the resonance curves
If damping is increased then the amplitude of the driven
system is decreased at all driving frequencies.
If damping is decreased then the sharpness of the peak
amplitude part of the curve increases.
The amplitude of the driven system tends to be:
- Equal to that of the driving system at very low
frequencies.
- Zero at very high frequencies.
- Infinity (or the maximum possible) when fA is
equal to f0 as damping is reduced to zero.
Phase difference
The driven system’s oscillations are always behind those
of the driving system.
The phase difference lag of the driven system depends
on:
1. The damping of the system.
2. The difference between the applied and natural
frequencies.
At the resonant frequency the
phase difference is π/2 (90°)
Phase difference curves
applied force frequency, fA
- 90°
more damping
less damping
- 180°
phase difference of driven system
compared with driving system
f0
Examples of resonance
• Pushing a swing
• Musical instruments (eg
stationary waves on
strings)
and TVs
• Orbital resonances of
moons (eg Io and Europa
around Jupiter)
wires or bridges (Tacoma
Narrows)
The Tacoma Narrows Bridge Collapse
An example of resonance
caused by wind flow.
Washington State USA,
November 7th 1940.
– 4 minutes – with commentary
– 3 minutes – newsreel footage
– 6 minutes - music background only
Motion in 2D - PhET - Learn about velocity and acceleration vectors. Move the ball with the
mouse or let the simulation move the ball in four types of motion (2 types of linear, simple
harmonic, circle). See the velocity and acceleration vectors change as the ball moves.
Pendulum Lab - PhET - Play with one or two pendulums and discover how the period of a simple
pendulum depends on the length of the string, the mass of the pendulum bob, and the
amplitude of the swing. It's easy to measure the period using the photogate timer. You can
vary friction and the strength of gravity. Use the pendulum to find the value of g on planet X.
Notice the anharmonic behavior at large amplitude
SHM & circular motion compared - NTNU
SHM (spring) and circular motion - netfirms
SHM with a spring & pendulum - Explore Science
Mass on spring - Fendt
Mass on a spring - NTNU
Mass on spring oscillations - NTNU
Spring oscillation with varying mass and k - NTNU
Spring oscillation showing graphs of s, v & a - NTNU
Simple Pendulum - Fendt
Pendulum in an accelerated car - NTNU
Pendulum with rotating suspension point - © 1998 Franz-Josef Elmer
Internet Links (Forced Oscillation & Damping)
Spring Oscillation - PhET - A realistic mass and spring laboratory.
Hang masses from springs and adjust the spring stiffness and
damping. You can even slow time. Transport the lab to different
planets. A chart shows the kinetic, potential, and thermal energy
for each spring.
Undamped and undriven pendulum - © 1998 Franz-Josef Elmer
Driven spring oscillator with one mass - Explore Science
Driven spring oscillator with two masses - Explore Science
Forced oscillation - Fendt
Coupled Penduala - Fendt
Pendulum driven by a periodic force - © 1998 Franz-Josef Elmer
Horizontally driven pendulum - © 1998 Franz-Josef Elmer
Vertically driven pendulum - © 1998 Franz-Josef Elmer
Resonance in a string - netfirms
RCL Resonance - netfirms
Core Notes from Breithaupt pages 34 to 49
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
What is an oscillation?
In the context of oscillations explain the meaning of the following: (a) equilibrium; (b)
displacement; (c) amplitude; (d) time period; (e) frequency; (f) angular frequency and
(g) phase difference.
Explain the conditions (in terms of acceleration and displacement) required for an
object to be undergoing simple harmonic motion.
State the defining equation for SHM shown on page 37.
Sketch the graphs shown on page 36 and explain the mathematical relationships
between them.
(a) State the equation for the time period for a mass-spring system. (b) Calculate the
mass required to give a mass-spring system of spring constant 25N/m a time period
of one second.
(a) State the equation for the time period for a simple pendulum. (b) Calculate the
length required to give a pendulum a time period of one second near the Earth’s
surface.
How, if at all would your answers to Q6(b) and Q7(b) be different if you were on the
Moon’s surface? (gMoon = 1.7 N/kg)
Explain how potential, kinetic and total energy vary during the oscillation of a simple
pendulum. (assume no resistive forces).
What is meant by ‘resonance’?
Sketch and explain the resonance curves on page 47.
Explain how damping affects the phase difference between the driving force and the
driven system.
Notes from Breithaupt pages 34 & 35
Oscillations
1.
2.
What is an oscillation?
In the context of oscillations explain the meaning of the
following: (a) equilibrium; (b) displacement; (c)
amplitude; (d) time period; (e) frequency; (f) angular
frequency and (g) phase difference.
3.
Try the summary questions on page 35
Notes from Breithaupt pages 36 & 37
The principles of simple harmonic motion
1.
2.
3.
4.
Explain the conditions (in terms of acceleration and
displacement) required for an object to be undergoing
simple harmonic motion.
State the defining equation for SHM shown on page 37.
Sketch the graphs shown on page 36 and explain the
mathematical relationships between them.
Try the summary questions on page 37
Notes from Breithaupt pages 38 & 39
1.
2.
Explain the ways in which circular motion and SHM
correspond with each other.
Try the summary questions on page 39
Notes from Breithaupt pages 40 to 43
Applications of simple harmonic motion
1.
2.
3.
4.
5.
(a) State the equation for the time period for a mass-spring system.
(b) Calculate the mass required to give a mass-spring system of
spring constant 25N/m a time period of one second.
(a) State the equation for the time period for a simple pendulum. (b)
Calculate the length required to give a pendulum a time period of
one second near the Earth’s surface.
How, if at all would your answers to Q1(b) and Q2(b) be different if
you were on the Moon’s surface? (gMoon = 1.7 N/kg)
Derive the equations for (a) the spring-mass system and (b) the
simple pendulum.
Try the summary questions on page 43
Notes from Breithaupt pages 44 to 46
Energy and simple harmonic motion
1.
2.
3.
4.
5.
Explain how potential, kinetic and total energy vary
during the oscillation of a simple pendulum. (assume no
resistive forces).
to Q1?
Repeat Q1, this time for a mass-spring system.
Describe with the aid of suitable graphs the different
types of damping.
Try the summary questions on page 46
Notes from Breithaupt pages 47 to 49
Forced oscillations and resonance
1.
2.
3.
What is meant by ‘resonance’?
Sketch and explain the resonance curves on page 47.
Explain how damping affects the phase difference
between the driving force and the driven system.
4.
Explain the demonstration ‘Barton’s Pendulum’ on page
48.
Give and explain an example of resonance with
bridges.
Try the summary questions on page 49
5.
6.
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