Test 2
Tuesday June 2 at 9:30 am
CNH-104
Energy, Work, Power,
Simple Harmonic Motion
Class will start at 11am
Physics 1B03summer - Lecture 7
Energy of a SHO
Physics 1B03summer - Lecture 7
Energy of a SHO
Recall that for a spring:
ETot = K + U
= 1/2mv2 + 1/2kx2
And we know that:
x=A cos(ωt+φ)
v=-A ω sin(ωt+φ)
Physics 1B03summer - Lecture 7
Energy in SHM
Look again at the block & spring
M
K  12 mv 2  12 m 2 A2 sin 2 (t   )
U  12 kx2  12 kA2 cos 2 (t   )
 k!
K  U  12 A2 m 2 sin 2 (t   )  k cos 2 (t   ) 
 12 kA2 sin 2 (t   )  cos 2 (t   ) 
 12 kA2  a constant (total mechanical energy)
Hence: ETot = ½kA2
We could also write E = K+U = ½ m(vmax )2
Physics 1B03summer - Lecture 7
Quiz
Suppose you double the amplitude of the motion, what
happens to the total energy?
a) Doubles
b) 4 x Larger
c) Doesn’t change
Physics 1B03summer - Lecture 7
Quiz
Suppose you double the amplitude of the motion, what
happens to the maximum speed ?
a) Doubles
b) 4 x Larger
c) Doesn’t change
Physics 1B03summer - Lecture 7
Quiz
Suppose you double the mass and amplitude of the object,
what happens to the maximum acceleration ?
a) Doubles
b) 4 x Larger
c) Doesn’t change
Physics 1B03summer - Lecture 7
Energy
Since we know the total energy of a SHM, we can
calculate the or displacement velocity at any point in
time:
ETot = ½ kA2 = K+U = ½ mv2 + ½ kx2
So, if x=0, all E is in kinetic, and v is at max
if x=A, all E is in potential, and v is zero
Physics 1B03summer - Lecture 7
Example 1
A 100g block is 5cm from the equilibrium position moving at 1.5m/s.
The angular frequency is 2 rad/s.
a) What is the total energy of the system ?
b) What is the amplitude of the oscillations ?
Physics 1B03summer - Lecture 7
Example 2
A 500g block on a spring is pulled 20cm and released.
The motion has a period of 0.8s.
What is the velocity when the block is 15.4cm from the
equilibrium ?
Physics 1B03summer - Lecture 7
Example 3
A 1.0kg block is attached to a spring with k=16N/m. While the
block is at rest, a student hits it with a hammer and almost
instantaneously gives it a speed of 40cm/s.
a)
b)
what is the amplitude of the subsequent oscillations ?
what is the block’s speed at the point where x=A/2 ?
Physics 1B03summer - Lecture 7
10 min rest
Physics 1B03summer - Lecture 7
SHM and Circular Motion
Uniform circular motion about in the
xy-plane, radius A, angular velocity  :
(t) = 0 +  t
(similar to, x=xo+vt)
A

and so
x  A cos  A cos( 0  t )
y  A sin   A sin( 0  t )
Hence, a particle moving in one dimension can be expressed
as an ‘imaginary’ particle moving in 2D (circle), or vice versa the ‘projection’ of circular motion can be viewed as 1D motion.
Physics 1B03summer - Lecture 7
x  A cos  A cos(o  t )
y  A sin   A sin(o  t )
Compare with our expression for 1-D SHM.
x  A cos(t  )
Result:
SHM is the 1-D projection
of uniform circular motion.
Physics 1B03summer - Lecture 7
Phase Constant, θo
For circular motion,
the phase constant
is just the angle at
which the motion started.
A

o
Physics 1B03summer - Lecture 7
Example
An object is moving in circular motion with an angular
frequency of 3π rad/s, and starts with an initial angle
of π/6. If the amplitude is 2.0m, what is the objects
angular position at t=3sec ?
What are the x and y values of the position at this time ?
Physics 1B03summer - Lecture 7
Simple Pendulum
Gravity is the “restoring force” taking the place of the
“spring” in our block/spring system.
L
Instead of x, measure the displacement as the arc length
s along the circular path.
θ
T
Write down the tangential component of F=ma:
Restoring force  mg sin 
d 2s
m 2  m at  m g sin( )
dt
But s  L
mg
s
mg sin θ
d 2
g
 2   sin 
dt
L
Physics 1B03summer - Lecture 7
Compare:
SHM:
Simple pendulum:
d 2x
2



x
2
dt
d 2
g
  sin 
2
dt
L
The pendulum is not a simple harmonic oscillator!
However, take small oscillations:
sin    (radians) if  is small.
Then
d 2
g
g
  sin    
2
dt
L
L
d 2
2




2
dt
Physics 1B03summer - Lecture 7
Simple Pendulum
Using sin(θ)~θ for small angles, we have the
following equation of motion:
d 2
g
 
2
dt
L
With:

g
L
------------------------------------------------------------------------Hence:
or:
gT 2
L 2 

4 2
g
2
4

L
2
g  L 
T2
Application - measuring height
- finding variations in g → underground resources
Physics 1B03summer - Lecture 7
For small  :
This looks like
d 2
g
 
2
dt
L
d 2x
2



x , with angle  instead of x.
2
dt
So, the position is given by  (t )   o cos(t   )
phase
constant
amplitude
(2 / period)
Physics 1B03summer - Lecture 7
Quiz:
A geologist is camped on top of a large deposit of nickel
ore, in a location where the gravitational field is 0.01%
stronger than normal. The period of his pendulum will be
a) longer.
b) same.
c) shorter.
(by how much, in percent?)
Physics 1B03summer - Lecture 7
Application
Pendulum clocks (“grandfather clocks”) often have a
swinging arm with an adjustable weight. Suppose
the arm oscillates with T=1.05sec and you want to
adjust it to 1.00sec. Which way do you move the
weight?
?
Physics 1B03summer - Lecture 7
Quiz
A simple pendulum hangs from the ceiling of an
elevator. If the elevator accelerates upwards, the period
of the pendulum:
a) Gets shorter
b) Gets larger
c) Stays the same
Question: What happens to the period of a simple
pendulum if the mass m is doubled?
Physics 1B03summer - Lecture 7
Example
A physicist wants to know the height of a building. He
notes that a long pendulum that extends from the
ceiling almost to the floor has a period of 12 s.
How tall is the building?
Physics 1B03summer - Lecture 7
Compare Springs and
Pendulum
Newton’s 2nd Law:
F  ma   kx
2
d x
k
 x
2
dt
m

k
m
F  m a   gmsin 
d 2
g
 
2
dt
L

g
L
Physics 1B03summer - Lecture 7
SHM and Damping – EXTRA !!!
SHM: x(t) = A cos ωt
Motion continues indefinitely.
Only conservative forces act,
so the mechanical energy is
constant.
Damped oscillator: dissipative
forces (friction, air resistance, etc.)
remove energy from the oscillator,
and the amplitude decreases with
time.
x
t
x
t
Physics 1B03summer - Lecture 7
A damped oscillator has external nonconservative
force(s) acting on the system. A common example
is a force that is proportional to the velocity.
f = bv where b is a constant damping coefficient
F=ma give:
dx
d 2x
 kx  b  m 2
dt
dt
For weak damping (small b), the solution is:
x
x(t )  Ae

b
t
2m
eg: green water
cos(t   )
A e-(b/2m)t
t
Physics 1B03summer - Lecture 7
10 min rest
Physics 1B03summer - Lecture 7
Wave Motion –
Chapter 20
•Qualitative properties of wave motion
•Mathematical description of waves in 1-D
•Sinusoidal waves
Physics 1B03summer - Lecture 7
A wave is a moving pattern. For example, a wave on a
stretched string:
v
time t
Δx = v Δt
t +Δt
The wave speed v is the speed of the pattern. No
particles move at this wave speed – it is the speed of
the wave.
However, the wave does carry energy and momentum.
Physics 1B03summer - Lecture 7
Transverse waves
The particles move
up and down
The wave moves this way
If the particle motion is perpendicular to the direction
the wave travels, the wave is called a “transverse wave”.
Examples: Waves on a string;
waves on water;
light & other electromagnetic waves;
some sound waves in solids (shear waves)
Physics 1B03summer - Lecture 7
Longitudinal waves
The particles move
back and forth.
The wave moves long distances
parallel to the particle motions.
Example: sound waves in fluids (air, water)
Even in longitudinal waves, the particle velocities
are quite different from the wave velocity. The
speed of the wave can be orders of magnitude
larger than the particle speeds.
Physics 1B03summer - Lecture 7
Quiz
B
A
C
wave motion
Which particle is moving at the highest speed?
A) A
B) B
C) C
D) all move with the same speed
Physics 1B03summer - Lecture 7
Non-dispersive waves: the wave always keeps the
same shape as it moves.
For these waves, the wave speed is determined
entirely by the medium, and is the same for all
sizes & shapes of waves.
eg. stretched string:
tension
T
v

mass/unit length

(A familar example of a dispersive wave is an
ordinary water wave in deep water. We will
discuss only non-dispersive waves.)
Physics 1B03summer - Lecture 7
Principle of Superposition
When two waves meet, the displacements add:
yobserved ( x, t )  y1 ( x, t )  y2 ( x, t )
(for waves in a “linear medium”)
So, waves can pass through each other:
v
v
Physics 1B03summer - Lecture 7
Quiz: “equal and opposite” waves
v
v
v
v
Sketch the particle velocities at the instant the
string is completely straight.
Physics 1B03summer - Lecture 7
Reflections
Waves (partially) reflect from any boundary
in the medium:
1) “Soft” boundary:
light string, or free end
Reflection is upright
Physics 1B03summer - Lecture 7
Reflections
2) “Hard” boundary:
heavy rope, or fixed end.
Reflection is inverted
Physics 1B03summer - Lecture 7
Quiz
A wave show can be described by the equation y=f(xvt),
as it travels along the x axis. It reflects from a fixed
end at the origin. The reflected wave is described by:
A)
B)
C)
D)
y=  f(xvt)
y=  f(x+vt)
y=  f( xvt)
y=  f( x+vt)
Physics 1B03summer - Lecture 7
The math: Suppose the shape of the wave at t = 0 is
given by some function y = f(x).
y
v
at time t = 0: y = f (x)
vt
at time t :
y
y = f (x - vt)
Note: y = y(x,t), a function of two variables;
f is a function of one variable
Physics 1B03summer - Lecture 7
Non-dispersive waves:
y (x,t) = f (x ± vt)
+ sign: wave travels towards –x
 sign: wave travels towards +x
f is any (smooth) function of one variable.
eg. f(x) = A sin (kx)
Physics 1B03summer - Lecture 7