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When a moving fluid is contained in a horizontal pipe, all parts of it
have the same elevation, i.e. y1
2
1
= y2, and Bernoulli’s equation simplifies to:
2
2
P1 + ρ v = P2 + ρ v
1
2
1
2
Thus, P + ½ρv2 remains constant, so that
If v increases à P decreases
If v decreases à P increases
Applications of Bernoulli’s Equation
Conceptual Example: Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.
Applications of Bernoulli’s Equation
Applications of Bernoulli’s Equation
Chapter 11
Vibrations and
Waves
The Ideal Spring and Simple Harmonic Motion
If you apply a force FxApplied to an ideal spring, it will compress or
expand by an amount x according to the relation,
Applied
x
F
= kx
spring constant
Units: N/m
The Ideal Spring and Simple Harmonic Motion
Example: A Tire Pressure Gauge
-- revisited.
The spring constant of the spring
is 320 N/m and the bar indicator
extends 2.0 cm. What force does the
air in the tire apply to the spring?
What pressure in psi does this reading
correspond to if the diameter of the
air chamber cross section is 6.38 mm?
The Ideal Spring and Simple Harmonic Motion
FxApplied = k x
= (320 N m ) ( 0.020 m ) = 6.4 N
FxApplied
6.4
6.4
5
P=
=
=
=
2.00
×10
Pa
2
−5
A
! 0.00638 $ 3.20 ×10
π#
&
" 2 %
Since 1 pound-per-square-inch (psi) ≅ 6895 Pa
1 psi
∴ P = 2.00 ×10 5 Pa ×
= 29.0 psi
6895 Pa
⇒ normal tire pressure
The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
The restoring force on an ideal spring is
Fx = − k x
A mass-spring system
undergoing Simple
Harmonic Motion
Simple harmonic motion is a type of periodic motion where the restoring
force on a mass is directly proportional to the displacement. The
displacement of the mass has a sinusoidal dependence on the elapsed
time, i.e. depends on time through a sine or cosine function.
Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
PE elastic = 12 kx 2
SI Unit of Elastic Potential Energy: joule (J)
Energy and Simple Harmonic Motion
Since the restoring force due to an ideal spring is a conservative force, the
total mechanical energy of a spring-mass system is conserved, and is
E = KE + PE = 12 mv 2 + 12 kx 2
Since v = 0 at ± xmax
2
⇒ E = 12 kxmax
2
∴ 12 kxmax
= 12 mv 2 + 12 kx 2
⇒ v=±
k
2
xmax
− x2
m
v = ±vmax = ±xmax
k
at x = 0
m
Simple Harmonic Motion and the Reference Circle
As an analogy with the sinusoidal dependence of displacement on time in SHM,
consider the projection on the x-axis of the position of a ball undergoing uniform
circular motion on a turntable. Record how x depends on time with a moving roll of film.
Simple Harmonic Motion and the Reference Circle
Assume that the turntable is rotating
at the constant angular speed ω and
the ball is at radius A.
Since θ = ω t,
⇒ x = A cosθ = A cosω t
Simple Harmonic Motion and the Reference Circle
amplitude A: the maximum displacement
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
1
f =
T
2π
ω = 2π f =
T
Simple Harmonic Motion and the Reference Circle
VELOCITY
vx = −vT sin θ = − !
Aω sin ω t
vmax
Simple Harmonic Motion and the Reference Circle
Example: The Maximum Speed of a Loudspeaker Diaphragm
The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm.
(a) What is the maximum speed of the diaphragm?
(b) Where in the motion does this maximum speed occur?
Simple Harmonic Motion and the Reference Circle
vx = −vT sin θ = − !
Aω sin ω t
vmax
(a)
(
) (
vmax = Aω = A(2π f ) = 0.20 ×10 −3 m (2π ) 1.0 ×10 3 Hz
)
= 1.3 m s
(b) The maximum speed
occurs midway between
the ends of its motion.
Simple Harmonic Motion and the Reference Circle
ACCELERATION
ax = −ac cosθ = − !
Aω 2 cosω t
amax
Simple Harmonic Motion and the Reference Circle
Find a relationship between vx and x for the reference circle:
x = A cosω t vx = −Aω sin ω t
2
2
x
v
cos2 ω t + sin 2 ω t = 1 ⇒ 2 + 2 x 2 = 1
A Aω
∴vx = ±ω A 2 − x 2
Compare this equation with the equation for v as a function of x that we got
earlier for the spring-mass system from conservation of energy, i.e.
k
2
v=±
xmax
− x2
m
This suggests that the sinusoidal equations of the reference circle for x, v
and a are also valid for the spring-mass system if we set
k
ω=
and A = xmax
m
Simple Harmonic Motion and the Reference Circle
Another way to extract the frequency of vibration of the spring-mass system:
Use the reference circle equations and Newton’s 2nd law.
x = A cos ω t
a x = − Aω 2 cos ω t
∑ F = −kx = ma
− kA = −mAω 2
k
ω=
m
x
Simple Harmonic Motion and the Reference Circle
Example: A Body Mass Measurement Device
The device consists of a spring-mounted chair in which the astronaut
sits. The spring has a spring constant of 606 N/m and the mass of
the chair is 12.0 kg. The measured period is 2.41 s. Find the mass of the
astronaut.
Simple Harmonic Motion and the Reference Circle
k
ω=
mtotal
mtotal = k ω 2
ω = 2π f =
mtotal =
mastro
k
= mchair + mastro
2
(2π T )
k
=
− mchair
2
(2π T )
2
(
606 N m )(2.41 s )
=
− 12.0 kg = 77.2 kg
4π 2
2π
T
Example: Simple Harmonic Motion of a vertical spring-mass system.
A 0.20 kg ball is attached to a vertical spring. The spring constant is 28 N/m.
Find the equilibrium position of the mass with respect to the unstrained position
StickiesSHM about this Stickies
of the spring and find the amplitude of the
equilibrium position
after the ball is dropped from the unstrained position.
Equilibrium position:
∑ F = ma ⇒ kd
0
− mg = 0
mg ( 0.20 ) ( 9.8)
∴d0 =
=
= 0.070 m
k
28
Amplitude of oscillations:
E f = E0
1
2
Stickies
2
kdmax
= mgdmax
0.20 ) ( 9.8)
mg
(
dmax = 2
=2
= 0.140 m
k
28
∴ A = dmax − d0 = 0.140 − 0.070 = 0.070 m
}
}
A
dmax
A
Stickies
The Pendulum
A simple pendulum consists of a mass attached to a frictionless
pivot by a cable of negligible mass.
FT = −mgsin θ
For small angles: sin θ ≈ θ
s
mg
Since θ =
⇒ FT ≈ −
s ↔ F = −kx
L
L
mg
k
mg
g
⇒k↔
⇒ω =
=
=
L
m
mL
L
g
∴ω =
Pendulum (small angles only)
L
The Pendulum
Example: Keeping Time
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2π
g
ω = 2π f =
=
T
L
T 2g
L= 2
4π
2
(
)
T 2 g (1.00 s ) 9.80 m s 2
L=
=
= 0.248 m
2
2
4π
4π
Damped Harmonic Motion
In simple harmonic motion, an object oscillates
with a constant amplitude.
In reality, friction or some other energy
dissipating mechanism is always present
and the amplitude decreases as time
passes.
This is referred to as damped harmonic
motion.
Damped Harmonic Motion
Different types of damped harmonic motion:
1)  simple harmonic motion – amplitude stays constant
2&3) underdamped – amplitude decreases but still oscillations
4)  critically damped – amplitude decreases to 0 without oscillations in
shortest possible time
5)  Overdamped – amplitude decreases to 0 without oscillations slower
than in critically damped case