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General Physics (PHY 2130)
Lecture 24
•  Solids and fluids
  Fluids in motion
•  Oscillations
  Simple harmonic motion
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Solids and fluids
  different states of matter; fluids
  density, pressure, etc.
Review Problem: A piece of metal is released under water. The volume of the
metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration?
(Note: when v = 0, there is no drag force.)
3
Example: A piece of metal is released under water. The volume of the metal is
50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note:
when v = 0, there is no drag force.)
Idea: apply Newton’s 2nd Law
to the piece of metal:
y
FBD for
the metal
FB
∑F = F
B
x
w
− w = ma
The magnitude of the buoyant force equals the
weight of the fluid displaced by the metal.
FB = ρ waterVg
Solve for a:
⎛ ρ waterV
⎞
ρ waterVg
FB
a=
−g =
− g = g ⎜
− 1⎟
⎜ ρ V
⎟
m
ρobjectVobject
⎝ object object ⎠
4
Example continued:
Since the object is completely submerged V=Vobject.
specific gravity =
ρ
ρ water
where ρwater = 1000 kg/m3 is the density of
water at 4 °C.
Given
ρ object
specific gravity =
= 5.0
ρ water
⎛ ρ waterV
⎞
⎛ 1
⎞
⎛ 1
⎞
⎜
⎟
a=g
− 1 = g ⎜
− 1⎟ = g ⎜
− 1⎟ = −7.8 m/s2
⎜ ρ V
⎟
⎝ S .G. ⎠
⎝ 5.0 ⎠
object
object
⎝
⎠
5
Fluid Flow
A moving fluid will exert forces parallel to the surface over which it
moves, unlike a static fluid. This gives rise to a viscous force that
impedes the forward motion of the fluid.
A steady flow is one where the velocity at a given point in a fluid is
constant. Steady flow is laminar; the fluid flows in layers.
An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity.
V1 =
constant
V2 =
constant
v1≠v2
Fluids in Motion: Streamline Flow
•  Streamline flow
•  every particle that passes a particular point moves exactly
along the smooth path followed by particles that passed the
point earlier
•  also called laminar flow
•  Streamline is the path
•  different streamlines cannot cross each other
•  the streamline at any point coincides with the direction of fluid
velocity at that point
Fluids in Motion: Turbulent Flow
•  The flow becomes irregular
•  exceeds a certain velocity
•  any condition that causes abrupt changes in velocity
7
Equation of Continuity
The amount of mass that flows though the cross-sectional area A1 is the
same as the mass that flows through cross-sectional area A2.
ΔV
= Av
Δt
is called the volume flow rate (units m3/s)
8
Δm
= ρAv
Δt
is the mass flow rate (units kg/s)
The continuity equation is
ρ1 A1v1 = ρ2 A2v2
If the fluid is incompressible, then ρ1= ρ2.
9
Example: A garden hose of inner radius 1.0 cm carries water at 2.0 m/s. The
nozzle at the end has radius 0.20 cm. How fast does the water move through
the constriction?
A1v1 = A2v2
⎛ A1 ⎞
⎛ πr12 ⎞
v2 = ⎜⎜ ⎟⎟v1 = ⎜⎜ 2 ⎟⎟v1
⎝ A2 ⎠
⎝ πr2 ⎠
2
⎛ 1.0 cm ⎞
= ⎜
⎟ (2.0 m/s) = 50 m/s
⎝ 0.20 cm ⎠
Oscillations
Simple Harmonic Motion
Recall: Hooke’s Law
•  Fs = - k x
•  Fs is the spring force
•  k is the spring constant
•  It is a measure of the stiffness of the spring
•  A large k indicates a stiff spring and a small k indicates a soft spring
•  x is the displacement of the object from its equilibrium
position
•  The negative sign indicates that the force is always directed
opposite to the displacement
•  The force always acts toward the equilibrium position
•  It is called the restoring force
Hooke’s Law Applied to a
Spring – Mass System
•  When x is positive (to the
right), F is negative (to the left)
•  When x = 0 (at equilibrium), F
is 0
•  When x is negative (to the left),
F is positive (to the right)
Example: two springs
The springs 1 and 2 in Figure have spring
constants of 40.0 N/cm and 25.0 N/cm,
respectively. The object A remains at
rest, and both springs are stretched
equally. Determine the stretch.
Motion of the Spring-Mass System
•  Assume the object is initially pulled to x = A and
released from rest
•  As the object moves toward the equilibrium
position, F and a decrease, but v increases
•  At x = 0, F and a are zero, but v is a maximum
•  The object’s momentum causes it to overshoot
the equilibrium position
•  The force and acceleration start to increase in the
opposite direction and velocity decreases
•  The motion continues indefinitely
Simple Harmonic Motion
•  Motion that occurs when the net force along the
direction of motion is a Hooke’s Law type of force
•  The force is proportional to the displacement and in the
opposite direction
•  The motion of a spring mass system is an example
of Simple Harmonic Motion
•  Not all periodic motion over the same path can be
considered Simple Harmonic motion
•  To be Simple Harmonic motion, the force needs to
obey Hooke’s Law
Amplitude, Period and Frequency
•  Amplitude, A
•  The amplitude is the maximum position of the object relative to the
equilibrium position
•  In the absence of friction, an object in simple harmonic motion will
oscillate between ±A on each side of the equilibrium position
•  The period, T, is the time that it takes for the object to
complete one complete cycle of motion
•  From x = A to x = - A and back to x = A
•  The frequency, ƒ, is the number of complete cycles or
vibrations per unit time
Acceleration of an Object in Simple
Harmonic Motion
•  Newton’s second law will relate force and
acceleration
•  The force is given by Hooke’s Law
F=-kx=ma
or
a = -kx / m
•  The acceleration is a function of position
•  Acceleration is not constant and therefore the uniformly
accelerated motion equation cannot be applied
Elastic Potential Energy
•  The energy stored in a stretched or compressed
spring or other elastic material is called elastic
potential energy
PEs = ½kx2
•  The energy is stored only when the spring is
stretched or compressed
•  Elastic potential energy can be added to the
statements of Conservation of Energy and WorkEnergy
Energy in a Spring Mass System
• 
Consider a situation:
1.  A block sliding on a
frictionless system
collides with a light
spring
2.  The block attaches
to the spring
Energy Transformations
•  The block is moving on a frictionless surface
•  The total mechanical energy of the system is the
kinetic energy of the block
Energy Transformations, 2
•  The spring is partially compressed
•  The energy is shared between kinetic energy and elastic
potential energy
•  The total mechanical energy is the sum of the kinetic energy
and the elastic potential energy
Energy Transformations, 3
•  The spring is now fully compressed
•  The block momentarily stops
•  The total mechanical energy is stored as elastic
potential energy of the spring
Energy Transformations, 4
•  When the block leaves the spring, the total mechanical
energy is in the kinetic energy of the block
•  The spring force is conservative and the total energy of the
system remains constant
Velocity as a Function of Position
•  Conservation of Energy allows a calculation of the
velocity of the object at any position in its motion
k 2
v=±
A − x2
m
(
)
•  Speed is a maximum at x = 0
•  Speed is zero at x = ±A
•  The ± indicates the object can be traveling in either
direction
A Vertical Mass and Spring
When a mass-spring system is oriented vertically, it will exhibit SHM with
the same period and frequency as a horizontally placed system.
At equilibrium position (b)
∑F
y
= +kd − mg = 0
At (c), displaced the equilibrium by y
Fspring , y = k (d − y )
∑F
y
= k (d − y) − mg = kd − ky − mg = −ky
∑F
y
= −ky
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Example: oscillator
A block of mass 1.00 kg is attached to a spring with a spring constant of
30.0 N/m, which is stretched 0.200 m from its equilibrium position. How
much work must be done to stretch it an additional 0.100 m? What
maximum speed will the block attain if the system is then let go?
Simple Harmonic Motion and Uniform
Circular Motion
•  A ball is attached to the rim of
a turntable of radius A
•  The focus is on the shadow
that the ball casts on the
screen
•  When the turntable rotates
with a constant angular speed,
the shadow moves in simple
harmonic motion
Period and Frequency from Circular Motion
•  Period
m
T = 2π
k
•  This gives the time required for an object of mass m
attached to a spring of constant k to complete one cycle of
its motion
•  Frequency
1 1 k
ƒ= =
T 2π m
•  Units are cycles/second or Hertz, Hz
•  The angular frequency is related to the frequency
k
ω = 2πƒ =
m
Motion as a Function of Time
•  Use of a reference circle
allows a description of
the motion
•  x = A cos (2πƒt)
•  x is the position at time t
•  x varies between
+A and -A
Graphical Representation of Motion
•  When x is a maximum
or minimum, velocity is
zero
•  When x is zero, the
velocity is a maximum
•  When x is a maximum in
the positive direction, a
is a maximum in the
negative direction
Verification of Sinusoidal Nature
•  This experiment
shows the sinusoidal
nature of simple
harmonic motion
•  The spring mass
system oscillates in
simple harmonic
motion
•  The attached pen
traces out the
sinusoidal motion
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Example: The period of oscillation of an object in an ideal mass-spring
system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the
equilibrium point?
Idea: let’s use energy conservation: at equilibrium x = 0:
1
1
1
E = K + U = mv 2 + kx 2 = mv 2
2
2
2
Since E=constant, at equilibrium (x = 0) the KE must be a maximum.
Thus, v = vmax = Aω.
ω=
2π
2π
=
= 12.6 rads/sec
T
0.50 s
and v = Aω = (5.0 cm )(12.6 rads/sec ) = 62.8 cm/sec