Download SIMPLE HARMONIC MOTION (1)

SIMPLE HARMONIC MOTION (1)
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Consider an air track with a cart of mass m attached to
a spring of force constant k
When the spring is at equilibrium length (neither
stretched or compressed), the cart is at position x = 0
If the cart is displaced from equilibrium by a distance x,
the spring exerts a restoring force given by F = -kx
The spring exerts a restoring force whose magnitude is
proportional to the distance it is displaced from
equilibrium
The force exerted by a spring is opposite in direction to
its displacement from equilibrium; this accounts for the
minus sign in F = -kx
A restoring force is one that always points toward the
equilibrium position
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SIMPLE HARMONIC MOTION (2)
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If the cart is released from rest at the location x = A, the
spring exerts a force on the cart to the left, causing the
cart to accelerate toward the equilibrium position
When the cart reaches x = 0, the net force acting on it
is zero, but its speed is not zero so it continues to move
to the left
As the cart compresses the spring, it experiences a
force to the right, causing it to decelerate and come to
rest at x = -A
The spring continues to exert a force to the right; thus
the cart immediately moves to the right until it comes to
rest again at x = A, completing one oscillation in time T
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SIMPLE HARMONIC MOTION (3)
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If a pen is attached to the
cart, it can trace its
motion on a sheet of
paper moving with
constant speed
On this chart, we
obtain a record of
the cart’s motion as a function of time
The motion of the cart looks like a sine or cosine
function
The position of the cart as a function of time can be
represented by a sine or cosine function
The position of the mass oscillates between x = +A and
x = -A, where A is the amplitude of the motion and is
the extreme displacement in each direction
The amplitude is one half the total range of motion,
which repeats with a period T
The position of the cart is the same at time t + T as it is
at time t
Thus the mathematical description of position versus
time in simple harmonic motion is
x = Acos(2πt/T) (ensure calculator set to radians)
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SIMPLE HARMONIC MOTION:
EXAMPLE
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An air track cart attached to a spring completes one
oscillation every 2.4s. At t = 0 the cart is released from
rest at a distance of 0.1m from its equilibrium position.
What is the position of the cart at 0.3s, 0.6s, 2.7s and
3.0s?
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Equilibrium
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CONNECTIONS BETWEEN
UNIFORM CIRCULAR MOTION AND
SIMPLE HARMONIC MOTION
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A peg is placed at the rim of
a turntable that rotates with
constant angular velocity
Viewed from above, the
peg exhibits uniform
circular motion
Viewed from the side
it appears to move
back and forth in a
straight line, as we
can see by shining a light to cast a shadow of the peg
onto a back screen
The shadow of the peg moves with simple harmonic
motion
Thus there is a connection between uniform circular
motion and simple harmonic motion
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Equilibrium
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SIMPLE HARMONIC MOTION:
POSITION AS FUNCTION OF TIME
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Recall that angular position is simply θ = ωt
Imagine drawing a radius vector of length A to the
position of the peg
When we project the shadow of the peg onto the back
screen, the shadow is at a location x = Acosθ, which is
the x component of the radius vector
Thus for an object undergoing a simple harmonic
motion, its position as a function of time is given by
x = Acosθ = Acos(ωt) = Acos(2πt/T)
When dealing with simple harmonic motion, ω is called
the angular frequency
ω = 2πf = 2π/T rad/s
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VELOCITY IN SIMPLE HARMONIC
MOTION
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Recall that the velocity of an object in uniform circular
motion of radius r is v = rω
The velocity is tangential to the object’s circular path,
as indicated above
When the peg is at the angular position θ, the velocity
vector makes an angle θ with the vertical
Thus the x component of the velocity is –vsinθ
Thus vx = -vsinθ = -rωsinθ
Since we are only concerned with movement in the x
direction, that r = A and θ = ωt, the velocity in simple
harmonic motion is v = -Aωsin(ωt) m/s
Plotting x and v for SHM shows that when the
displacement from equilibrium is a maximum, the
velocity is zero
When x = +A and x = -A, object is momentarily at rest
The speed is a maximum when the displacement from
equilibrium is zero, i.e. vmax = Aω (since largest value of
sinθ =1)
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ACCELERATION IN SIMPLE
HARMONIC MOTION
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The acceleration of an object in uniform circular motion
is given by acp = rω2
Direction of acceleration: toward centre of circular path
When the angular position of the peg is θ, the
acceleration vector is at an angle θ below the x-axis,
and its x component is –acpcosθ
Setting r = A and θ = ωt the acceleration in simple
harmonic motion is a = -Aω2cos(ωt) m/s2
Position and acceleration for SHM are plotted below
Acceleration and position vary with time in the same
way, but with opposite signs
When the position has a maximum positive value, the
acceleration has a maximum negative value
After all, the restoring force is opposite to the position,
hence the acceleration a = F/m must also be opposite
to the position, thus a = -ω2x
Maximum acceleration occurs when –cosθ = 1, thus the
maximum acceleration amax = Aω2
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SIMPLE HARMONIC MOTION:
EXAMPLE
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A 747 is hit by turbulence. Upon landing, data from the
airliner’s black box indicates that the plane moved up
and down with an amplitude of 30.0m, and a maximum
acceleration of 1.8g. Treating the up and down motion
of the plane as simple harmonic, find the time required
for one complete oscillation. Also find the plane’s
maximum vertical speed.
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THE PERIOD OF A MASS ON A
SPRING
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Net force exerted by a spring with constant k, acting on
a mass m at position x is F = -kx
Since F = ma, it follows that ma = -kx
Substituting: m[-Aω2cos(ωt)] = -k[Acos(ωt)]
Which yields ω2 = k/m or ω = √(k/m)
Noting that ω = 2π/T, it follows that the period of a
mass on a spring is T = 2π√(m/k) seconds
The period increases as the mass increases, and
decreases with the spring’s force constant
A larger mass has greater inertia, and thus takes longer
for the mass to move back and forth through an
oscillation
A larger value of the force constant k indicates a stiffer
spring, which means a mass would complete an
oscillation is less time that it would on a softer spring
Example: A 0.120kg mass attached to a spring
oscillates with an amplitude of 0.075m and a maximum
speed of 0.524 m/s. Find the force constant and the
period of motion.
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A VERTICAL SPRING
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Up to now, we have considered springs that are
horizontal
When a mass m is attached to a vertical spring, it
causes the spring to stretch
The vertical spring is in equilibrium when it exerts an
upward force equal to the weight of the mass, i.e the
spring stretches by an amount y0
Thus, for vertical springs: ky0 = mg
Or y0 = mg/k
A mass on a spring oscillates around the equilibrium
point y = y0, and also exhibits simple harmonic motion
Example: A 0.26kg mass is
attached to a vertical spring.
When the mass is put into
motion, its period is 1.12s.
How much does the mass
stretch the spring when it
is at rest in its
equilibrium position?
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THE SIMPLE PENDULUM
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A simple pendulum consists of a mass m suspended by
a light string or rod of length L
The pendulum has a stable equilibrium when the mass
is directly below the suspension point, and oscillates
about this position if displaced from it
Consider the potential energy of the system
When the pendulum is at an angle θ with respect to the
vertical, the mass m is above its lowest point by a
vertical height L(1 – cosθ)
If the potential energy is 0 at θ = 0, then the potential
energy for general θ is: U = mgL(1 – cosθ)
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FORCES ACTING ON A PENDULUM
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The forces acting on the mass are due to gravity and
the tension in the string (both are vectors)
The tension acts in the radial direction, and keeps the
mass moving along its circular path
The net tangential force acting on m is just the
tangential component of its weight: F = mgsinθ
The direction of the net tangential force is always
toward the equilibrium point, thus F is a restoring force
For small angles sinθ ≈ θ, so the arc length s = Lθ
Equivalently θ = s/L
If the mass is
displaced from
equilibrium by a small
arc length s, the force
is F = mgsinθ ≈ mgθ
= (mg/L)s
Comparing to a mass
on a spring, F = kx
Letting x = s and
k = mg/L
Period of a pendulum
is the same as that of
a mass on a spring
T = 2π√(m/k) = 2π√(m/(mg/L)) = 2π√(L/g) seconds
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PENDULUM EXAMPLE
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A pendulum is constructed from a string 0.627m long
attached to a mass of 0.25kg. When set in motion, the
pendulum completes one oscillation every 1.59s. If the
pendulum is held at rest and the string is cut, how long
will it take for the mass to fall through a distance of
1.0m?
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DAMPED OSCILLATIONS (1)
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Up to this point, oscillating systems in which no
mechanical energy is gained or lost (due to friction, air
resistance) have been considered
In most physical systems, there are energy losses
As the mechanical energy of a system decreases, its
amplitude of oscillations decrease as well
This type of motion is called a damped oscillation
An oscillating mass may lose its mechanical energy to
a force such as air resistance that is proportional to the
speed of the mass
r and
r opposite in direction, which can
be defined as F = −bv
The damping constant is b and is a measure of the
strength of the damping force (kg/s)
If the damping constant is small, the system will
continue to oscillate, but with a continuously decreasing
amplitude – called underdamped motion
In underdamped motion, the amplitude decreases
exponentially with time: A = A0e-bt/2m where A0 is the
initial amplitude, m is the mass, time t
As the damping is increased, a point is reached where
the system no longer oscillates, but simply relaxes back
to the equilibrium position – critically damped
The system is overdamped if the damping is increased
beyond this point, and the system still returns to
equilibrium without oscillating, but takes longer
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DAMPED OSCILLATIONS (2)
Shock absorbers are
designed to be critically
damped, so that when a
car hits a bump, it returns
to equilibrium quickly
without oscillations
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DRIVEN OSCILLATIONS
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Applying an external force to a
system furnishes it with energy
– positive work is done
If you hold the end of a
string from which a small
weight is suspended, and
the weight is set in motion,
holding your hand still will
cause it to stop oscillating
If you move your hand back and forth, the weight will
oscillate indefinitely
This motion is called a driven oscillation
Moving your hand at the wrong frequency (too high or
too small) will cause the weight to oscillate erratically
Oscillating your hand at an intermediate frequency will
result in large amplitude oscillations for the weight
This frequency is called the natural frequency, f0, of
the system
Natural frequency of a pendulum: f0 = 1/T = (1/2)√(g/L)
For a mass on a spring: f0 = 1/T = (1/2)√(k/m)
Generally driving any system near its natural frequency
results in large oscillations
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RESONANCE
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Consider a plot of amplitude,
A, versus driving frequency
f for a mass on a spring
There are large amplitudes
for frequencies near f0
This type of large response
is known as resonance
The curves are referred
to as resonance curves
Systems with small damping
have a high narrow peak on their resonance curve
Tuning a radio changes the resonance frequency of an
electrical circuit (made up of capacitors, inductors and
resistors) in the tuner
When its resonance frequency matches the frequency
being broadcast by a station, that station is picked up
Bridges affected by high winds may swing with large
amplitudes, especially if the frequency of the wind
matches the natural frequency of the bridge as a
system (recall the Millennium Bridge fiasco)
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