Transcript Document

Copyright Sautter 2003
Vibrating
Tuning fork
A weight on
a spring
200
grams
A boy on
a swing
Simple Harmonic Motion
• Simple harmonic motion (SHM) is a repeated
motion of a particular frequency and period.
• The force causing the motion is in direct relationship
to the displacement of the body. (Hooke’s Law)
• The displacement, velocity, acceleration and force
characteristics are specific a various points in the
cycle for SHM.
• SHM can be understood in terms of the
displacement, velocity, acceleration and force
vectors related to circular motion.
• Recall that the displacement vector for circular
motion is the radius of the circular path. The
velocity vector is tangent to the circular path and the
acceleration vector always points towards the center
of the circle.
Elongation of spring
\
F
O
R
C
E
(N)
200
grams
400
grams
600
Slope = spring constant
ELONGATION (M)
grams
Simple Harmonic Motion
• SHM motion can be represented as a vertical
view of circular motion. Using this concept, we
can see the variations in the vector lengths and
directions for displacement, velocity and
accelerations as those values for SHM.
• Use the following slide showing a mass on a
spring, vibrating in SHM to examine the
variations in these three vectors as the reference
circle rotates.
• See if you can decide which trig functions (sine,
cosine or tangent) govern each to the three
vectors in SHM
Top of cycle
Mid cycle
Bottom of cycle
CLICK
HERE
Displacement = +max
Velocity = 0
Acceleration = - max
Kinetic Energy = 0
Net Force = - max
Displacement = 0
Velocity = max
Acceleration = 0
Kinetic Energy = max
Net Force = 0
Displacement = - max
Velocity = 0
Acceleration = + max
Kinetic Energy = 0
Net Force = + max
The velocity vector (black)
is always directed tangentially
to the circular path.
The acceleration vector (red)
is always directed toward
the center of the circular path
Displacement Vector of Circular motion
& Displacement in SHM
y = +max
200
grams
y=0
y=0
200
200
grams
grams
200
grams
y = -max
Displacement vector
on Reference Circle
Vertical
View
Simple Harmonic
Position
Note that the vertical
view of the
displacement vector
is 0 at 00, 100 % upward
at 900, 0 at 1800, 100 %
downward at 2700
and finally 0 again at 3600
y = +max 900
y=0
1800
y=0
00
What trig function is
0 at 00, 1.0 (100%) at 900,
0 at 1800, -1.0 (100% and
pointing down) at 2700,
and 0 again at 3600
y = -max 2700
Displacement vector
on Reference Circle
Vertical
View
The SINE
y = Amp x sin θ
The velocity vector is
always tangent to the
circular path
The reference circle is
turned sideways
and viewed vertically.
This shows the velocity
vector of a body in
Simple
Harmonic Motion.
Velocity Vector of Circular motion
& Velocity in SHM
V=0
200
V = + max
V = -max
grams
200
200
grams
grams
200
grams
V=0
Velocity vector on
Reference Circle
Vertical
View
Simple Harmonic
Position
V=0
900
V = + max
00
V = -max
1800
Velocity vector on
Reference Circle
Note that the vertical
view of the
velocity vector
is 100 % upward at 00,
0 at 900, 100% downward
at 1800, 0 at 2700
and finally 100% again
at 3600
What trig function is
1.0 (100%) at 00, 0 at 900,
1.0 (100%) at 1800,
0 at 2700, and
1.0 again at 3600
V=0
2700
Vertical
View
The COSINE
V = Vmax x cos θ
Acceleration Vector of Circular motion
& Acceleration in SHM
200
a = -max
a=0
grams
200
200
grams
grams
a=0
200
grams
a = +max
Acceleration vector
on Reference Circle
Vertical
View
Simple Harmonic
Position
Note that the vertical
view of the
acceleration vector
is 0 at 00, 100 % downward
at 900, 0 at 1800,
100 % upward at 2700
and finally 0 again at 3600
a = -max, 900
a=0
00
a=0
1800
What trig function is
0 at 00, -1.0 (100%) at 900,
0 at 1800, +1.0 (100% and
pointing down) at 2700,
and 0 again at 3600
a = +max, 2700
Acceleration vector
on Reference Circle
Vertical
View
The - SINE
a = amax x ( -sin θ)
200
grams
200
grams
90 o
200
200
grams
grams
200
0o
180 o
Y = Amplitude x Sin 
V = Velocity max x Cos 
grams
360 o
270 o
Acc. = Acc. max x (-Sin )
Amp
Y = Amp x Sin 
Displacement
y
t
dy/dt = Vmax Cos 
Velocity
dy/dt
t
d2y/dt2 = Amax (-Sin )
Acceleration
d2y/dt2
t
Simple Harmonic Motion
• Recall the following relationships pertaining to circular
motion. R = the radius of the reference circle.
= o t
=2f
=2/T
Vlinear =  R
acentripetal = V2 / r
COMBINING THESE EQUATIONS WE GET:
=2ft
V=2fR
V=2R/T
ac = 4 2 f 2 R
ac = 4 2 R / T2
Expanded Displacement Equations
The amplitude (A) = the radius of the reference circle (R)
y = A x sin θ
y = A x sin o t
y = A x sin 2  f t
Expanded Velocity Equations
V = Vmax x cos θ
V = 2  f A x cos θ
V = 2  f A x cos o t
V = 2  f A x cos 2  f t
V = 2  A/ T x cos 2  f t
Expanded Acceleration Equations
The amplitude (A) = the radius of the reference circle (R)
a = amax x ( -sin θ)
a = V2 / A x ( -sin θ)
a = 4 2 f 2 A x ( -sin θ)
a = 4 2 A / T2 x ( -sin θ)
a = amax x ( -sin 2  f t )
Deriving the Equation for Period of a
Weight – Spring System
• Fspring = Facceleration of mass
• -kx = ma
• The elongation of the spring equals the amplitude (A) of the
vibration.
• a = 4 2 A / T2 x ( -sin θ)
• -kA =m 4 2 A / T2 x ( -sin θ)
• The amplitude of vibration is reached when θ = 90 degrees.
Sine of 900 = 1.0
• Substituting 1.0 for sin θ and rearranging:
• T2 = 4 2 m A / kA = 4 2 m/k
• Solving for T gives:
π
Small masses vibrate with
shorter periods
Large masses vibrate with
longer periods
π
Springs with larger constants
vibrate with shorter periods
π
Springs with smaller constants
vibrate with longer periods
T = 2
l/g
θ
PEmax = mg h
T
W
(tension)
centripetal
force
KE =0
h
KEmax = ½ mv2
PE =0
Fr
Restoring
Force
Fc
θ (mg cos θ)
Fr
Vector
Diagram
Fcentrifugal
mg
(weight)
Deriving the Equation for Period of a
Simple Pendulum
• The pendulum experiences centripetal acceleration as
it swing in an arc. The component of the weight
causing tension in the string which supplies the
centripetal force is given by m x g cos θ.
• mg cos θ = mac
• Centripetal acceleration is given by: a = 4 2 R / T2
• mg cos θ = m 4 2 R / T2
• The radius of the circle is the pendulum length (L).
Rearranging the equation gives:
• T2 = 4 2 L / g cos θ, for small angles cos θ  1.0 and:
T2 = 4 2 L / g , taking square roots gives:
π
SHORT PENDULUMS
HAVE A SHORT
PERIOD OF
OSCILLATION
π
LONG PENDULUMS
HAVE A LONG
PERIOD OF
OSCILLATION
STRONGER GRAVITY FIELDS
RESULT IN SHORTER PERIODS
OF OSCILLATION
GRAVITY ON EARTH
9.8 M/S2
GRAVITY ON MOON
1.6 M/S2
π
π
π
Torsional (Twist) Pendulums
act because of Hooke’s Law
applied in rotational form
τ=-k θ
I (Moment of Inertia)
replaces mass in the
Weight- spring system equation
π
π
Solving SHM Problems
• What is the maximum velocity of an object in SHM
with a amplitude of 5.0 cm and a frequency of 10
hertz? What is its maximum acceleration ?
• Solution: (a) the maximum velocity occurs at 00 and
1800 (the midpoint of the cycle)
• V = 2  f A x cos θ
• V = 2  10 x 5.0 x cos 00 = 100  (1.0) = 313 cm/sec
• (b) the maximum acceleration occurs at 900 and 2700
(the endpoints of the cycle)
• a = 4 2 f 2 A x ( -sin θ)
• a = 4 2 (10)2 x 10 x ( -sin 900) = 4000 2 ( -1.0)
• a = 39,500 cm/s2 or 39.5 m/s2
Solving SHM Problems
An object vibrates with an amplitude of 10 cm and a period of 2.0
seconds. Find the velocity and acceleration when the displacement
is 5.0 cm.
A
y
θ
y = 5.0 cm, A = 10 cm
Sin θ = 5.0 / 10 = . 50
Sin –1 (0. 50) = 300
Reference circle
•
•
•
•
(a) V = 2  A/ T x cos θ, V = 2  10 / 2.0 x cos 300
V = 10  x 0.866 = 27.2 cm/s
(b) a = 4 2 A / T2 x ( -sin θ), a = 4 2 10 / 22 x ( -sin 300)
a = 4 2 10 / 2.0 2 x ( -0.50) = 5.0 2 = 49.3 cm/s2
Solving SHM Problems
A spring with a constant of 40 lbs/ft has a 5 lb object suspended
from it. Find its elongation and period of vibration when set into
motion.
F = - kx
5
lbs
π
• (a) F = - kx, x = -F /k
• x = - (-5 lbs/ 40 lbs/ft) = 0.125 ft or 1.5 inches (note that the
negative – 5 lbs means that the weight acts downward)
• (b) T = 2 π ((5/ 32) / 40)1/2 = 0.393 seconds ( note that the 5/32
converts weight in pounds to mass in slugs)
Solving SHM Problems
A spring has a 0.30 second period of vibration when a 30 N
weight is hung from it. How much will it stretch when a 50 N
weight is hung from it ?
F = - kx
30
Nts
π
• To find the the stretch of the spring we must use Hooke’s
Law and know the spring constant.
• Using T = 2 π (m/k)1/2 , we can solve for k as:
• k = 4 π2 m / T2 = 4 π2 (30/9.8) / 0.302 = 1343 N/ m (note 30 / 9,8
gives the mass of the object)
• Now using Hooke’s Law : F = - kx, x = -F / k
• X = 50 / 1343 = 0.0372 m or 3.72 cm
Solving SHM Problems
A pendulum 1.00 meters long oscillates at 30 times a minute.
What is the value of gravity ?
θ
T = 2
l/g
• 30 oscillations per minute = 30 / 60 = 0.50 oscillations per
second
• f = 1/ T, f = 1 / 0.50 = 2.0 sec
• T = 2 (l / g)1/2 , g = 4 2 l / T2
• g = 4 2 x 1.00 / 2.02 = 9.87 m/s2
Solving SHM Problems
A torsion pendulum consists of a 2.0 Kg solid disk of 15 cm
radius. When a 1 N-m torque is applied it displaces 150.
What is its frequency of vibration?
τ=-k θ
π
• The moment of inertia of a disk is given by: I = ½ mr2
• I = ½ (2.0) (.15)2 = 0.0225 Kg-m2 (note: 15 cm = .15 m)
• τ = - k θ, k = τ / θ, (150 in radians = 0.262 rad)
• k = 1 / 0.262 = 3.82 N-m / radian
• T = 2 π (0.0225 / 0.262)1/2 = 0.482 seconds
• f = 1 / T = 1 / 0.482 = 2.07 hertz
A spring 60 mm long is stretched by 5.0 mm when a 200 gram
mass is suspended from it. What is its spring constant in N/m ?
(A) 1.6 (B) 40 (C) 196 (D) 392
Click
here for
answers
An object in SHM has an amplitude of 12 mm and a period of 0.40 seconds.
What is its maximum velocity in cm per second ?
(A) 3.0 (B) 19 (C) 38 (D) 43
A swing moves back and forth every 4.0 seconds. How long is the
swing in meters?
(A) 2.5 (B) 4.0 (C) 6.2 (D) none of these
A 24 kg ball with a radius of 20 cm is suspended from a wire. It is
Rotated through 100 when a torque of 0.50 N-m is applied. Find the period.
(A) 2.30 sec (B) 1.15 sec (C) 5.3 sec (D) 1.11 sec
In order to change the frequency of a mass – spring system by a factor
of 2 the mass must be multiplied by a factor of
(A) 4 (B) 2 (C) ½ (D) ¼