Transcript vibrations and waves

Vibrations and Waves
Topics
13.01
13.02
13.03
13.04
Hooke’s Law
Elastic Potential Energy
Comparing SHM with Uniform Circular Motion
Position, Velocity and Acceleration
as a Function of Time
13.05 Motion of a Pendulum
Vibrations and Waves (3 of 33)
Hooke’s Law
If an object vibrates or oscillates back and forth over the
same path, each cycle taking the same amount of time, the
motion is called periodic (T).
m
x=0
We assume that the surface is frictionless. There is a point
where the spring is neither stretched nor compressed; this
is the equilibrium position. We measure displacement
from that point (x = 0 ).
Vibrations and Waves (4 of 33)
Hooke’s Law
m
x=0
F
m
x
The restoring force exerted by the spring depends on the
displacement:
Fs  kx
[13.1]
The minus sign on the force indicates that it is a restoring
force – it is directed to restore the mass to its equilibrium
position.
Vibrations and Waves (5 of 33)
Hooke’s Law
F
m
F  kx
x
(a) (k) is the spring constant
(b) Displacement (x) is measured from the equilibrium point
(c) Amplitude (A) is the maximum displacement
(d) A cycle is a full to-and-fro motion
(e) Period (T) is the time required to complete one cycle
(f) Frequency (f) is the number of cycles completed per second
Vibrations and Waves (6 of 33)
Hooke’s Law
If the spring is hung vertically, the only
change is in the equilibrium position,
which is at the point where the spring
force equals the gravitational force.
F  kxo
xo
m
Equilibrium
Position
mg
Vibrations and Waves (7 of 33)
Hooke’s Law
Any vibrating system where the restoring force is
proportional to the negative of the displacement
F  kx
moves with simple harmonic motion (SHM), and is often
called a simple harmonic oscillator.
Vibrations and Waves (8 of 33)
Elastic Potential Energy
Potential energy of a spring is given by:
kx 2
PEs 
[13.3]
2
The total mechanical energy is then:
mv 2 kx 2
Etotal 

2
2
The total mechanical energy will be conserved
Vibrations and Waves (9 of 33)
Elastic Potential Energy
m
A
If the mass is at the limits of its motion, the
energy is all potential.
kA 2
PE 
2
vma
x
m
2
mv max
KE 
2
x=0
If the mass is at the equilibrium point, the
energy is all kinetic.
Vibrations and Waves (10 of 33)
Elastic Potential Energy
kA 2
Etotal 
2
The total energy is, therefore
And we can write:
kA 2 mv 2 kx 2


2
2
2
This can be solved for the velocity as a function of position:

k 2
v
A  x2
m
where

[13.6]
 
k 2
k
v max 
A A
m
m
Vibrations and Waves (11 of 33)
Elastic Potential Energy
The acceleration can be calculated as function of displacement
F
m
x
F  kx
ma  kx
k
a    x
m
k
amax    A
m
Vibrations and Waves (12 of 33)
Comparing Simple Harmonic Motion with Circular Motion
vma
If we look at the projection onto the
x axis of an object moving in a circle
of radius A at a constant speed vmax,
we find that the x component of its
velocity varies as:
v  v max 1 
v  v max sin θ
v x
A
q
x2
A2  x2
x
A2

k  A 2  x 2 
v  A


m 
A




k 2
v
A  x2
m
This is identical to SHM.
v max  A
k
m
A2  x2
sin θ 
A
Vibrations and Waves (13 of 33)
Comparing Simple Harmonic Motion with Circular Motion
Therefore, we can use the period and frequency of a particle
moving in a circle to find the period and frequency of SHM:
2 πA
k
 2fA

v max  A
T
m
k
v max  A
m
m
T  2π
k
1 1 k
f 
T 2π m
Vibrations and Waves (14 of 33)
Comparing SHM with Uniform Circular Motion (Problem)
A mass m at the end of a spring vibrates with a frequency
of 0.88 Hz. When an additional 680 g mass is added to m,
the frequency is 0.60 Hz. What is the value of m?
Frequency of oscillation
f
1 k
2 m
2
mf 
(11 - 7b)
k is constant regardless
of the mass attached to
the spring
k
4 2
mf 2  constant
m1f12  m 2f 22
m 0.88 Hz 2  m  0.680 kg 0.60 Hz 2
 m  0.59 kg
Vibrations and Waves (15 of 33)
Vibrations and Waves 11-01
A mass on a spring undergoes SHM. When the mass
passes through the equilibrium position, its instantaneous
velocity
(A) is maximum.
(B) is less than maximum, but not zero.
(C) is zero.
(D) cannot be determined from the information given.
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
Vibrations and Waves 11-02
A mass is attached to a vertical spring and bobs up and
down between points A and B. Where is the mass located
when its kinetic energy is a minimum?
(A) at either A or B
(B) midway between A and B
(C) one-fourth of the way between A and B
(D) none of the above
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
Comparing SHM with Uniform Circular Motion (Problem)
A 0.60 kg mass at the end of a spring vibrates 3.0 times per
second with an amplitude of 0.13 m. Determine the velocity
when it passes the equilibrium point,
vmax  2fA
vmax  23 Hz 0.13 m
(11 - 6)
 2.45 m/s
Vibrations and Waves (18 of 33)
Comparing SHM with Uniform Circular Motion (Problem) con’t
A 0.60 kg mass at the end of a spring vibrates 3.0 times per
second with an amplitude of 0.13 m. Determine the velocity
when it is 0.10 m from equilibrium,
vmax  2.45 m/s
v  v max 1 
v  2.45 m/s
x2
A
2
(11 - 5)
0.10 m 2
1
0.13 m 2
 1.6 m/s
Vibrations and Waves (19 of 33)
Comparing SHM with Uniform Circular Motion (Problem) con’t
A 0.60 kg mass at the end of a spring vibrates 3.0 times per
second with an amplitude of 0.13 m. Determine the total
energy of the system,
vmax  2.45 m/s
2
mv max
E
2
(11 - 4b)
0.60 kg 2.45 m/s 2
E
2
 1.8 J
Vibrations and Waves (20 of 33)
Vibrations and Waves 11-03
A mass is attached to a vertical spring and bobs up and
down between points A and B. Where is the mass located
when its potential energy is a minimum?
(A) at either A or B
(B) midway between A and B
(C) one-fourth of the way between A and B
(D) none of the above
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
Vibrations and Waves 11-04
Doubling only the amplitude of a vibrating mass-andspring system produces what effect on the system's
mechanical energy?
(A) increases the energy by a factor of two
(B) increases the energy by a factor of three
(C) increases the energy by a factor of four
(D) produces no change
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
Comparing SHM with Uniform Circular Motion (Problem)
A mass of 2.62 kg stretches a vertical spring 0.315 m. If the
spring is stretched an additional 0.130 m and released, how
long does it take to reach the (new) equilibrium position
again?
F  kx
(11 - 1)

F mg 2.62 kg 9.8 m/s 2
k 

x
x
0.315 m
T  2
m
k

 81.5 N/m
(11 - 7a)
2 2.62 kg
T 2 m


4 81.5 N/m
4 4 k
 0.282 s
Vibrations and Waves (23 of 33)
Vibrations and Waves 11-05
Doubling only the spring constant of a vibrating massand-spring system produces what effect on the system's
mechanical energy?
(A) increases the energy by a factor of three
(B) increases he energy by a factor of four
(C) produces no change
(D) increases the energy by a factor of two
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
The Simple Pendulum
A simple pendulum consists of a mass at the end of a
lightweight cord. We assume that the cord does not
stretch, and that its mass is negligible.
Vibrations and Waves (25 of 33)
The Simple Pendulum
F  mg sinq 
q
L
 mg 
F  
x
 L 
Small angles x  s
 mg 
F  
s
 L 
m
T  2
k
x
sinq  
L
m
 2
mg
L
k for
SHM
x
m
F
s
q
mg
L
 T  2
g
Vibrations and Waves (26 of 33)
Vibrations and Waves 11-06
A simple pendulum consists of a mass M attached to a
weightless string of length L. For this system, when
undergoing small oscillations
(A) the frequency is proportional to the amplitude.
(B) the period is proportional to the amplitude.
(C) the frequency is independent of the length L.
(D) the frequency is independent of the mass M.
~M1 ~M2 ~M31 ~M4 ~M5 ~M6 ~M7 ~M8 ~M9 ~M10
~M11~M12~M13~M14~M15~M16~M17~M18 ~M19~M20
2
~M21~M22~M23~M24~M25~M26~M27~M28 ~M29~M30
3
~M31~M32~M33~M34~M35~M36~M37~M38
~M39~M40
4
~M41~M42~M43~M44~M45~M46~M47~M48
~M49~M50
~M51~M52~M53~M54~M55~M56~M57~M58 ~M59~M60
0%
20%
40%
60%
80%
100%
Comparing SHM with Uniform Circular Motion (Problem)
The length of a simple pendulum is 0.760 m, the pendulum
bob has a mass of 365 grams, and it is released at an angle
of 12.0° to the vertical. With what frequency does it
vibrate? Assume SHM.
1 g
f
2 L
1 9.8 m/s 2
f
2 0.76 m
(11 - 11b)
 0.572 Hz
Vibrations and Waves (28 of 33)
Comparing SHM with Uniform Circular Motion (Problem) con’t
The length of a simple pendulum is 0.760 m, the pendulum
bob has a mass of 365 grams, and it is released at an angle
of 12.0° to the vertical. What is the pendulum bob’s speed
when it passes through the lowest point of the swing?
Conservation of energy
L
KE  PE  0
q0
L cosq
(6 - 11a)
mv 2
 mgh  0
2
mv 2
 mgL1  cos q   0
2
h  L  L cosq
v  2gL1  cos 



v  2 9.8 m/s 2  0.760 m  1  cos 12o

 0.571 m/s
Vibrations and Waves (29 of 33)
Comparing SHM with Uniform Circular Motion (Problem) con’t
The length of a simple pendulum is 0.760 m, the pendulum
bob has a mass of 365 grams, and it is released at an angle
of 12.0° to the vertical. Assume SHM. What is the total
energy stored in this oscillation, assuming no losses?
L
2
mv max
E
2
(11 - 4b)
0.365 kg 0.571 m/s 2
E
2
E  0.0595 J
q0
L cosq
h  L  L cosq
vmax  0.571 m/s
Vibrations and Waves (30 of 33)
Summary of Chapter 11
For SHM, the restoring force is proportional to the
displacement.
F  kx
The period is the time required for one cycle, and the
frequency is the number of cycles per second.
Period for a mass on a spring:
m
T  2π
k
During SHM, the total energy is continually changing
from kinetic to potential and back.
mv 2 kx 2
Etotal 

2
2
Vibrations and Waves (31 of 33)
Summary of Chapter 11
A simple pendulum approximates SHM if its
amplitude is not large. Its period in that case is:
L
T  2π
g
The kinematics of a mass/spring system:
Velocity

k 2
v
A  x2
m
Acceleration
k
a    x
m

 
k 2
k
v max 
A A
m
m
k
amax    A
m
Vibrations and Waves (32 of 33)