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Vibrations and Waves
13.01 Hooke’s Law
13.02 Elastic Potential Energy
13.03 Comparing SHM with Uniform Circular Motion
13.04 Position, Velocity and Acceleration as a Function of Time
Topics
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
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 ).
x = 0
Vibrations and Waves (4 of 33)
m
x = 0
Hooke’s Law
The minus sign on the force indicates that it is a restoring force – it is directed to restore the mass to its equilibrium position.
The restoring force exerted by the spring depends on the displacement:
m
x
F
[13.1] kxFs
Vibrations and Waves (5 of 33)
m
x
F kxF
(a) (k) is the spring constant
(b) Displacement (x) is measured from the equilibrium point
(c) Amplitude (A) is the maximum displacement
(e) Period (T) is the time required to complete one cycle
(f) Frequency (f) is the number of cycles completed per second
(d) A cycle is a full to-and-fro motion
Hooke’s Law
Vibrations and Waves (6 of 33)
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.
m
xo okxF
mgEquilibrium
Position
Hooke’s Law
Vibrations and Waves (7 of 33)
Any vibrating system where the restoring force is proportional to the negative of the displacement
moves with simple harmonic motion (SHM), and is often called a simple harmonic oscillator.
kxF
Hooke’s Law
Vibrations and Waves (8 of 33)
Potential energy of a spring is given by:
The total mechanical energy is then:
The total mechanical energy will be conserved
2kx
2mv
E22
total
Elastic Potential Energy
[13.3] 2
kxPE
2
s
Vibrations and Waves (9 of 33)
If the mass is at the limits of its motion, the energy is all potential.
m
A 2kA
PE2
m
x = 0
vmax
If the mass is at the equilibrium point, the energy is all kinetic.
2
mvKE
2max
Elastic Potential Energy
Vibrations and Waves (10 of 33)
This can be solved for the velocity as a function of position:
The total energy is, therefore
And we can write: 2
kAE
2
total
2kx
2mv
2kA 222
where mk
AAmk
v 2max
Elastic Potential Energy
[13.6] xA mk
v 22
Vibrations and Waves (11 of 33)
The acceleration can be calculated as function of displacement
m
x
F
kxF
kxma
xmk
a
Amk
amax
Elastic Potential Energy
Vibrations and Waves (12 of 33)
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:
This is identical to SHM.
A
vmax
x
θsinvv max
22 xAmk
v
22 xA
mk
Avmax AxA
θsin22
AxA
mk
Av22
v
Comparing Simple Harmonic Motion with Circular Motion
Vibrations and Waves (13 of 33)
2
2
maxA
x1vv
Therefore, we can use the period and frequency of a particle moving in a circle to find the period and frequency of SHM:
mk
Avmax mk
Avmax T
Aπ2
km
π2T
mk
π21
T1
f
Comparing Simple Harmonic Motion with Circular Motion
fA2
Vibrations and Waves (14 of 33)
7b)-(11 mk
21
f
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?
Comparing SHM with Uniform Circular Motion (Problem)
Frequency of oscillation
22
4
kmf
constantmf 2 222
211 fmfm
22 Hz 60.0kg 680.0mHz 88.0m kg 59.0m
k is constant regardless of the mass attached to the spring
Vibrations and Waves (15 of 33)
Vibrations and Waves 11-01
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
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.
Vibrations and Waves 11-02
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
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
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,
Comparing SHM with Uniform Circular Motion (Problem)
6)-(11 fA2vmax
m 13.0Hz 32vmax m/s 45.2
Vibrations and Waves (18 of 33)
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,
Comparing SHM with Uniform Circular Motion (Problem) con’t
5)-(11 A
x1vv
2
2
max
2
2
m 13.0
m 10.01m/s 45.2v m/s 6.1
m/s 45.2vmax
Vibrations and Waves (19 of 33)
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,
Comparing SHM with Uniform Circular Motion (Problem) con’t
4b)-(11 2
mvE
2max
2
m/s 2.45 kg 60.0E
2 J 8.1
m/s 45.2vmax
Vibrations and Waves (20 of 33)
Vibrations and Waves 11-03
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
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
Vibrations and Waves 11-04
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
Doubling only the amplitude of a vibrating mass-and-spring 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
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?
Comparing SHM with Uniform Circular Motion (Problem)
1)-(11 kxF
xF
k x
mg
m 315.0m/s 9.8 gk 62.2 2
N/m 5.18
7a)-(11 km
2T
km
42
4T
N/m 5.81kg 62.2
42
s 282.0
Vibrations and Waves (23 of 33)
Vibrations and Waves 11-05
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
Doubling only the spring constant of a vibrating mass-and-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
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.
The Simple Pendulum
Vibrations and Waves (25 of 33)
mg
F
s
x
sinmgF
Lx
sin
x L
mgF
Small angles x s
s L
mgF
k forSHM
km
2T
Lmgm
2gL
2T
L
m
The Simple Pendulum
Vibrations and Waves (26 of 33)
Vibrations and Waves 11-06
~M1 ~M2 ~M3 ~M4 ~M5 ~M6 ~M7 ~M8 ~M10~M9~M11~M12~M13~M14~M15~M16~M17~M18 ~M20~M19~M21~M22~M23~M24~M25~M26~M27~M28 ~M30~M29~M31~M32~M33~M34~M35~M36~M37~M38 ~M40~M39~M41~M42~M43~M44~M45~M46~M47~M48 ~M50~M49~M51~M52~M53~M54~M55~M56~M57~M58 ~M60~M590% 20% 40% 60% 80% 100%
4
3
2
1
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.
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.
Comparing SHM with Uniform Circular Motion (Problem)
11b)-(11 Lg
21
f
m 76.0
m/s 8.921
f2
Hz 572.0
Vibrations and Waves (28 of 33)
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?
L0
cosh L L
cosL
Comparing SHM with Uniform Circular Motion (Problem) con’t
Conservation of energy
11a)-(6 0PEKE
0cos1mgL2
mv2
cos1gL2v
o2 12cos1m 0.760 m/s 9.8 2v m/s 571.0
0mgh2
mv2
Vibrations and Waves (29 of 33)
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?
Comparing SHM with Uniform Circular Motion (Problem) con’t
4b)-(11 2
mvE
2max
2
m/s 0.571 kg 365.0E
2
J 0.0595E
L0
cosh L L
cosL
m/s 571.0vmax
Vibrations and Waves (30 of 33)
Summary of Chapter 11
For SHM, the restoring force is proportional to the displacement. kxF
The period is the time required for one cycle, and the frequency is the number of cycles per second.
km
π2T Period for a mass on a spring:
2kx
2mv
E22
total
During SHM, the total energy is continually changing from kinetic to potential and back.
Vibrations and Waves (31 of 33)
A simple pendulum approximates SHM if its amplitude is not large. Its period in that case is:
gL
π2T
The kinematics of a mass/spring system:
mk
AAmk
v 2max
Amk
amax
22 xAmk
v Velocity
xmk
a
Acceleration
Summary of Chapter 11
Vibrations and Waves (32 of 33)