Download Key words: Vibrations, Waves, Vibrational Motion, Periodic Motion

13. VIBRATIONS AND WAVES.
Key words: Vibrations, Waves, Vibrational Motion, Periodic Motion,
Simple Harmonic Motion, Restoring Force, Hook’s Law, Amplitude,
Period, Frequency, Phase Angle, Angular Frequency, Simple Pendulum,
Damped Motion, Critically Damped Systems, Force Vibrations, Natural
(or Resonant) Frequency, Resonance, Mechanical waves, Pulse, Wave
Velocity, Transverse Wave, Longitudinal Wave, Interference,
Superposition Principle, Standing Waves, Nodes, Antinodes, Normal
Mode, Constructive Interference, Destructive Interference.
13.1 Vibrations
We have studied two types of motion – translational and rotational. Now we
will begin to analyze a third basic type of motion – Vibrational Motion.
Vibrational motion occurs very often in technology and nature. For
example, all atoms in solids execute vibrational motion around the
equilibrium positions. We will pay attention to a very common type of
vibrational motion: periodic motion. Periodic Motion is motion that repeats
itself in a definite cycle. It occurs whenever an object has a stable
equilibrium position and is subject to a restoring force that acts when the
object is displaced from equilibrium. When the restoring force is directly
proportional to the displacement (we are familiar with a good example of
this situation – the spring force from Hooke’s Law), which is often the case
for small displacements, the resulting motion is called Simple Harmonic
Motion or SHM. The oscillating object undergoes SHM if the Restoring
Force follows the Hook’s Law: Restoring Force is proportional to the
displacement.
F = -kx
(13-1)
The maximum displacement from equilibrium is called the Amplitude, A.
The Period, T, is the time required for one complete cycle (back and forth).
The Frequency, f, is the number of cycles per second. They are related by
T
1
f
(13-2)
The period of oscillations for a mass m on the end, of an ideal massless
spring is given by
T
m
k
2
(13-3)
To find the position of an object as a function of time in SHM is not easy.
We can write Newton ’s second law of motion for an object that performs
SHM (sometimes it is called simple harmonic oscillatior or SHO) as follows:
F
ma
kx
ma
k
x
m
a
(13-4)
You can see from Equation (13-4), that acceleration in SHM is not constant
and we can get the dependence x(t) using calculus. The result shows that
SHM is sinusoidal. The general solution can be written as:
x
(13-5)
A cos wt
Where A is the amplitude, Φ is the Phase Angle, and w is the Angular
Frequency.
w
2 f
k
m
(13-6)
The values of A and Φ depend on the initial conditions (x and v at t = 0).
During SHM, the total mechanical energy
E
1
mv x 2
2
1 2
kx
2
1 2
kA
2
(13-7)
const
is continually changing from potential energy to kinetic energy and back
again. From the basic energy relation in equation (13-7), we can determine
the velocity of an object in SHM as a function of position.
v
k 2
A
m
x2
1/ 2
(13-8)
According to equation (13-8), the maximum speed vmax occurs at x = 0.
k
A
m
vmax
(13-9)
A Simple Pendulum is an idealized model consisting of a point mass
suspended by a weightless unstretchable string in a uniform gravitational
field. A simple pendulum of length L approximates SHM if its amplitude is
small and friction can be ignored. In this case, its period is given by
T
2
L
g
(13-10)
Equation (13-10) shows that for small oscillations, the period of a small
pendulum for a given value of g is independent of its mass and amplitude
and is determined entirely by its length.
When friction is present (for all real springs and pendulums), the motion is
said to be Damped Motion. The amplitude decreases in time, and the
mechanical energy is eventually completely transformed to thermal energy.
If the friction is very large, so no oscillations occur, the system is said to be
overdamped. If the friction is small enough that oscillations occur, the
system is underdamped, and the displacement is given by
x
Ae
Lt
cos w1t
(13-11)
Where L and w’ are constants. For Critically Damped Systems, no
oscillation occurs and equilibrium is reached in the shortest amount of time.
If an oscillating force is applied to a system capable of vibrating the
amplitude of these Force Vibrations can be very large if the frequency of
the applied force is near the Natural (or Resonant) Frequency of the
oscillating object. This phenomenon is called Resonance. All mechanical
structures – such as buildings, bridges, and airplanes – have one or more
natural resonant frequencies. It can be disastrous to subject the structure to
an external driving force at one of those frequencies.
EXAMPLE 13-1. A spring scale stretches 2.8 cm when a 3.7 kg object
hangs from it. (a) What is the spring constant? (b) What will be the
amplitude? (c) What will be the frequency of vibration if the object is pulled
down 2.5 cm more and released so that it vibrates up and down?
X1= 2.8 cm = 0.028m
m =3.7kg
X2= 2.5cm = 0.025m
a. k = ?
b. A = ?
c. f = ?
(a) According to equation (13-4), Newton’s second law of motion for this
case can be written as follows.
F
k
ma
kx1
1.30 10 3
mg
k
3.7kg 9.80
mg
x1
m
s2
0.028m
N
m
(b) Because the object will oscillate about the equilibrium position, the
amplitude A will be the distance the object was pulled down from
equilibrium.
A
x1
A 0.025m
© According to (13-6), the frequency of oscillation will be
f
1
2
k
m
1.30 10 3
f
1
2
N
m
3.7 kg
ƒ = 3.0 s¬¹ = 3.0Hz
EXAMPLE 13-2.
A 0.60kg mass at the end of a spring vibrates 3.0 times per second with an
amplitude of 0.13 m. Determine: (a) the velocity when it passes the
equilibrium position; (b) the velocity when it is 0.10m from equilibrium;
(c) the total energy of the system, and (d) the equation describing the
motion of the mass assuming that displacement from the equilibrium
position x was maximal at t = 0.
m = 0.60kg
f = 3.0Hz = 3.051
A = 0.13m
t = 0, x = xmax
(a) x = 0, v = ?
(b) x = 0.10m, v = ?
(c) E = ?
(d) x = x(t) = ?
(a) According to (13-8), the maximum velocity is at equilibrium position
x = 0.
k 2
A
m
v
k
A
m
v max
v max
v2
(b)
v
k 2
A
m
vmax 1
x2
x2
1/ 2
, f
k 2
A 1
m
x2
2.45
m
s
x2
vmax 2 1
A2
m
0.245
s
A2
k
m
2 fA
2 3.0 s _ 1 0.13m
x2
1
2
1
0.10m
0.13m
A2
2
x = 0.10m
v 1.6
(c)
v = vmax, x = 0 so, according to (13-7)
E
(d)
m
s
1
mvmax 2
2
1
m
0.60kg 2.45
2
s
1.80 J
According to (13-5) the general solution for x=x(t) is x=A cos (wt+Φ)
Because the object has maximal displacement at t = 0, we can deduce that
Φ=0. Therefore, in this case, the position will be described by the function
x = (0.13m) cos[(6.0π 1/s) t]
13.2 Mechanical waves.
A Mechanical Wave is any disturbance from the equilibrium condition that
propagates from one region of space to another through some material called
the medium. Vibrating objects act as sources of waves that travel outward
from the source. Waves on a water surface and on a string are examples. The
wave may be a Pulse (a single crest) or it may be continuous (many crests
and troughs). In a periodic wave, the motion of each particle of the medium
is periodic.
The wavelength λ of a continuous wave is the distance between two
successive crests (or any two identical points on the wave shape). The
frequency f is the number of full wavelengths (or crests) that pass a given
point per unit time. The Wave Velocity v (how fast a crest moves) is equal
to the product of wavelength λ and frequency f or the ratio of wavelength
and period:
v
f
T
(13-12)
The Amplitude of a wave is the maximum height of a crest, or depth of a
trough, relative to the normal (or equilibrium) level. In a Transverse Wave
the oscillations are perpendicular to the direction in which the wave travels.
An example is a wave on a string. In a Longitudinal Wave, the oscillations
are along (parallel to) the line of travel; sound is an example.
The velocity of both longitudinal and transverse waves in matter is
proportional to the square root of an elastic force factor divided by an inertia
factor (or density). For a string, with mass per unit length in μ, and tension
FT, the speed of transverse waves is
v
FT
(13-13)
When two waves pass through the same region of space at the same time, the
Interference occurs. From the Superposition Principle, the resultant
displacement at any point and time is the sum of their separate
displacements. When a sinusoidal wave is reflected at a stationary [not
movable] or free end, the original and reflected wave combine to make, at
certain frequencies, Standing Waves in which the waves seem to be
standing still rather than traveling. In this case, a medium is vibrating as a
whole. This is a resonance phenomenon and the frequencies at which
standing waves occur are called Resonant Frequencies. The points of
destructive interference (no vibration) are called Nodes. Points of
constructive interference (maximum amplitude of vibrations) are called
Antinodes. The simplest case is standing waves on a string of length L; they
can have only certain specific frequencies. When both ends are held
stationary, the allowed frequencies fn are
fn
n
v
2L
nf1 n 1,2,3...
(13-14)
A Normal Mode is a motion in which all particles on the string move
sinusoidally with one of the frequencies fn. Each value of fn corresponds to a
different normal mode. Standing waves on a string are a simple example of
an Interference effect - wave phenomena that occur when two or more
waves overlap in the same region of space. According to the principle of
superposition, if two periodic waves are in step at a point, then their
amplitudes add together. This phenomenon is called Constructive
Interference. If the waves are a half-cycle out of step at a point, the
resulting amplitude is smaller and the phenomenon is called Destructive
Interference.
EXAMPLE 13-3.A sinusoidal wave travels along a string. The time for a
particular point to move from maximum displacement to zero is 0.170s.
What are the (a) period and (b) the frequency of an oscillation? (c) The
wavelength is 1.40m, what is the wave speed?
t = 0.170s
λ = 1.40m
(a) T = ?
(b) f = ?
(c) v = ?
(a) The motion from the maximum displacement to zero
(equilibrium position) is one-fourth of a cycle. Therefore
T= (4)•(0.170s)=0.680s
(b) f = 1/T = 1.471 1/s
(c) v
f
1.40m
0.680s
T
2.06
m
s
EXAMPLE 13-4. (a) AM radio signals have frequencies between f1AM =
550kHz and f2AM = 1600 kHz and travel with a speed of 3.00 108
m
. What
s
are the wavelengths of these signals
(b) On FM, the frequencies range from f1FM = 88.0MHz to f2FM =
108MHz and travel at the same speed. What are their
wavelengths?
f1AM = 550 kHz
f2AM = 1600kHz
f1FM = 88.0 MHz
f2FM = 108 MHz
v
3.00 108
m
s
a. λ1AM = ?
b. λ1FM = ?
v
a.
f
λ2AM = ?
λ2FM = ?
v
f
v
1 AM
f1 AM
v
2 AM
b.
f 2 AM
v
1FM
f1FM
v
2 FM
f 2 FM
m
s
3
550 10 Hz
3.00 10 8
m
s
3
1600 10 Hz
545m
3.00 10 8
188m
m
s
6
88.0 10 Hz
3.41m
m
s
6
108 10 Hz
2.78m
3.00 10 8
3.00 10 8