This preview shows pages 1–10. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MECHANICAL WAVES Part I In Physics 11, you used ?=mZ to solve for the motion of a few particles.
By combining f=ma with the relation Etorques = change in angular momentum,
you solved for the motion of a rigid body in which all the particles that make
up the body were restrained to move together because the object was rigid. In reality, there are no rigid bodies. Especially, the particles that
make up a liquid or gas move relative to one another. The study of these
general motions is called continuum mechanics. One special and important
motion of the particles in a continuous material is the transport of a
disturbance called a wave. We will now develop the equations and terminology
that describe wave motion. Our treatment here is limited to waves in linear material in which all
waves travel at the same constant speed; I. A Wave Pulse , The waves shown in Figs. 1 and 2 are moving to the right. The motion in
Fig. 1 is called a transverse wave because the string particles move
perpendicularly to the direction that the wave is moving. The motion in
Fig. 2, on the other hand, is called a longitudinal wave because the particles
of the spring are moving right and left, which are parallel to the direction
of the wave motion. HHHIHHH .—.> HHHHHHI ¢— —> IOIHHIHHI Z‘—— ‘———> IOIOIHHHH
Fig. 1. Transverse Wave Fig. 2. Longitudinal Wave Ia. The Equation for a Wave
Consider, as an example, the motion in Fig. 1 when the end of the string
at x = 0 is moved up and down so the displacement_y0(t) of the string at x = 0 varies in time as
2 y0(t) = Ace'“’ . (1) When we say that the wave is traveling to the right with a velocity v, we
mean that the displacement of the string at x = Ax = vt is x 2
e “(t v) , y(x,t) = A0 (2) as illustrated in Fig. 3. That is, the displacement at a distance x = Ax
has the same form as at x = 0 except that it is delayed by the time it
takes to travel from x = 0 to X = Ax. If the wave travels in the negative x—direction, replace (t — g) in Eq. (2) by (t + €9, y(0,t) (displacement) y(Ax,t) \ K —— —— t Fig. 3a. The displacement at x = 0. Fig. 3b. The displacement at x = Ax. Note that the displaCement of a one—dimensional wave, as in Eq. (2), is
a function of two independent variables, here x and t. Also, if we set x = 0
in Eq. (2) we obtain y0(t) in Eq. (1), and so we will write y(0,t) in place
of y0(t). Example I. At x = 0, the string in Fig. 1 is moved up and down so that
y(0,t)= 2.0 sin(80t) in SI units. What is the displacement y(x,t) for any
value of x if the wave travels to the right at 40 m/s? Solution: Replace t by (t — £5) in the expreSsion for y(0,t); that is
y(x,t) = 2.0 sin[80(t  x/40)] = 2.0 sin(80t ~ 2x). lb. The Velocity of a Wave. Experiment: In Experiment 5 of Physics 22, you measured the speed of a
wave pulse as it traveled back and forth along a guitar wire. Data from
that experiment is shown in Fig. 4. We will explain later in these notes why
the pulse is reflected up—sidedown when it reaches the ends of the wire.
Thus, the time between two up pulses is the time to travel the length of the
wire twice, about 2.0 meters. 2.4983 mend—‘04: 2.543' hIIHtIHIHllnllhllﬂllldllullnIIHIIHIIdIIHIIHIIH11[HIIMIIHIIldllnllnlllnlluld
8 sec .81 .82 .83 .84 .85 .86 .8? .88 .89 8.1 Fig. 4. Data from experiment 5 in Physics 22. When the wire moves up and down
as the pulse passes through a magnetic field, a Faraday voltage is induced in
the wire. From Fig. 4, we determine that the time for the pulse to travel 2.00 m (the time between two up pulses) is 0.00593 5, and so the velocity of the wave
is v _ 2.00
‘ 0.00593 When we return to the study of electromagnetic waves (consisting of = 337 m/s. changing electric and magnetic fields), we will describe how the Velocity of these waves is measured. The result is about 3.0 X 108 m/s. ? Formulas for Wave Velocities: Using F = ma and a knowledge of the contact
forces between neighboring particles that make up the material in which
mechanical waves travel, we can determine the velocity of waves in terms of
the properties and states of these materials. To illustrate this procedure,
consider the pulse traveling in the string shown in Fig. 1. A drawing of the
pulse is shown in Fig. 5 as observed by someone traveling With the wave. To this observer, the wave is stationary and the string is moving in the opposite
direction along the shape of the pulse. enlarged \ Fig. 5. A pulse on a string as Fig. 6. Force diagram for the piece
as seen by an observer moving of string at the top of the pulse.
with the wave. The string is moving with a speed v relative to the moving observer,
where v is also the wave speed In order to use F = ma to analyse the motion of the string, we need F,
m and a. Ignoring the weight of the piece of string shown in Fig.6, the sum
of forces on this piece is 27 sin(A0/2) down, where A0 is the angle subtended
_ from the center of curvature for the arc of the piece of string and T is the tension in the string. For small angles, sin(A0/2) m A0/2, and so the force
is 7(A0) down. The mass of the piece of string is Am = p(length) = p(RAO), where p is
the mass (kg) per length (meter) of the string material. For the small piece of string shown in Fig. 6, we consider it to be a circle of material moving with a speed v, which is also the wave speed that we
wish to determine. For circular motion, the acceleration is a = vz/R. We now have expressions for F, Am, and a. By substituting into F = ma,
2
we obtain T(A0) = (pRAO)%— or v = (3) for the speed of the wave, where T is the tension and p is the mass per
length of the string material. Expressions for the speeds of different mechanical waves have been
obtained in similar ways using F = ma. In general the result is of the form v _ restoring force inertial term The velocity of sound waves in air with pressure as the restoring force was
first analysed by Newton, who incorrectly assumed that the pressure was given
by p = nRT/V, with the Kelvin temperature T constant. Others recognized that
the motions are so fast that no heat flow takes place and so the work changes
the air thermal energy. By assuming an adiabatic process to determine the
pressure from va = constant, where 7 = (specific heat at constant pressure)/(specific heat at constant V)
they obtained an expression for the speed of sound waves in a gas that can be written "‘
“HE, (4)
with M = molecular mass of the gas molecules. M s 28.8 x 10"3 kg/mole. For air, 7 N 1.4 and Example 2: What is the speed of sound in air when the temperature is 300 K? Solution: V = v 1.4(8.314)300)/(28.8 x 10'3) = 348 m/s. II. Sinusoidal Waves In example 1, the end of a string was vibrated so that y(0,t) = A sin(wt),
with w = 80 rad/s. The wave traveled in the xdirection with a speed v = 40
m/s and the resulting wave was y(x,t) = A sin[w(tx/v)]. (5)
We consider such waves in detall because (1) therenmany common and important: examples of such waves and (ii) wave trains of other shapes can be analysed as a sum of Sinusoidal waves. To start, set x = 0 in Eq. (5) and plot y(0,t) = A sin(wt) in Fig. 7 as a
function of time and set t = 0 in Eq. (5) and plot y(x,0) as a function of x
Note in Fig. 7 that y(0,t) oscillates with a frequency f = w/(ZW) I/f. in Fig.8. x — meters
Fig. 8. y(x,0) = — A sin(wx/v) Time ~ seconds
Fig. 7. y(0,t) = A sin(wt) In Fig. 8, we notice that the function y(x,0) oscillate in x and repeats its form in a distance A = vT : v/f. The variable A is called the wavelength. For easy reference, let us summarize the above relations with the expressions v = A/T = fA, Eq.(6a)
T = I/f, Eq (6b)
w = 2rf = ZW/T Eq.(6c)
and k = ZW/A Eq.(6d)
and rewrite Eq. (5) as
y(x,t) = A sin[w(t—x/v)] Eq (7a)
= A sin[2w(ft  x/A)] Eq.(7b)
= A sin(wtkx). Eqs.(7c) III. Energy Transported by Waves The string in Fig.1 would exert a force on your hand if you were holding
the right end of the string when the pulse arives. As your hand reacts and
moves up, work is done. That is, the wave transports energy. We limit our analysis of this energy transport to sinusoidal waves.
Consider the wave on a string given by Eq.(7c). ’The derivative with respect
to time of Eq.(7c) is the velocity of the piece of string at the point x and
the time I. If p is the mass per length (meter) of the string, then p(dx)
is the the mass of a piece of string of length dx. Continuing, the kinetic energy of this piece is 2
dk = %(’dm)v2 = %(pdx)[%;v] = %(pdx)A2w2[cos(wtkx)]2 =‘ k(dx), (8)
where k(x,t) = %p[Aw cos(wtkx)]2 (9) is the kinetic energy per length at time t. A graph of Eq.(9) is shown in Fig.9 for one wavelength A. >< — METERS Fig. 9. Distribution of the kinetic energy of pieces of the string at t = 0
k(x,0) (see Eq.(9)), platted for one wavelength. The total kinetic energy in one wavelength is equal to the area under this
curve, K = IAk(x,0)dx 0 Each piece of string oscillates up and down in simple harmonic motion, which you studied in Physics 11. The sum of the kinetic and potential
energies is conserved as each piece of string moves with the kinetic energy a
maximum at the middle of the oscillation where the potential energy is zero.
The kinetic energy goes to zero at the maximum displacement where the
potential energy is equal to what the maximum of the kinetic energy was at the
midpoint. Thus, the potential energy of one wavelength of the string is equal
to the kinetic energy in a wavelength, as given by Eq. (10). Consequently, the sum of the kinetic and potential energies in a wavelength is épAzsz. (10) II épAszIA[cos(kx)]2dx
0 K + V = 50,42sz . (11) During the time of one period of oscillation, this energy travels one
wavelength A, or, using wave velocity v = A/T, K+V122A 122 '
Average power <P> _ T  EpA w T — EA w v (12) is the average power transported by the wave,
We will refer to this expression when we determine the power transported by an electromagnetic wave, such as sunlight. Mechanical Waves — Part II Sum of Waves I. Three—dimensional Waves Before considering the sum of waves, we generalize the treatment of
of onedimesional waves treated in Part I, to waves that spread
out in three dimensions, as is the case shown in Fig. 1 for a sound wave
generated by an oscillating point source S.
The circles, called wavefronts in the
figure, represent the maximums of the waves
that are spreading out in all directions. Because The power transported by the
wave is the same for all distances r from
the source, and the area of a wave front varies as r2 and the powar depends on the
amplitude squared (see Eq.13 of Part I),
we conclude that the amplitude of the
three—dimensional wave varies as A/r,
where A is a constant. Thus, the
three—dimensional sinusoidal wave has the form Fig.1. Sound wave in 3—D. y(r,t) =§ (1) Compare this expression to Eq.(7c) for a 1—D wave in Part I of these notes. sin(wt — kr). II. Interference of Two Sound Waves. Consider, as an example, the two sound sources in Fig. 2 that oscillate
in phase with an angular frequency w. A listener located as shown in the
figure hears the sum of both sounds. For simplicity, assume that the
intensity of both received sounds are equal. Then the received wave is y(t) = A sin(wtkr1) + A sin(wtkr2). (2) Referring to the trigonometric identity sin a + sin ﬂ = 2cos[a—;é]sin[a—;£] , Eq.(2) can be rewritten as
y(t) = 2A cos[0.5k(r2ri)]sinﬁut—0.5k(r1+r2)].(3) From the argument of the sine term, we  loudspeakers conclude that this received signal
oscillates at the same angular
frequency w_as the frequency of the
sources. However, the amplitude of
this signal (the part of Eq.(3)
before the sine term, (r 'r )
2A cos[0.5k(r2r1)] = 2A cos{:—§rJ;{}, (4) varies depending on the difference (rzrl) Estener Fig.2. The listener hears
the sum of the two waves. in the distances from the two sources to
the listener. In particular. the net amplitude is zero if
(rZrl) is A/Z, 3A/2, 5A/2, etc. because cos(ﬂ/2) COS(3W/2) = On the other hand, the amplitude of the received signal is 2A if
(rZrl) is 0, A, 2A, etc. because cos(0) = —cos(w) = cos(2r) = etc. = 1. etc. = 0 . In summary, the received signal is zero if the distances from the two
signals differ by (2n+1)A/2 and is a maximum if the difference in distances is
nA, where n = 0,1,2,3,4,etc. Example 1: A listener in Fig.2 standing on the center line would hear a maximum
sound. The intensity of the sound decreases as the listener moves to either
side, reaching the first minimum at the position in the figure. What is the
frequency of the sound? The speed of sound is 340 m/s. Solution: V
Refer to Fig.3 and use the Pythagorean/ loudspeﬂkas
theorem to determine r1 and r2 :
r F """"" “a """"" '5
2 = «(.75 + .40) + (3.0) = 3.2129 m
r F """" '5 """"" ‘5
1 = V(.75  .40) + (3.0) = 3.0203 m
and so r2 — r1 = 0.1926 m. For this point
to be the first minimum rZr1 = A/Z, or
A = 0.385] m. Then, f = (340 m/s)/A = 883 Hz.
Fig.3. Example 1.
II. Reflection of Waves When a wave reaches a boundary of the media in which it is traveling,
part of the wave is reflected and part is transmitted into the adjoining material. In this section, we limit our analysis to situations in which the
wave is totally reflected. In the example of a wave traveling
in a string that is fixed at its end, as shown in Fig. 4, the wave is totally
reflected upside down. To explain why the reflected wave is
inverted if the end is fixed, let yi(x,t) be the incident wave and yr(x,t) be the
reflected wave and the total wave is y(x,t) = y,(x,t) + y,(x,t)
Since the string is fixed at x = L, y(L,t) = 0
or yr(L,t) = — yi(L,t). That is, the reflected wave starts upside down and travels in the opposite direction to the incident direction,
as shown in Fig. 4. Fig.4. Reflection from a
fixed and Another situation in which a wave is
totally reflected occurs when a wave reaches
a free end, as is illustrated in Fig.5. At
a free end, no force is applied, and so
the slope of the relected wave at the
end is the same as that of the incident
wave there. We conclude that the wave
is not inverted as shown in Fig. 5. .
i
t
%
. m.. Fig. 5. Reflection from a free end. Hstener (L401n III. Standing Waves
As an example of reflection from a fixed end, let the incident wave yi(x,t) = A sin(wt+kx) (incident) (5) be a wave traveling in the negative x—direction toward a fixed end at x = 0.
Assume that the wave reaches the end at t = 0. Then the reflected wave is yr(x,t) = —A sin(wtkx) (6) and the total displacement is y(x,t) = yi(x,t) + yr(x,t) = A[sin(wt+kx)  sin(wtkx)]
y(x,t) = 2A sin(kx)cos(wt) . (7) We used the trigonometric identity
sin a  sin ,6 E 25in[g;2'§]cos to obtain Eq. (7). To interpret Eq. (7), we identify the term 2A sin(kx) as the amplitude of
the oscillations of the piece of string at the position x. The angular
frequency of these oscillations is w, as specified in the cosine term. The
plot in Fig.6 of i 2Asin(kx) is called the envelope of the so—called standing
wave in Eq.(7) because it shows the limits of the amplitude of the oscillations of each piece of string.
2A 0.00 —2A 0.00 2A X  METERS
Fig. 6. The amplitude of the standing wave, 2A sin(kx), in Eq.(7). The locations where the envelope is zero are called nodes; the string does
not move at the nodes. The string on either side of a node are oscillating
180 degrees out of phase, i.e., when one side is moving up, the string on the
other side is moving down. This motion is called a standing wave because the
envelope pattern in Fig.6 does not move or change with time. The motion given by Eq.(7) is unrealistic because it requires that the
reflected wave never reaches the other end of the string where it would be
reflected again. That is, Eq.(7) applies to strings that are infinitely long.
However, it also applies to a finite string whose length iS‘SHCh that its
other end is fixed at a node of the pattern in Fig. 6, which does not move.
Since the nodes in Fig. 6 are a half wavelength A/Z apart, this standing wave
pattern can occur if the length of the string is nA/Z, where n = 1, 2, 3, etc. Three standing wave patterns for a string fixed at both ends are shown in
Fig. 7. For these standing waves, the wavelengths are 2L, L, and 2L/3, where
L is the length of the string. The speed of the waves on a string, as given
by Eq. (3) of part I of these notes, depends on the tension and the mass
density of the string. Assuming these are specified, the speed of the waves
are the same for all the patterns in Fig. 7 From f = v/A, we conclude that the three lowest frequencies of vibration
are v/2L, v/L, and f = 3v/2L. The lowest frequency of vibration, v/2L, is free and Fig.7. Three Standing Waves for a
string fixed at both ends. Fig.8. Standing Waves with one
end fixed and one end free. called the fundamental or the first harmonic. The frequencies of the higher
harmonics are multiples of the fundamental, as given in Fig.7. In Fig.8, we show three standing waves for a system with one end fixed
and the other end free. In this case, the wavelengths are 4L, 4L/3,
and 4L/5, respectively. The frequencies of the three lowest modes are given
in Fig.8. An example of such a system is the sound vibrations in an organ
pipe with one end open and the other end closed. Example 2: A steel piano wire that is 1.0 m long has a mass per unit length of 0.006 kg/m. It is fixed at both ends. The tension in the wire is adjusted so
that the fundamental frequency is 500 Hz. (a) What is the wave speed on the,
wire? (b) What is the tension in the wire? (c) What is the frequency of the
second harmonic? (d) The wire vibrating at its fundamental produces a sound
wave that travels at 340 m/s in air. What is the wavelength of this sound
wave? '
Solution: (a) Referring to the top drawing in Fig.7, we conclude that the wavelength
is 2.0 m. From v = fA = (500 m/s)(2.0 m) = 1,000 m/s. (b) According to Eq.(3) of part I of these notes, v = V;7E = 1,000 m/s,
r = (1000)2(0.006 kg/m) = 6,000 Newtons. (c) From Fig. 7, f2 = 2f1 = 1,000 Hz. (d) Since v = fA = 340 m/s = (500 Hz)A, or, A = 0.68 m. 10 ...
View
Full
Document
This note was uploaded on 02/29/2008 for the course PHYSICS 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .
 Spring '08
 Hickman

Click to edit the document details