(CHAPTER-1)
WAVE
WAVE:
Wave is a
period disturbance in a medium or in space. We are all familiar with water
waves. There are also sound waves, as
well as light waves, radio waves and other electromagnetic waves. In all cases,
the disturbances consequently progress through the medium or space. The
properties and behavior of waves are very important in physics. The chief
characteristics of a wave are its speed of propagation, its frequency, its
wavelength and its amplitude. The speed of propagation is the distance covered
by the wave per unit time. The frequency is the number of complete disturbance
(cycle) in unit time usually expressed in Hertz. The wavelength is the distance
in meters between two successive points of equal phase in a wave. The amplitude
is the maximum difference of the disturbed quantity from its mean value.
Mechanical
wave:
They originate in the displacement of some
portion of an elastic medium
from its normal position, causing it to oscillate about an equilibrium
position. Because of the elastic properties of the medium, the disturbance is
transmitted from one layer to the next. This disturbance or wave consequently
progresses through the medium. Energy can be transmitted over considerable
distances by wave motion. Mechanical waves are characterized by the transport
of energy through matter by the motion of a disturbance in that matter without
any corresponding bulk motion of the matter itself. Sound waves, water waves etc. are the examples of
mechanical waves.
Transverse
wave:
When the
motions (vibrations) of the matter properties conveying the wave are
perpendicular to the direction of propagation of the wave itself, the wave is
called a transverse wave. For example, light
waves (electromagnetic waves) are of this kind, with electric and magnetic
fields varying in a periodic way at right angles to each other and to the
direction of propagation.
Longitudinal
wave:
When the
motion of the particles conveying
a mechanical wave back and forth along
the direction of propagation, the wave
is called a longitudinal wave. For example,
when a vertical spring under tension is set oscillating up and down at one end,
a longitudinal wave travels along the spring; the coil vibrates back and forth
in the direction in which the disturbance travels along the spring. Sound waves
in a gas are longitudinal waves. Some waves are neither pure longitudinal nor
purely transverse. For example,in waves on the
surface of water the particles of water move both up and down and back and
forth, tracing out elliptical paths as the water waves move by.
Equation
of a travelling wave:
Let us
consider a long stretched in the x-direction along which a transverse wave is
traveling. At some instant of time, say t=0, the shape of the string can be
represented by,
y=f(x), at t=0.
where y is the transverse displacement of the
spring at the position x. In fig (a), we show a possible wave form (a plus) on
the string at t=0.At sometime t later the wave
has traveled a distance vt to the right where v is the magnitude of the wave velocity. The equation
of the curve at the time t is therefore,
y=f(x-v t), at t=t.
This gives us the same wave form about the point x= vt at time t as we had about x=0 at the time t=0
(fig -b)
Let us now consider a wave train along the (+x)
direction at the time t=0, the wave train is given by, Y=Ymax Sin
The wave shape is a sine curve (fig-c). The maximum displacement Ym is the amplitude of the sine curve. The
value of the transverse displacement y is the same at x as it is at (x + λ),
(x+2λ), etc. The symbol is called the wavelength of the wave train and
represents the distance
between two adjacent points in the wave having the same phase. As time goes on, let the wave train travel
to the right with a phase velocity V. The wave travels the distance x= vt in the time t.
Hence, the replacing x by (x-vt) we have the desired equation of the wave at the
time t giving y in terms of both x and t.
Y=Ym Sin
(x-vt).
This is the required equation of a traveling
wave.
From the equation show that,
Or,
=
We have,
Y=ymSin
(x - vt)
=
= - k2y………. (1) [where, k =
]
Again,
Y=ymSin
(x - vt)
Or, Y=ymSink(x
- vt)
Or,
= - ym
kvCosk(x – vt)
From equation (1) and (2) we see that,
s =
=
[Proved]
The Superposition Principle:
It is an experimental fact that for many kinds
of waves two or more waves can transverse the same space independently of one
another means that the displacement of any particle at a given time is simply
the sum of the displacements that the individual waves alone would give it.
This process of vector addition of the displacements of a particle is called
superposition.
For
example,
Radio waves
of many frequencies pass through a radio antenna; the electric currents set up
in the antenna by the superposed action of all these waves are very complex.
Nevertheless, we can still tune to a particular station, the signal that we
receive from it being in principle the same as that which we would receive if
all other stations were to stop broadcasting.
Standing
Wave:
Let us consider two wave trains of the same
frequency, speed and amplitude which are travelling in opposite directions
along a string. Two such waves may be represented by the equations,
Y1=ym sin (kx +wt)
Y2=ym sin (kx-wt)
Hence, the resultant may be written as,
y=y1 + y2
y= ym
sin (kx-wt)+ym sin(kx + wt)
= ym {2sin
(kx-wt +kx+wt) –cos
(kx+wt-kx+wt)}
=2ym
sinkxcosωt
……………………..(1)
Equation (1) is the equation of a standing wave.
Notice that a particle at any particular point x executes simple harmonic
motion as time goes on, and that all particles vibrate with the same frequency. In a travelling wave each
particle of the string vibrates with the same amplitude.
Characteristic of a standing wave, however, is
the fact that the amplitude
is not the same for different particles but
varies with the location x of the particle. In fact, the amplitude, 2y sinkx
has a maximum value of 2y at positions where,
kx =
,
,
, etc.
Or, x =
,
,
, etc .
Fig. The envelope of a standing wave
corresponding to a time exposure of the motion and showing the patters of nodes
(N1 N2 N3 N4 ) and antinodes (A1
A2 A3).
These points are called antinodes and are spaced
one-half wavelength apart. The amplitude has a maximum value of zero at
position where,
kx = π
,2π,3π,etc.
Or, x =
, λ,
, 2 λ , etc.
These points are called nodes and are spaced
one-half wavelength apart. The separation between a node and an adjacent anti
node is one-quarter wavelength. It is clear that energy is not transported
along the string to the right or to the left, for energy cannot flow past the
nodal points in the string which are permanently at rest. Hence, the energy
remains standing in the string.
Wave Speed:
Let us derive the velocity of a pulse in a
stretched string by a mechanical analysis. In Fig. we show a wave pulse
proceeding from right to left in the
string with a speed v. We consider a
small
(Figure Derivation of wave speed by
considering the forces on a section of string of length ∆l)
section of the pulse of length ∆l to form an arc of a
circle of radius R, as shown in the diagram. If µ
is the mass per unit length of the string, the so-called liner density,
then µ ∆l
is the mass of this element. The tension F in
the string is a tangential pull at each end of this small segment of the
string. The horizontal components cancel and the vertical components are each
equal to Fsinө. Hence, the total vertical
force is 2Fsinө. Since, ө is small,
we can take,
Sinө =~ө
and 2Fsinө =2Fө
=2F.
=F.
This gives the force supplying the centripetal
acceleration of the string particles directed toward 0.Now
the centripetal force acting on a mass µ∆l moving
in a circle of radius R with speed V is µ∆l .Notice that
the tangential velocity V of this mass element
along the top of the arc is horizontal and is
the same as the pulse phase velocity. Combining the equivalent expressions just
given we obtain,
F
=
Or, v =
is
the wave speed.s
Resonance:
In general, whenever a system capable of
oscillating is acted on a periodic series of impulses having an oscillation
of the system equal or nearly equal to
one of the natural frequencies of oscillation of the system, the system is set
into oscillation with relatively large amplitude. Thus phenomenon is called
resonance and the system is said to resonate with the applied impulses.
Sound Waves (Audible, Ultrasonic and
Infrasonic):
Sound waves are longitudinal mechanical waves.
They can be propagated in solids, liquids and gases. The material particles
transmitting such a wave oscillate in the direction of propagation of the wave
itself. Sound waves are confined to the frequency range which can stimulate the
human ear and brain to the sensation of hearing. This range is from about 20 cycles/ second to about 20,000
c/s and is called the audible range. A longitudinal mechanical wave
whose frequency is below the audible range is called an infrasonic waves and
one whose frequency is above the audible rage is called an ultrasonic or
supersonic waves.
Doppler Effect:
When there is a relative motion between the
source and the observer the pitch of the sound as heard by the listener or
observer is in general not the same as when there is no relative motion between
them. This phenomenon is called Doppler Effect.
Doppler Effect Equation:
Let us consider a reference frame at rest in the
medium through which the sound travels. Figure (a) shows a source of sound S at rest in this frame and an observer O moving toward the source at a speed V. The circle represents wave fronts, spaced one wavelength
apart traveling through the medium. If the observer were at rest in the medium
he would receive waves in time t where V is the speed of the sound in the medium and λ is the wavelength. Because of his motion toward
the source, however, he receives addition waves in this same time t. The
frequency ʋ´ that he hears is the number of waves received per unit time or,
ʋ’ =
=
=
= Ʋ.
=ʋ
(1+vₒ/v)
The frequency
heard by the observer is the ordinary frequency heard at rest plus the increase ʋ (
) arising from the motion of the observer. When the observer is in
motion away from the stationary source, there is a decrease in frequency ʋ (
) corresponding to the waves
that do not reach the observer each unit of time because of his reaching
motion. Then
ʋ` = Ʋ.
=ʋ (1-vₒ/v)
Hence, the general relation holding when the
source is at rest with respect to the medium but the observer is moving
through it is ʋ’= Ʋ.
where the
plus sign holds for motion toward the source and the minus sign holds for
motion away from the source.
When
source is motion but observer is in rest:
When the
source is in motion toward a stationary observer the effect is a shortening of
the wave length for the source is following after the approaching waves and the
crests therefore come closer together. If the frequency of the source is ʋ and its speed is VS then during each
vibration it travels a distance and
each wavelength is shortened by this amount. Hence, the wavelength of the sound
arriving at the observer is not,
λ =
λ’ =
λ-
=
─
=
Fig. (b)
Therefore, the frequency of the sound heard by
the observer is increased being,
ʋ` =
=
=ʋ (
)
If the source moves away from the observer the
wavelength emitted is
greater than λ i.e. (That
means)
λ’ =
λ+
=
+
=
So, that the observer hears decreased frequency namely,
ʋ’ =
=
=ʋ (
)
Hence, the general relation holding when the
observer is at rest with respect to the medium but the source is moving through
it is,
ʋ’=
ʋ (
)
Where the minus sign holds for motion toward the
observer and the plus sign holds for motion away from the observer. If both
source and observer move through the transmitting medium, the observer hears a
frequency,
ʋ’= Ʋ.
Where the
upper signs (+ numerator ,- denominator ) correspond to the source and observer
moving along the line joining the two
in the
direction toward the other and the lower signs in the direction away
from the other.
PROBLEMS:
(1)A
transverse sinusoidal wave is generated at one end of a long horizontal string by a bar which moves the
end up and down through a distance of 1/2ft. The motion is continuous and is
repeated regularly twice each second.
(a)If the string has a liner density of O.OO5O
slug/ft and is kept under a tension of 2.Olb, find the speed, amplitude,
frequency and wavelength of the wave motion.
(b)Assuming the wave moves from left to right
and that at t=Othe end of the string described by x=O, write the equation of
the wave.
Solution:
(a)The end moves 1/4ft away from the equilibrium
position first above it, then below it, for a total displacement of 1/2ft
.Therefore, the amplitude y is 1/4ft.The entire motion is repeated twice each
second so that the frequency is 2.O vibrations per second. The wave speed is
given by,
Here,
F=2.0 lb
µ=0.005 slug/ft
V
= ?
Or, v =
=
=
2Oft/sec
The wavelength is given by,
λ =
=
=1Oft
(b)The general expression for a transverse
sinusoidal wave moving from left to right is y=ym Sin (kx-wt-ɸ)
requiring that y=O for the condition x=O and t=O yields,
O = ym sin(-ɸ)
This means that the phase constant ɸ must be
zero. Hence for this wave
y=ymsin
(kx- ωt)
And with the values just found,
ym =1/4ft
=o.25ft
λ =1Oft
Or, k =
=
ft -1
=0.2
ft -1
V=20ft/sec
ω =vk
=(20ft/sec).( 0.2
ft -1)
=4
sec -1
We obtain as the equation for the wave,
y=O.25 sin (0.2π -4πt)
Where x and y are in feet and t in seconds.
(2) The equation of a transverse wave
traveling in a rope is given by,
y=10sin π
(0.01x-2.00t).
where y and x are expressed in
centimeters and t in seconds.
(a)Find the amplitude, frequency,
velocity and wavelength of the wave.
(b)Find the maximum transverse speed of
a particle in the rope.
SOLUTION:
The equation of a transverse wave is given by,
Y=10 Sin π (0.01x-2.00t)
Or,
y=10 Sin (0.01π-2.00πt)
Comparing this equation with the general equation
of a transverse sinusoidal wave,
Y=ym Sin (kx –
wt)
Amplitude, ym =10cm.
k =
,
Wavelength
, k= 0.01 π
Or, 0.01π =
Or, 0.01=
Or, λ =200 cm
Velocity, V =
=
=200 cm/sec
Frequency, ʋ =
=
=1 cm
(b)The transverse displacement of a particle at x
= x0 in the rope at any time is,
Y=10 sin π (0.01x0
– 2.00t)
Or, y=10 Sin
(0.01πx0 -2πt)
Or, y= - 10 Sin
(2πt – ζ0)
[where ζ0 =0.01 π x0 ]
= -10x2π Cos (2πt –ζ0)
Or, Vy = -
62.8cos (2πt – ζ0)
Vy will be maximum when Cos (2πt –ζ0)
= ±1.
Thus maximum transverse speed of a particle in
the rope is 62.8 cm/sec.
(3)The
equation of a transverse wave traveling in a rope is given by,
Y = 6Ocos
(0.0050x – 8.0t - 0.57)
in
which x and y are expressed in centimeters and t in seconds.
(a)Find
the amplitude, frequency, velocity and wavelength of the waves.
(b)Find
the maximum transverse speed of a particle in the rope.
SOLUTION:
The equation of a transverse wave traveling in a
rope is given by,
Y= 60 Cos (0.0050x - 8.0t – 0.57)
Or, y = 60Cos (0.0025π -
4.0πt - )
Or, y = 60Sin (0.0025πx –
4.0t -
Comparing
this equation with the general equation of a transverse sinusoidal wave,
Y=ym
Sin (kx – wt - ɸ)
Amplitude, ym = 60cm.
K=0.0050π
Wavelength, λ=
=
= 800 cm
Velocity, V = =
=
= 1600 cm/sec
Frequency, ʋ =
=
=2.0 Hz
(b)The
transverse displacement of a particle at x = x0 in the rope at any
time is,
Y=60 Cos
(0.0050x0 - 8.0t – 0.57)
Or, y=60Cos(0.0025πx0 -
4.0πt -
)
Or, y =60 Cos [
(0.0025πx0 -
Y = 60Cos (ζ0 - 4.0πt ) [Where ζ0 =0.0025π x0]
Y= -60cos(4.0πt- ζ0)
= - 60x4π Sin (4πt –ζ0)
Or, Vy
= - 754 Sin (4πt – ζ0)
Vy
will be maximum when Sin (4πt –ζ0) = ±1.
Thus
maximum transverse speed of a particle in the rope is 754 cm/sec or 7.5m/s.
(4) What is the speed of a transverse wave is a
rope of length 2.0 meters and mass 0.060kg under a tension of 500N?
SOLUTION:
Here,
L = 2.0m
F = 500N
µ =
=0.030 kg/m
We know
that,
V =
=
= 130 m/s
Answer: 130 m/s.
(5)With
what tension must a rope with length 2.50m and mass 0.12kg be stretched for
transverse waves of frequency 40.0Hz to have a wavelength of 0.70m?
SOLUTION:
Here,
L =2.50m
ʋ =4o.oHz
λ =0.70m
µ =
=
= 0.048 kg/m
T =?
We know that,
V =
Or, F = v2µ
= (ʋλ) 2 µ
= (40.0x0.70)2x0.048
= 37.632N
(6)
A string is stretched with a force of 100N. Its linear mass density is 0.1kg/m.
One end of the string is oscillating with amplitude of 0.01m and frequency
400Hz so that traveling waves are set up in the positive x-direction. Calculate
the velocity and wavelength of the wave.
SOLUTION:
Here,
T=100N
µ=0.1kg/m
ʋ=400 Hz
λ=?
We know that,
V=
=
= 31.62 m/s
Again, λ =
=
s
= 0.079056m
Answer:
31.62 m/s and 0.079056m.
(7) A siren emitting a sound of frequency
1000vib/sec moves away from you toward a cliff at a speed of 10 m/s.
(a)
What is the frequency of the sound you hear coming directly from the siren?
(b)
What is the frequency of the sound you hear reflected off the cliff?
Take
the speed of sound in air as 330m/s.
SOLUTION:
Here,
ʋ=1000vib/sec
vS =10m/s
v =330 m/s
(a) Ʋ´ =?
(b) Ʋ₺ =?
(a)We know that,
Ʋ´= ʋ (
)
= 1000.(
)
=
970.58vib/sec
(b) The frequency of sound reflected off the
cliff is,
Ʋ”= ʋ (
)
= 1000.(
)
=1031.25vib/sec
Answer: (a) 970.58vib/sec.
(b) 1031.25vib/sec.
(8) A source of sound waves of frequency
1080vib/sec moves to the right with a speed of 108ft/sec relative to the
ground. To its right is a reflecting surface moving to the left with a speed of
216ft/sec relative to the ground. Take the speed of sound in air to be
1080ft/sec and find
(a)
The wavelength of the sound emitted in air by the source
(b)
The number of waves per second arriving at the reflecting surface
(c)
The speed of the reflected waves
(d)
The wavelength of the reflected waves.
SOLUTION: Here,
Ʋ =
1080vib/sec
Vs =
108ft/sec
V =
1080ft/sec
(a) λ =?
(b) Ʋ´=?
(c) V =?
(d) λ” =?
(a)We know that,
λ´ =
=
[since Ʋ´= Ʋ x (
) ]
=
=
= 0.90ft
(b) We know
that,
V= Ʋ’ λ’
Or, Ʋ’ =
=
= 1200vib/sec
(c) We know that,
V= Ʋ’ λ’
= 1200x0.90
= 1080 ft/sec
(d) We know that,
λ´’ =
=
[since Ʋ´= Ʋ x (
) ]
=
=
=
=0.8ft
Answer: (a) 0.90ft
(b) 1200vib/sec
(c) 1080ft/sec
(d) 0.8ft
(9) A police siren emits a sinusoid
waves with frequency 300Hz. The speed of sound is 340m/s.
(a)
Find the wavelength of the waves if the siren is at rest in the air.
(b)
If the siren is moving at 30m/s , find the wavelength of the waves ahead of and
behind the source
(c)
If a listener L is at rest and the siren is moving away from L at 30m/s, what
frequency does the listener hear?
SOLUTION: (a)
Here,
V=340m/s
Ʋ=300Hz
Vs=30m/s
(a) λ =?
(b) λ´=?
(c) Ʋ’=?
(a)We know that, when the source is at rest,
λ =
=
=1.13m
(b) In
front of the siren the wavelength is,
λ´=
0r, λ´=
[since Ʋ´= Ʋ x (
) ]
=
=
= 1.03m
Behind the siren, the wavelength is,
λ´ =
=
[since Ʋ´= Ʋ x (
) ]
=
=
= 1.23m
(c) Ʋ’ =
=
= 275.67Hz
Answer: (a) 1.13m
(b) 1.03m and 1.23m
(c) 275.67Hz
(10)
A police car with its 300Hz siren is moving toward a warehouse at 30m/s,
intending to crash through the door.
(a)What
frequency does the driver of the police car hear reflected from the warehouse?
The speed of sound is 340m/s.
(b)
What is the frequency of the sound heard by a stationary listener in the
warehouse?
SOLUTION:
Here,
V=340m/s
Vs=30m/s
Ʋ’=300Hz
(a) Ʋ’=?
(b) Ʋ’=?
We know that,
ʋ`=
= 300 x(
=358.0645Hz
(b)We know that,
ʋ`=
= 300 x (
=329.0322Hz
Answer:
(a) 358.064Hz
(b) 329.0322Hz
Answer: 37.632 N
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