Crackiitjee.in.Phy.ch21
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Wave & Wave Motion Wave: A wave is simple disturbance which propagates energy (momentum) from one place to the other without the transport of matter. The quantities like amplitude, wavelength, frequency & phase are used to characterize wave which have no meaning for a particle.
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The most important characteristic of a wave that differentiates it from a particle is diffraction. Diffraction is a convincing proof of wave nature. Wave Motion:
The process of energy transmission in which disturbance propagate
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energy by affecting the medium without the actual transport of matter.
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Characteristic of Wave Motion: (i)
In wave motion, medium particle do not leave their position
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but vibrate along their equilibrium position.
(ii)
Medium does not travel.
(iii)
Transfer of energy & momentum in a medium takes place.
(iv)
Phase of different particles of medium charges continuously.
Essential property for transmission of wave in a medium: (i)
Medium must have the property of inertia, i.e. the property to oppose the change in the position.
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(ii)
The property of elasticity after replication of force displacement must take place and after removal body should recover the initial position).
(iii)
Resistance of the medium should be small.
(iv)
Displacement of medium particle takes place for an in.
The characteristic of elasticity involves that if after application of force
regain its initial position.
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displacement takes place then it a removal of force the particles should
The motion of particles remark simple harmonic motion (S.H.M.) that is restore force remains directly proportional to displace from mean
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position and always directed toward mean position.
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Difference between wave motion & SHM:
In wave motion all the particles performing. S.H.M. but phase of different particles remain different.
(i)
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Various types of waves and differences between them: Depending on the necessity of medium– Wave are classified as– (a)
Mechanical &
(b)
Non-mechanical waves
(a)
Mechanical or elastic waves like sound waves:
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A wave may or may not require a medium for its propagation. The waves which do not require medium for their propagation are non-mechanical e.g., light, heat & radio-waves are non-mechanical as they propagate throw vacuum. Infact all electro-magnetic waves such as -rays X-rays or micro-waves are non-mechanical.
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On the other hand the waves which require medium for their propagation are called mechanical waves. Ex.
Waves on strings & springs, seismic waves or sound waves. One can not listen to his companion on the moon, or sound from the
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propagation.
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sun does not reach the earth because there is no medium for its
Apart from mechanical (elastic) and non-mechanic (electro magnetic
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waves) waves there is also another kind of waves called 'matter waves'. The represent wave like properties of particles and a governed by the laws of quantum physics. Waves are of two types: (i)
Longitudinal waves.
(ii)
Transverse waves.
Transverse wave:
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If the particles of the medium vibrate at right angle to the direction of wave motion or energy propagation the wave is called transverse wave. These are propagated as crests and troughs. Waves on strings are always transverse.
Longitudinal waves:
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diagram here
If the particles of the medium vibrate in the direction of waves motion, the wave is called longitudinal. These are propagated as compressions
iit
and ????????? and also known as pressure or compressional waves.
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waves.
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Waves on springs or sound waves in air are examples of longitudinal
diagram here
The transverse or longitudinal nature of a wave is decided by polarization, as transverse wave can be polarized while a longitudinal wave can not be polarized. Number of dimensions in which they propagate energy: One-Dimensional
Two-Dimensional
Three-Dimensional
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Waves tord
propagated string
are
in Waves formed on the Waves propagated in sky one- water surface are two- in the form of light and
dimensional waves.
dimensional waves.
sound wave are threedimensional waves.
When propagation of wave takes place in one-dimension, twodimension & three-dimension then waves are said to be onetwo-dimensional
and
three-dimensional
waves
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dimensional, respectively. Note:
All non-mechanical waves are transverse, this ??????? implies that–
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If a wave is longitudinal it is mechanical but if a wave is mechanical it
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may or may not be longitudinal.
If a wave is non-mechanical it is transverse but if a wave is transverse it
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may or may not be non-mechanical. Mechanical waves in different media– (i)
In strings, mechanical waves are always transverse that to when string is under a tension.
(ii)
In gases and liquids mechanical waves are always longitudinal. For example:
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Sound waves in air or water. This is because fluids can not sustain sheer. (iii)
In solids mechanical waves can be either transverse or longitudinal depending on the mode of excitation. he speeds of two waves in the same solid are different longitudinal wave travels faster than the transverse wave. For example:
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If we strike a rod at right angle, figure (a the waves in the rod will be transverse while if the rod is struck at the side or is rubbed with a cloth. The waves in the rod will be longitudinal
Audible or Sound Waves:
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as shown in figure (b).
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These are longitudinal mechanical waves which lie with in the range of sensitivity of human ear typically 20 hertz to 20 kilo hertz. These are
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generated by vibrating modes. Such as tuning for vocal cords, stretch strings or membranes. Infrasonic waves: The longitudinal mechanical waves whose frequencies lie before 20 hertz are called infrasonic waves, l is greater than 16.6 meter. Infrasonic waves are created by earth quakes, p-waves i.e. pressure waves, volcanic eruptions, ocean waves and elephants and wells. Perodic motion such as of a pendulum at frequency lesser than 20 hertz. 20 hertz also produces these waves.
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Ultrasonic waves: Longitudinal mechanical waves having frequencies greater than 20 kilo hertz are called ultrasonic waves. Here is less than 1.66 cm. Though human ear can not detect these waves certain creatures such as mosquito, fish, dog and bat show response to these. These waves can be produced by the high frequency vibrations of a quartz crystal under an alternating electric field, peso electric field.
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Velocity of Sound:
When a sound wave travels through a medium such as air, water or steel it will set particles of medium into vibrations as it passes through them. For this to happen the medium must position both inertia i.e.
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mass density so that kinetic energy may be stored and elasticity so that
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potential ones is stored. These two properties of matter determine the velocity of sound i.e. velocity of sound the characteristic ??? the
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medium in which wave propagates. In one medium velocity of wave remains constant. Though velocity of particles making simple harmonic motions are different though velocity of particular, performing simple harmonic motion during propagation of wave remains different at time but the velocity of wave always remains constant in one medium. When medium change then only velocity of wave changes. Velocity of sound in medium of elasticity e and density given by: V=
e
…(i)
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(a)
As solids are most elastic while gases at least that is modulus of elasticity for solid remains more than for liquids remains more than for gases. E s > Ee > E g So, velocity of sound is maximum in solids and maximum or minimum in gases. Se >
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Velocity of sound in a steel is more than velocity of sound in a water is more than velocity of sound in air. Normally velocity of sound in steel is approximately 5000 m
iit
while velocity of sound in water approximately 1500 m/s
(ii)
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while velocity of sound in air is approximately 330 m/s. Propagation of sounds in solids:
Y
cr
Vsolid =
and Vliquid =
Vliquid or Vgas =
B B
where B = Bulk Modulus. (iii)
Newton's formula:
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When sound propagates through air temperature remain constant i.e. process is isothermal Bulk modulus of elasticity = E = P. So velocity of sound in air, according to Newton Vair =
P
= 1.3 kg/m3
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P = 1.01 × 105 N/m2
so velocity of sound in air
iit
1.01 105 279 m / s = 1.3
ac k
While experimental value is 332 m/s. Laplace's correction:
cr
According to Laplace, propagation of sound in air is not isothermal, but it is a adiabatic process that means quantity of heat remains constant. So bulk modulus of elasticity for a adiabatic process is = P.
Where is ratio of specific heat to constant pressure. Ratio of specific heat at constant pressure to specific heat at constant volume.
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so
CP 1 2 CV F
Where F is degree of freedom. For monoatomic gases F = 3, so = 1.67 For diatomic gases F = 5,
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so = 1.4 and For polyatomic gases F = 6, so = 1.33
P .
ac k
iit
In this way, we can find velocity of sound in gases V = In case of gases
P
cr
V=
=
PV Mass
=
RT Mass
=
RT M
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mass molar mass
so Vrms =
3RT M
so ratio of V and Vrms i.e.
V Vrms
3
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So V < Vrms, i.e. velocity of sound in a gas is of the order of the speed of gas molecules and lesser than it. (vi)
Velocity of gas at constant temperature depends on nature
iit
of gas that is atomicity and molecular weight. Lighter is the
Effect of temperature:
RT M
In a gas V =
cr
(viii)
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gas greater will be the velocity.
V T
so
V' V0
T' T
Here temperature T is taken to be in Kelvin =
273 T 273
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1 t = 273
1/2
When 't' is very small, we can write V' = V0 + 0.61 T Here T is in °C. If temperature in increases by 1°C. Velocity of sound
Effect of humidity: Since, V=
P
iit
(viii)
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increases by approximately 0.6 m/s.
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with the increase in humidity.
density decreases, so speed of sound increases, with rise
cr
in humidity velocity of sound decreases. So with rise in humidity velocity of sound increases. Therefore sound travels faster in humid are in rainy season than in dry air in summer season at the same temperature.
(ix)
Effect of pressure: Since, V=
P
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=
RT M
Pressure as no effect on velocity of sound in a gas as long as temperature remains constant. Equation of plane progressive wave: The velocity of the oscillating particles are different indifferent states of oscillations but the velocity of disturbance remains constant and
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depends only upon the nature of the medium. If we produce waves in a medium continuously the particles of medium oscillates continuously with a constant amplitude. In this condition the disturbance produce in the medium is called a plane progressive wave.
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When a plane progressive wave propagates a medium then at any
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instant all the particles of the medium oscillates in the same way but the phase of oscillation changes from particle to particle. If on the
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propagation of a wave in a medium the particles of the medium perform simple harmonic motions than the wave is called a simple harmonic progressive wave. Figure shows the y – x curve for the wave, when wave is propagating in x-direction and 'y' is the displacement of medium particles at different positions. Particle location are also shown 1, 2, 3, 4, 5, 6 at different positions.
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Let the time be counted from the instant when the particle-1 situated at the origin starts oscillating. If 'y' is the displacement of this particle after time 't' seconds then: i = R sin wt The displacement of the particle 6 at a time 't' will be the same as that of the particles 1 at that time
x second earlier. V
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Since V is the speed of disturbance which remains constant. So can write:
x V
ac k
where 2 2n
iit
Y = R sin wt
= angular frequency
cr
so y = A sin 2nt since n
x V
V
2 Vt x l
so y = A sin
V n
T
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2 t x T
y = A sin
t x T
y = A sin 2
This is the equation of plane progressive wave propagating in +X direction. Now we can write the equation of plane progressive wave
t x T
y = A sin 2
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propagating in –X direction, as
If be the phase difference between the above wave travelling along
iit
the +X direction and an other wave then the equation of that wave will be:
t x T
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y = R sin 2
cr
If at any point X in the medium we draw a plane perpendicular to the X-axis then all the particles on the plane will have the same displacement 'y' at the same instant 't'. Such a wave is called a plane wave. Phase of particle: The quantity which expresses at any instant the position of the particle and its direction of motion is called the phase of particle. Relation between phase difference and path difference:
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Since equation of a simple harmonic progressive wave travelling in +X direction instantaneous displacement of a medium particle at a distance x from the origin.
t x T
y = A sin 2
t x . T
Here, phase of particle is 2
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Suppose at any instant t, 1 and 2 are the phases of two particles whose distances from origin are x1 and x2 respectively then;
iit
t x 1 2 1 T
ac k
t x 2 2 2 T
2 x2 x1
cr
1 2
2 x
so, phase difference = path difference ×
2
if x then 2 Relation between phase difference and time difference:
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Suppose the phase of particle distance X from the origin is 1 at time t1 and 2 at time t2 then:
x t 1 2 1 T x t 2 2 2 T
2 t T
if t T
ac k
then 2
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so,
2 t1 t2 T
iit
So, 1 2
This means that after one time period the phase of oscillation of a
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particle become the same as in the beginning. Time displacement graph of a particle in progressive wave as shown in figure.
One side particle displacement y is shown and on other side time t is plotted. In second graph 'distance displacement' graph is shown. Distance is plotted in the direction or propagation of wave and displacement is on one axis as shown in figure. Example:
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A particular wave is given by y = 0.2 sin (0.5 x – 8.2 t), then find the amplitude wave propagation factor
K
wavelength,
frequency
time-period
and
value
displacement at x = 10 m and t = 0.5 sec. Solution: Amplitude (a) = 0.2 m
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2
K = 0.5
2 4 K w 2
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frequency (n) =
iit
K=
= 1.31 hertz
1 n
cr
time-period (T) = = 0.77 sec. Speed (V) =
w K
= 16.4 m/s displacement y = A sin (Kx – wt)
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of
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y = 0.156 m. Example: A progressive wave of frequency 500 hertz is travelling with a velocity of 360 m/sec. How far apart are two points 60° out of edge? Solution:
V n
2m x 2
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so, x
iit
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= 0.72 m
so, x
0.72 2 3
Example:
cr
= 0.12 m
The equation of a simple harmonic progressive wave y = 0.3 sin (314 t – 1.57 x) where t, x and y are in second, meter and cm respectively. Calculate the frequency and wave-length of the wave? y can be written as– Solution:
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y=
0.3 sin 314 t – 1.57 x meter 100
y = 0.003 sin 2 50t
x 4
so, time period t=
1 sec. 50
4m
2 K
iit
V n
ac k
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n = 50 sec–1
Relation between particle velocity and wave velocity:
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The velocity of particles of medium in which a sound wave is propagating is given by V particle. =
dy dt
= wA cos wt – Kx
…(i)
so, maximum value of particle velocity is wa. The wave velocity of the sound is however fixed for a given medium
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V n
w K
Strain in a medium: The strain in a medium in which compression rarefaction of sound wave are travelling is given by
y x
= kA cos wt – kx
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strain =
Vparticle strain
w K
ac k
y max. k x
iit
Thus, the maximum strain is
= wave velocity
cr
Particle velocity = wave velocity × strain in the medium. Wave equation for a sound wave is represented by y = A sin (wt – kx) Condition: For any function to represents the wave
2y 2 2y V t2 x2
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where V =
w k
= wave velocity Characteristics of Sound:
(i)
Loudness
(ii)
Pitch
(iii)
Quality
(i)
Loudness:
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Sound has three characteristic–
iit
The loudness which differentiates a faint sound with an
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intense sound is known as loudness yellow color in this wood.
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Differentiate between loudness and Intensity:
Loudness
Loudness is
related
Intensity to Intensity is not related to
sensitivity of ear.
sensitivity of ear. For
intensity its physical
For loudness its physical measurement can be done. measurement can not be done.
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loudness intensity. loudness density of medium. loudness amplitude. loudness (ii)
1 . square of distnace of point
Pitch:
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This property differentiates a sharp sound with a grave sound. Buzzing of mosquito ha higher pitch than barking of dog. Pitch provides sensation of sharp and grave sound while
(iii)
Quality:
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frequency gives its measurement.
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Quality is the property which differentiate between the two sounds of same frequency and same intensity.
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Pitch of a sound is that sensation by which be differentiate a buffalo voice, a male voice and a female voice. This sensation depends on the dominant frequency present in the sound. Higher the frequency higher will be the pitch and vice-versa. The dominant frequency of a buffalo voice is smaller than that of a male voice which intern is smaller than that of a female voice. Echo:
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Sharp sounds created in front of distance walls or clips that reflected and after sometime we hear back hour own sound as reflected from these obstacles. These reflected sounds are called echo's. For hearing echo, two important factors must be taken into a account– (i)
The reflecting body or the obstacles must be quite large is compared to the wavelength of sound.
(ii)
Hour persisting of hearing. The sound persists 0.1 sec. even
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after the sounding body has stopped vibrating. This time of 0.1 sec. is called the persistence of hearing. So, if want to here a distant echo the reflected sound must come after
echo's are possible.
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the 0.1 sec. in a small room whose walls are less than 16.5 m away no
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In this case original sound merges with the reflected sound and sustain sound reverberation is heard.
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Power and Intensity:
If a sound wave is given by y = A sin wt – kx is propagating through a medium the particle velocity V particle =
dy = Rw cos wt – kx dt
so, If is the density of medium. Kinetic energy of the particle per unit volume
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1 dy = 2 dt =
2
1 w2a2 cos2 wt kx 2
Kinetic Energymax. = Potential Energymax.
U=
1 w2a2 2
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= Eenergy density
E = U + V
1 2 2 a w Sx 2
ac k
=
iit
So, the energy associated with a volume Sx
So, power rate of transmission of energy
E T
cr
P=
=
1 Vw2a2 S 2
In case of sound wave displacement amplitude is related to pressure amplitude as– P0 = VRw So, intensity can be written as–
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P02 I= 2V When sound wave propagate in air the amplitude, displacement 'y' of the particles of air is of the order 10–6 m or less. Hence, in order to draw the yx curve for sound wave a very large scale is selected for y. Intensity of wave:
through oscillations.
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In wav motion the energy is propagated from one place to anther place
Thus the direction of propagation of a wave in a medium is the
iit
direction of flow of energy. The amount of energy flowing per unit time through unit area perpendicular to the direction of propagation of the
second.
ac k
wave is called the intensity (I) of the wave. Its unit is Joule/meter2
cr
I = 22n2R2V
where R is amplitude of oscillations of the particles of the medium for a given medium. The density is constant and for a given wave the speed V is constant.
I a2 and I n2 Relation between Intensity and distance from the source:
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Intensity is inversely proportional to the square of distance from the source for a point source. The intensity (I) at a point due to waves coming from a point source is inversely proportional to the square of distanced 'r' from the source. So I
1 r2
Provided no energy is absorbed by the medium. If the wave source is
I
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cylindrical then:
1 r2
U=
E V
ac k
E S T
cr
I=
iit
Intensity is also called energy flux or power density.
so,
I E V U ST E
so,
I E ST U ST E
V ST
L V T
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I=U×V so, Intensity = Energy density × Velocity. In case of electro-magnetic waves that is light waves
EV 0
I=
E c V 1 . 0
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c=
where µ0 is absolute magnetic permeability of free space or vacuum
ac k
E2 I= c 0
iit
and 0 is absolute electric permeability of free space or vacuum.
cr
I = c0 E2
Remember P0 = VRw the pressure wave is 90° out of edge with respect to displacement wave. i.e. when displacement will be maximum then pressure will be minimum and vice-versa. A sound sensor detect pressure changes. The phase diagram for pressure wave and displacement wave are shown in figure. For displacement wave, if
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y = A sin t – kx then, for pressure wave P = P0 cos wt – kx which indicates that pressure wave is 90° out of edge with respect to displacement wave. Loudness: A normal human ear can just hear a sound of frequency 1 kilo hertz
je e. in
having the intensity of 10–12 watt/m2.
i.e. threshold of hearing. Sensation of loudness L remains directly
L log I
ac k
L = K log I
iit
proportional to logarithm of intensity of sound.
This is known as waver phasional equation.
cr
Intensity of sound is nothing to do with the sensitivity of the ears. Intensity of sound defines quantity and can be measured. On the other hand the loudness of sound is as sensation of hearing produce own our ear. Musical interval: The ratio of the frequencies of two pure sound nodes is known as musical interval between the two nodes.
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Whenever two nodes are sounded in succession then they produces a different effect smoothing or jerking to the ear. This effect depends upon the musical interval between the nodes, whatever be there frequencies. Those intervals whose effect on the ear is pleasant or soothing have been given special name– For unison – 1:1
For fifth – 3:2 For fourth – 4:3 For major third – 5:4
ac k
For major six – 5:3
iit
For minor third – 6:5
je e. in
For octave – 2:1
For minor six – 8:5
cr
For major tone – 9:8 For minor tone 10:9 For seventh – 15:8
For semi tone – 16:15 Decibels: The loudness of a sound is heard by the human ear is not directly proportional to the intensity of the sound but rather is proportional to
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the logarithm of the intensity. The human ear can hear sounds of intensities as low. I = 10–12 watt/m2. This value is called the threshold of hearing. The human ear can hear sounds of intensities as high as 1 W/m2. This is known as threshold of pain.
log10
I Io
je e. in
so, sound level is 10
Sound level is measured in decibel (dB).
iit
1 of Bel (B which was named to honour Alexander. 10
ac k
Decibel is
At an intensity of 20 decibel (dB), a frequency of 1000 hertz can easily be heard but a frequency of 100 hertz count not be heard at all. To
cr
hear 100 hertz the intensity level is to be increased to 35 decibel to hear is sound of frequency 2000 hertz. The intensity level would have to be increase to 40 decibel. Ultrasound: Ultrasounds are sounds with frequencies above 20,000 hertz. Birds and dogs can hear ultrasounds and bats use them for navigation. The S.I. unit of intensity is watt/m2. Since human ear response to sounds intensities over a white range that is from 10–12 watt/m2 and so
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instead of specify intensity of sound in watt/m2. We use a logarithm scale of intensity called the sound level defined as sound level. SL = 10 log10
I Io
Where Io is threshold of human ear = 10–12 watt/m2. A sound of intensity Io as
I Io
je e. in
dB = 10 log10 = 0 decibel.
While sound at the upper range of human hearing called threshold of
iit
pain, as a intensity of 1 watt/m2.
ac k
For, sound level (SL) = 10 log10
1 1012
= 12 decibel.
cr
We also use dB as a relative measure to compare different sounds with another rather than with reference intensity as For two intensities I1 and I2 as SL1 – SL2 = 10 log10
I1 I 10log10 2 I0 I0
So, SL1 – SL2 = 10 log10
I1 I2
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If,
I1 2 I2
then, SL1 – SL2 = 3 dB. Remember ratio of two intensities co-responds to the difference in their sound levels. Example: What is the maximum possible sound level in decibel of sound waves
je e. in
in air. Given that density of air equal to 1.3 kg/m3. V = 332 m/s and atmospheric pressure = 1.01 × 105 N/m2. Solution:
iit
For maximum possible sound intensity pressure amplitude of wave will
P0 = P
ac k
be equal to the atmospheric pressure.
cr
= 1.01 × 105 N/m2
P02 I= 2V
1.01 10 = 5
2
2 1.3 332
= 1.16 × 107 watt/m2
Sound level = 10 log10
I Io
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107 = 10log10 1012 = 190 dB. Example: The power of sound from the speaker of a radio is 20 miliwatt by turning the knobe of volume control. The power of sound is increased to 400 milliwatt. What is the power increased in decibel as compare to the
je e. in
(a)
original power. (b)
How much more intense is an at decibel sound then at a 20
iit
dB whisper.
SL2 – SL1 = 10 log10
= 10 log10
P2 P1
= 10 log10
400 20
cr
(a)
ac k
Solution:
I2 I1
So, change in sound level (SL) = 13 decibel. (b)
SL2 – SL1 = 10 log10
I2 I1
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so, 80 – 20 = 10 log10
6 = log10
so,
I2 I1
I2 I1
I2 106 I1
Example:
je e. in
A dog while barking delivers about 1 miliwatt of power, if this power is uniformly distributed over a hemispherical area. What is the sound level at a distance of 5m. What would the sound level be if instead of one dog, five dogs start barking at the same time each delivering 1
iit
milliwatt of power?
=
P S
cr
I
ac k
Solution:
P
1 4r2 2
103 = 2 2 = 6.37 microwatt/meter2 Sound level (SL) = 10log10
I Io
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6.37 106 = 10log10 1012 = 10 log 6.37 + 6 = 6.8 decibel If there are five dogs barking at the same time and same level.
SL2 – SL1 = 10 log10
I2 I1
SL2 – SL1 = 10 log10
5I1 I1
iit
SL2 – SL1 = 10 log10 5
je e. in
I2 = 5I1
ac k
so, SL2 = 68 + 10 × 0.7 75 decibel.
cr
Principle of Superposition of Waves:
According to principle superposition of waves, when two or more than two sound waves propagates in a medium then resultant, displacement of a medium particle at a given is equal to vector sum of displacement of that particle due to individual independent waves at that time. y = y1 + y2 + y3 ……
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This is known as principle of superposition of wave. Propagation of wave characteristics like frequency, wave l etc. do not get change i.e. each wave produces its effect independent way. Beats: When two sounds wave of nearly same frequency propagates in a medium in same direction then intensity of resultant wave at a particular place varies periodically with.
je e. in
The variation in an intensity of sound in known as ????? one waxing and one manning of sound forms one beat an number of beats heard per second is known as beat frequency.
Let us assume that two sound waves of amplitude a are propagating in
iit
same direction having frequencies n1 and n2 respectively and
ac k
displacement produces by then at time 't' are y1 and y respectively. y1 = A sin w1t
cr
= A sin 2n1t
y2 = A sin w2 t = A sin 2n2 t n 1 > n2
From principle of superposition y = y1 + y2 y = A sin 2n1t + A sin 2n2t
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y = A2 sin (n1 + n2) t cos (n1 – n2) t = 2A cos (n1 – n2)t sin (n1 + n2)t if, A = 2A cos (n1 – n2) t then, y = A sin (n1 + n2) t So, maximum amplitude = + 2A
A = 2A cos (n1 – n2) t
je e. in
minimum amplitude = 0
So, maximum value of amplitude
(n1 – n2) t = k
K n1 n2
cr
so t =
ac k
where, K = 0, 1, 2, 3
iit
cos (n1 – n2) t + 1
1 2 3 , , etc. n1 n2 n1 n2 n1 n2
The time interval between any two consecutive maximum intensity sound is
1 sec. n1 n2
Hence, in one second we hear maximum intensity sound for (n 1 – n2) times.
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For minimum value of amplitude cos (n1 – n2) t = 0
n1 n2 t
k 2
where K = 1, 3, 5…. t=
1 3 5 , , , ... 2 n1 n2 2 n1 n2 2 n1 n2
je e. in
t=
k 2 n1 n2
So, the time-interval between any two consecutive minimum intensity
iit
1 sec. n1 n2
ac k
sound is
Therefore, in one second minimum intensity sound a heard (n 1 – n2)
cr
times.
In one second both minimum and maximum intensity sound are heard for (n1 – n2) times.
So, beat frequency is n1 – n2. Application of Beats: To find the frequency of unknown timing fork: The method ?????? is either load a tuning fork or filing a tuning fork. On loading a tuning for after putting some wax on the tuning fork its
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frequency decreases and after filing a tuning fork its frequency increase. How to calculate the frequency of unknown tuning fork by waxing a tuning fork: n1 is known and n2 is unknown. After sounding two tuning fork let x beats are Hz per second. So frequency of unknown tuning fork must be either n1 + x or n1 – x.
je e. in
Since after waxing frequency must decrease, if after waxing, number of beats Hz. fter waxing be once again sound to two tuning fork together if after waxing number of beats per second increases than beats frequency is n1 – x. If after waxing number of beats Hz per second
iit
decreases then beats frequency is n1 + x. In this way we can find the
ac k
frequency of unknown tuning fork.
Comparison between progressive and stationary waves: These waves advance in a medium with a definite velocity
cr
(i)
while stationary waves remains stationary between two boundary in the medium.
(ii)
In progressive waves all particles of the medium vibrate and the amplitude of vibration is same for all of them. While in stationary waves except nodes all other part of the medium vibrate but the amplitude of vibration different from one particle to the other particle. The amplitude is zero at the nodes is maximum at the ???????.
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(iii)
At any instant, the phase of vibration varies continuously from one particle to other particle in progressive waves in case of stationary waves at any instant, the phase of all particles between two succession nodes is the same. In stationary waves at any instant the phase of all particle between two successive nodes is the same but phase of particles on one side of a node is opposite to the ??? of
(iv)
je e. in
particle on he other side of the node. In progressive waves at no instant all the particles of ???? medium pass through their mean position simultaneously while in case of stationary waves all particles of the medium
(v)
ac k
period.
iit
pass the their mean position simultaneous twice in lack time-
In longitudinal progressive waves all the particles of the ??? suffer in succession the same variation in pressure and
cr
density. ???? in case of stationary waves in longitudinal stationary wave the variation in pressure and density is maximum at ?????? and minimum at antinodes.
(vi)
In case of progressive crest and trough in transverse progressive wave a centres of compression and rarefactions in longitudinal progressive waves advance with a definite velocity who in case of stationary waves crest and trough in transverse stationary waves and centres of compression and
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rarefactions
in
longitudinal
stationary
waves
occur
alternately at definite place and do not advance. (vii)
In case of progressive waves, these waves transmit energy in the medium while stationary waves do not transmit energy in the medium.
Reflection of pulse from rigid and free support: Whenever any displacement wave gets reflected from rigid end its
je e. in
suffers by a phase change of as shown in figure, but when displacement wave is reflected from the free end phase change do not take place i.e. from free end compression returns in the form of compression and rarefactions returns in he form of rarefactions. While
iit
from rigid and compression returns in the form of rarefactions and
ac k
rarefaction returns in the form of compression. Formation of Stationary Waves:
cr
When two waves of same frequencies and same amplitude travels in a opposite direction and same speed their superposition gives rise to a new type of waves called stationary waves or standing waves. Equation of progressive wave propagation in + Y direction is y1 = A sin (wt – kx) If this wave is reflected from the free and then the equation of reflected displacement wave is y2 = A sin (wt + kx)
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The resultant equation is y = y1 + y2 y = 2A cos kx sin wt y = Ax sin wt where A is amplitude of standing wave = 2A cos kx Which is not a constant but varies periodically with position. Node with
je e. in
time as in beats. Nodes:
Nodes are the point where amplitude is maximum. cos kx = 0
3 5 , , are the positions of form of nodes. Remember 4 4 4
ac k
So, x =
iit
i.e. kx =
cr
displacement node remains the ????? of pressure antinodes i.e. at nodes displacement is ???? and pressure change is maximum. Position of Antinodes: Antinodes are point where amplitude that is displacement amplitude maximum. the point where displacement amplitude is maximum pressure remains minimum. So pressure node correspond to displacement antinode and pressure antinode correspond to displacement node. So position of antinode can found by rating cos kx = + 1. So kx = , , 2.
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So x = 0,
2 3 , , , ...... are the positions ????? antinodes which 2 2 2
are shown in the figure. Example: The standing wave y = 2A sin (kx cos wt) in a ??? organ pipe is the resultant of superposition of y1 = A sin wt – kx and y2 = …… Find y2? Solution:
so, y = 2a sin (kx cos wt)
je e. in
2 sin A cos B = sin (A + B) + (sin A – B)
y = y 1 + y2
ac k
y1 = A sin (wt – kx)
iit
= A sin (kx + wt) + A sin (kx – wt)
= – R sin (kx – wt)
cr
y2 = y – y1
= A sin (kx + wt) + 2A sin (kx – wt) y2 = A sin (wt + kx) – 2A sin (wt – kx) Example: Sound waves of frequency 660 Hz. Force normally ???? perfectly reflecting wall the shortest distance from wall at which the air particles have maximum amplitude of vibration is …… Find this value? (V = 330 m/s)
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Solution: At rigid and node is formed and free end antinode is formed. So, at the distance between consecutive antinode and normal is
. 4
330 660
= 0.5 metre.
which is the distance between consecutive antinode 4
and node. Vibrations of air columns:
iit
0.125 m i.e.
je e. in
So, the shortest distance where particle have minimum distance is
ac k
An organ pipe is a cylindrical ???? of uniform diameter which is used to produce sound.
(i)
cr
Organ pipes are of two types:
(ii)
Open organ pipe.
Closed organ pipe.
Open organ pipe is open at both the ends while closed organ is open only at one end. In organ pipe the close and reflects the longitudinal wave like a rigid boundary i.e. the state of compression in the form of compression and the state of rarefaction in the form of rarefaction, because the rigid boundary for pressure wave phase change do not
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take place. The open end of the pipe reflect the longitudinal wave like a free boundary i.e. its reflects state of compression in the form of rarefaction and the state of rarefaction in the form of compression. In organ pipe longitudinal stationary waves are formed. Vibrations of air column in closed organ pipe: Always remember that is the ????? end displacement node is formed and at the free displacement antinode is formed.
je e. in
In terms of pressure node or pressure antinode at rigid end, pressure antinode is formed while at free end pressure node is formed but we are looking figures only in the terms of displacement wave. The three
Overtones:
ac k
second over tone.
iit
figures have been shown for fundamental node, first on tone and
Tones of frequencies higher than the function mental tone are known
cr
as overtones.
Fundamental Frequency: The minimum frequency ??????? which can be produce in a ????? is known as fundamental frequency. Any frequency which is n times of the fundamental frequency is ???? name nth harmonic. In fundamental mode of vibration at rigid end node will ???? formed and at free end antinode will be formed.
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Length of the organ pipe is l and the distance between antinode and node is
n1 =
=
. So, fundamental mode vibration 4
V 1
V 4L
je e. in
This n1 is the fundamental tone or first harmonic of ??? pipe.
L
32 4 4L 3
n2
V 2
cr
ac k
2
iit
Calculation for first overtone:
n2 =
3V 4L
n2 = 3n1 This n2 is known as first over tone of the closed pipe or third harmonic because n2 is the three times of the fundamental frequency. Similarly in a closed organ pipe for 2nd overtone–
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n=
5 3 4 4L 5
so, 3
n3 =
5V 4L
= 5n1
je e. in
This is known as second over tone or fifth harmonic. n1 : n2 : n 3 = 1 : 3 : 5
A closed organ pipe produces only odd harmonic.
iit
Vibrations of air columns in open organ pipe:
ac k
In free ends antinodes are formed. An organ pipe produces both the even and odd harmonic.
If temperature of air is increases than speed of sound in air is also
cr
increases. Therefore frequency of nod of the organ pipe also increases. Figure shows the diagram for fundamental mode of vibration, first overtone and second over for displacement waves. L=
1 2
1 2L n1 =
V 2L
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This is known as lowest frequency or fundamental tone ??? first harmonic of the open organ pipe. For 1st overtone– L=
2 2 2 2L 2
n2
V 2
n2 =
je e. in
2
2V 2L
iit
n2 = 2n1
ac k
This is known as 1st overtone or 2nd harmonic. For 2nd overtone–
3 3 2
cr
L=
3
2L 3
n3
V 3
n3 =
3V 2L
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n3 = 3n1 This is known as 2nd overtone or 3rd harmonic. So open organ pipe: n1 : n 2 : n3 = 1 : 2 : 3 End corrections: According to Lord Rayleigh, in an organ pipe the antinode is not formed exactly at the open end but slightly outside because the particles of ??????? completely free to move in a every direction. Therefore
je e. in
vibrations of the air particles in the pipe are not restricted to the open end but occur outside also. The antinode is formed at a place slightly outside the open. Thus the length of vibrating air columns become slight greater than the length of the pipe. The distance of antinode
iit
from the open end is called the end correction. End is usually
ac k
represented by the letter 'e'.
In fundamental mode of vibration, for closed organ
V 4 L e
cr
n1 =
For open organ pipe– n1 =
V 2 L 2e
e = 0.6 r where, r is the radius of pipe. Resonance tube:
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Closed organ pipe having an air column of variable length. If the length of air column is varied until its natural frequency equals the frequency of the fork than the column resonates and emits a loud node. In this condition the air column of the tube is called the resonance column. The resonance tube is one metre long and five cm in diameter made up of brass or glass. When the pronk of tuning fork goes from A to B hen a wave of compression travels downward in the
je e. in
air column. This wave of compression is reflected from the water surface as a wave of compression and returns into the open and of the tube.
Where it is reflected as a wave rarefaction exactly at this moment
iit
pronk of fork after completing its half vibration begins to returns from
ac k
B to A. Adjacent in the tube a wave of rarefaction. The two waves of rarefactions be in same phase force each other.
cr
1st resonance length L1 =
4
and 2nd resonance length L2 =
L 2 L1
3 4
2
2 L2 L1 2 1 L2 L1
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=
2
= 2 × (L2 – L1) so, velocity (V) = n = 2n (L2 – L1) Velocity at 0°C = velocity at t°C – 0.60, if temperature variation is small.
4
L2 e
3 4
…(i)
…(ii)
iit
L1 e
je e. in
Now applying end correction.
e=
ac k
After solving these equation,
L 2 3L1 2
cr
IInd resonance length is more than three times of Ist resonance length, if end correction is considered. If end correction is not considered than IInd resonance length is three times of the Ist resonance length. Doppler Effect in Sound: The apparent change in the frequency of the source due to a relative motion between the source and the observer is know as Doppler effect.
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Case-I: when sound source is moving and observer is stationary. Let the source is emitting n waves per second if the source is stationary then these n waves will spread in the distance V, so
V , but, since source is n
moving so in one second distance covered by the source is Vs, so n
'
V Vs n
je e. in
waves will now spread in distance V – Vs.
V '
n'
V V Vs n
cr
n' > n
ac k
n'
iit
where ' is the apparent wave length.
' <
Always remember apparent wavelength will change only when source will move, if source is not moving than apparent wavelength will not change at all. Case-II: When observer is moving and source is stationary. Let source is emitting n waves per second and observation speed is Vo.
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n' n =n
Vo
Vo n V
n' n
V Vo V
'
je e. in
Case-III: Figure shows that both source and observer are moving towards each other with speeds Vs and Vo respectively.
iit
'V' the speed of sound.
Imagine that is only source were moving than the apparent frequency
V V Vs n
cr
n1 =
ac k
heard by observer would be n1,
Since observer is also moving, So, n' =
n' =
V Vo V n1
V Vo V Vs n
This is the ????????????????????????. Case:
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When source is moving towards the hill. Figure shows two observers at point A and point B and source is moving with speeds Vs. Let V is wave velocity and 'n' is the frequency of sound source. For the observer situated at A in between the source 'S' and wall. The observed frequency of the waves receipt directly from the source receding.
V V Vs n
je e. in
n' =
Observed frequency of reflected waves will also be n' =
V V Vs n
iit
but for the observer situated at location B behind the source, the
ac k
observed frequency of the waves received directly from the source will be source receding
V V Vs n
cr
n' =
but the observed frequency of the reflected wave n' =
V V Vs n
diagram here
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Remember some important points about doppler's effect: (i)
Doppler effect in sound in not symmetrical in nature.
(ii)
Doppler effect is not symmetrical with respect to the position of the observer and motion of the source. The change in apparent frequency is more when the source is moving.
(iii)
Transverse Doppler effect.
direction. (iv)
je e. in
The Doppler effect in sound does not take place in transverse
Then a speed of source is more than speed of sound then
(v)
ac k
place.
iit
source gets a hate of sound so Doppler effect do not take
If speed of observer is more than speed of sound than the sound waves will never reach the observer.
cr
So, once again Doppler effect will not take place. Also when both source and observer are moving in a same direction than also Doppler effect will not take place.
Find the speed of transverse wave in a stressed string: A figure shows a wave pulse of length dl. The direction of tension acting on the pulse have been shown: 2T sin = 2T where, is small
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2T sin = T
dl r
dl r
d dm = µdl
where µ is mass per unit length.
V2 Tdl dm r r
T
iit
V=
V2 Tdl r r
ac k
µdl
je e. in
For circular motion,
So, speed of transverse wave is not a wave function of wave
cr
characteristics. It is a function of tension in a string and mass per unit length.
V
T
For fundamental node or Ist harmonic we can write frequency. n=
1 2L
T
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In general, we can write possible frequency of vibration of string:
P
FP =
2L
T
where µ is mass per unit length of string. Vibrations of String: When a string capable of vibrating that is under tension is set into
je e. in
vibration transverse harmonic waves will propagate along it. If the length of string is finite reflected wave will also exists and so due to multiple reflection standing waves of large amplitude called resonant standing waves will be produced i.e. the waves in a tord string of finite
iit
length are transverse stationary. The strings will vibrate in such a way
ac k
that fixed. Points of string are nodes as the string a these point is not free to move while the points of plucking or free end is an antinode as
cr
here displacement will be maximum. Figure shows, the fundamental node, 1st overtone 2nd overtone, figure shows the 1st harmonic, 2nd harmonic and third harmonic figure. Vibrations corresponding to 1st harmonic, 2nd harmonic and 3rd harmonic have been shown in figure. In tord string: The fixed end are nodes than correspond to 1st harmonic we can write: n=
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2L n=
n
V V 2L
The higher possible frequencies are called overtones with f 2 be second harmonic or f3 be third harmonic or second overtone. Regarding
je e. in
frequency of a vibrating string it is birth noting that– (i)
As a string has many natural frequencies all integral multiples of fundamental frequency so when it is excited with a tuning fork or a vibrating body the string will be in
iit
resonance with the given body.
Meysan's Law of Vibration: F
1 L
cr
(ii)
ac k
If any one of its natural frequency coinside with that of body.
If T and µ are constant
F
T
If L and µ are constant
F
1
If T and L are constant,
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If M is the mass of string and length L than
M L
so, we can write
1
f=
T
2L
ac k
1 T 2 ML
iit
T M L
2L
=
je e. in
1
=
1
=
T r2
cr
2L
=
1 2Lr
T
Viscosity and Newton's law of viscous force: In case of steady flow of a fluid when a layer of fluid slips or tense to slip on a adjacent layer in contact the two layers adjust tangential force
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on each other, which tries to destroy the relative motion between them. The property of a fluid due to which it opposes the relative motion in its different layers is called viscosity or fluid friction or internal friction and the force between the layers opposing the relative motion is called viscous force. Viscosity:
je e. in
The property of a fluid due to which it opposes the relative motion between its different layers is called viscosity or fluid friction or internal friction. Newton found that viscous force 'F' acting on any layer of a fluid is directly proportional to it’s a and to the velocity
AdV dy
cr
So, F
dV at the layer. dy
ac ki it
gradient
so, F is .
AdV dy where is coefficient of viscosity or simply viscosity of fluid.
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Here negative sign (–) shows that viscous forces on a fluid layer X in a direction, which opposes the relative velocity i.e. relative motion. Figure shows the
dV that is velocity gradient. dy
Regarding viscosity of a fluid it is birth noting that it depends only on the nature of fluid and is independent of area considered or velocity gradient. Its dimensions are M L–1 T–1 and its unit is poisewile (PI) in S.I.
je e. in
(ii)
unit while in C.G.S. unit it is dyne sec/cm2 called poise (P). 1 poisewile = 10 poise (iii)
In case of liquid viscosity increases with density while for
ac ki it
gases it decreases with increase in density for with increase in temperature viscosity of liquids decreases while that for gases increases.
With increase in pressure the viscosity of liquid except water
cr
(iv)
increases while that of gases in practically independent of pressure.
Stoke's law and terminal velocity: When a body moves through a fluid, the fluid in contact with the body in dracked with it.
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This establishes relative motion in fluid layers near the body due to which viscous force starts operating the fluid exerts viscous force on the body to oppose its motion. The magnitude of the viscous force depends on the shape and size of the body is speed and the viscosity of the fluid. Stokes established that if a sphere of radius r moves with velocity v through a fluid of viscosity
F = 6 rv
je e. in
. The viscous force opposing the motion of the sphere is
This law is called Stoke's law.
W = mg
4 3 r 3
ac k
=
iit
Now if sphere is drop in a fluid its weight
X vertically downwards while upthrust
4 3 r g 3
cr
Th =
and viscous force 6 rv at vertically upwards. Initially v = 0 and > g,
> so, the body will be accelerated downwards because of the acceleration the velocity will increase and hence viscous force also starts increasing. At a certain stand hen viscous force F will balance the net downward force W – Th, then acceleration will become zero and
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the body will fall with constant velocity. This constant velocity is called terminal velocity. So if Vt is terminal velocity. Find the Terminal velocity in the following way: Figure shows the spherical ball, weight of the ball is acting in vertically downward direction, upthrust force is acting in vertically upward direction and viscous force is also acting in vertically upward direction in equilibrium.
je e. in
Thrust + viscous force = W So in equilibrium, F = W – Th
4 3 r g 3
ac k
=
iit
so, 6rvt = W – Th
Terminal velocity (Vt)
2 2 r : g 9
cr
=
Terminal velocity r2
( – ) g and graph shows the velocity with time or distance of the fall of spherical ball.
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Reynold Number: NR =
Vcr
Reynold number is dimensionless
is density of fluid Vc is critical velocity
je e. in
r is radius of tube
is coefficient of Visocity. Critical velocity:
Critical velocity is the maximum velocity upto which flow of fluid
iit
remains study in nature.
ac k
Example:
Two equal drops of water are falling through air with a steady
Solution:
cr
velocity, if the drops coalesce than find the velocity of new drop?
2 r2 Vt g 9
…(i)
diagram here
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the drops are coalesce. Final radius is R, so volume of one bigger drop will e equal to volume of two smaller drops. So,
4 3 4 3 4 3 R r r 3 3 3
r = R = 21/3 × r
2 21 / 3r2 g 9
Vt ' 2 /3 2 Vt
ac k
Vt' = (2)2/3 × V
…(ii)
iit
=
je e. in
Final terminal velocity (Vt)
Final terminal velocity will become (2)2/3 times of the initial terminal
Example:
cr
velocity.
An air bubble of radius 1 mm is allowed to rise through a long cylindrical column of a viscous liquid of radius 5 cm and travels a steady rate of 2.1 cm/sec. If the density of liquid is 1.47 gm/cm 3. Find its viscosity? (g = 980 cm/sec2) Solution:
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Here, due to force of buoyant the bubble will move up and so viscous force which opposes the motion will act downward and as weight of bubble is zero in dynamic equilibrium. Thrust = viscous force
i.e.
4 3 r g 6rVt 3
je e. in
diagram here
ac k
2 r2g 9 Vt
iit
where Vt is terminal velocity.
2 1.47 0.1 980 9 2.1
cr
2
= 1.524 poise. Example:
A spherical ball of radius 1 × 10–4 m and density 104 kg/m3 falls freely under gravity through a distance h before entering at tank of water. If after entering the water, the velocity of the ball does not change. Find h the viscosity of water is 9.8 × 10–6 N-s/m2. Solution:
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After falling a height h, the velocity of the ball will become V=
2gh
After entering the water, this velocity does not change so this velocity is equal to terminal velocity (Vt) i.e.,
2gh
2 2 r g 9
2gh
h=
20 20 2 9.8
2 104 9
2
104 103 9.8 106
ac k
so,
iit
2gh
je e. in
diagram here
cr
= 20.41 m.
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