Wave Motion in One Dimension

Wave is one of primary subjects of Physics. This topic is quite familiar to you, and revision of concepts that you have already studied in XII will help you to understand things better. Importance of study of waves is highlighted in brief below. Waves arise in a wide range of physical phenomena. In our common day experience light radiation from sun, radio transmission from broadcasting station, sound/speech reaches us as waves. Wave occurs as ripples on a pond and as seismic waves following an earthquake. Furthermore, we communicate with each other through a variety of different waves.(Radio waves, sound) At the microscopic level, the particles of matter have a wave nature as expressed by quantum wave mechanics. Therefore it may be said that waves are at the heart of many branches of the physical sciences including optics, electromagnetism, quantum mechanics and acoustics.

Most of us experienced waves as kid when we dropped a pebble into a pond. At the point where the stone/pebble dropped in water, waves are created. The waves move outward towards shore from the point of creation. What really moves? Have you played, Hello-hello using two ice-cream cups in your childhood? Is there any science behind it?

Remark concerning wave phenomena (A. Einstein & L. Infeld, The Evolution of Physics, 1961)A bit of gossip starting in Washington reaches New York [by word of mouth] very quickly, even though not a single individual who takes part in spreading it travels between these two cities. There are two quite different motions involved, that of the rumor, Washington to New York, and that of the persons who spread the rumor. The wind, passing over a field of grain, sets up a wave which spreads out across the whole field. Here again we must distinguish between the motion of the wave and the motion of the separate plants, which undergo only small oscillations... The particles constituting the medium perform only small vibrations, but the whole motion is that of a progressive wave. The essentially new thing here is that for the first time we consider the motion of something which is not matter, but energy propagated through matter.

Interesting questions from Halliday and Resnik*1) In a long line of people waiting to buy tickets, the first person leaves and a pulse of motion

occurs as people step forward to fill the gap. As each person steps forward, the gap moves through the line. Is the propagation of this gap transverse or longitudinal?

2) Consider the “wave” at a baseball game: people stand up and shout as the wave arrives at their location, and the resultant pulse moves around the stadium. Is this wave transverse or longitudinal?

Important Books*Fundamentals of Physics by Halliday, Resnik and Walker,Fundamentals of vibration and waves by S.P. Puri

Dr. Devidas Gulwade, Department of Physics, VES CollegeFeel free to get back to me for any queries/suggestions

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Wave at time t

Wave at t=0

A wave is generated by a particle of a body, which is executing periodic vibrations that due to elasticity of medium gets transmitted to other adjoining (neighbouring) particles. As a consequence, each successive particle acquires kinetic energy due to its displacement. Therefore, the wave motion is accompanied by energy through the medium.

Have a look at the equation that you have studied and revise the meaning, definitions of each parameter and significance. [Refer your school level books or any other std. book]

-------------------------------------------------------------- (Equation I)Few points about the above equation

Let the wave be represented by equation at time t=0.

y(x)=A.sin(2 πλ

x ¿ (Why ?)

Let the velocity of wave is v. After time t what will be the distance travelled by the wave? ---------------

What is the displacement of particle at x after time t(or at any instant of time t)

y(x,t)= --------(Equation II)*

*We are discussing the special case for sinusoidal wave and we will use this analogy for general equation of wave.Rearrange and do manipulations so as to convert Equation II to Equation I. [Hint: a) You will have to substitute some terms in equation I. [Use relation between λ & k; What is relationship between λ, T and velocity of wave?; relation between T and ω]What is expression for angular velocity (ω) and wave vector (k). Understand its significance.

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What is the velocity of particle and velocity of wave? [Understand the difference between these two quantities, v=d(?)/d(?). Let u be velocity of particle (also called as transverse velocity) then u=d(?)/d(?) Write expressions for u. Comment on Phase difference between x (particle displacement) and u (particle velocity), What is maximum value of u and acceleration (du/dt)]

Differential equation of wave motion (Equation of wave)a) Start with sinusoidal wave equation (Equation I)b) Calculate dy/dt, dy2/dt2 and dy/dx, d2y/dx2. Establish relationship between second derivative

w.r.t. t and x)

c) Construct expression ∂2ψ∂ x2 =

1v2

∂2 ψ∂ t2 [This is differential equation of wave motion and

sinusoidal wave is one of the solutions of the equations. We will study general solution of this equation in the next section]

Problems1. The string is driven at a frequency of 5.00 Hz. The amplitude of the motion is 12.0 cm, and

the wave speed is 20.0 m/s. Determine the angular frequency and angular wave number k for this wave, and write an expression for the wave function. Calculate the maximum values for the transverse speed and transverse acceleration of any point on the string. [Ans: ω=31.4 rad/s, k=1.57 rad/m, y=(0.120 m) sin(1.57x -31.4t ), 3.77 m/s; 118 m/s2.]

2. A wave travelling in a string is given by a) y=10cos(0.1x-4.0t) b) 0.00327sin(72.1x-2.72t) where x and y are in cm and t is in seconds. Calculate following parametersAmplitude, T, frequency, velocity and wavelength Max. Transverse speed and acceleration of the particle in the string

One dimensional wave motion

The differential equation of wave equation is,

∂2ψ∂ x2 =

1c2

∂2ψ∂ t 2 (Eqn. 1)

A function ψ is function of space and time and c=√ ρT

. It will be shown later that c is velocity with

which the wave travels without change of the form. The wave variable ψ may represent transverse displacement in case of wave on string, pressure in sound waves or the electric field in a radio wave. It is function of x for plane waves but will become function of two coordinates (say x and y) for

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water waves i.e. two dimensional waves. Similarly it will be function of three coordinates for three dimensional waves, ex. sound waves in gas.

Rearranging the terms is Eqn 1.

( ∂2

∂ x2 −1c2

∂2

∂t 2 )ψ=0

( ∂∂ x

−1c

∂∂ t )( ∂

∂ x+ 1

c∂

∂ t )ψ=0 --------------------------------------(Eqn 2)

c velocity of wave is not a function of x or t therefore the differential operator has been put in two factors.

Let,ξ=x−ctη=x+ct

Now, ξ+η=2 x and ξ−η=2 ct∂ x∂ ξ

=12

∂ t∂ ξ

=−12c

∂∂ ξ

= ∂∂ x

. ∂ x∂ ξ

+ ∂∂ t

. ∂ t∂ ξ

∂∂ ξ

=12

∂∂ x

− 12c

∂∂t

∂∂ ξ

=12 ( ∂

∂ x−1

c∂∂t ) (Eqn 3)

Similarly it can be shown that

∂∂ η

=12 ( ∂

∂ x+ 1

c∂∂t ) (Eqn 4)

Putting values of ∂

∂ x+1

c∂

∂ t and ∂

∂ x−1

c∂

∂ t in Eqn 2

Therefore wave equation (Eqn 1) becomes,∂2ψ

∂ ξ . ∂ η=0

Integrating this equation with respect to ξ, we get

∂ψ∂ η

=F (η)

Where F (η) is an arbitrary function. Integrating once more w.r.t. ηwe get,

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ψ=f 1 ¿

Where f1 and f2 are arbitrary functions. Inserting the values of We have,ψ= f 1 ¿ .................................................(Eqn 5)

This is called as D’Almberts’ solution of the wave equation representing on-dimensional wave.Assume for the sake of simplicity let f2=0, therefore,

ψ= f 1(x−ct )This solution implies that in each plane x=constant, the field amplitude changes with time and at a particular moment, the field is different for different values of x. Thus, the field has the same value for all those values of x and t, which give same value of the argument x-ct ; that is x-ct=constant orx=constant+ctCalling two points (x1, t1) and (x2, t2) where ψ has the same values, we get,x1 - ct1 = x2 - ct2

c = (x1-x2)/(t1-t2) -------------------------------(Equation 6)Therefore, point moves with velocity c. accordingly f1(x-ct) represents a wave moving in the +ve x direction with velocity c. Analogously, the solution f2(x+ct) represents plane wave moving with velocity in –x direction.

1. Verify whether following equations represents one dimensional wave a) y =Ax+Bt b) y = x2+c2t2 c) y = 10sinkx.sinct d) y = ekx-ct

Classification of wavesThe solution of wave discussed in previous section represents a wave which progresses with time. These types of waves are called as travelling waves. The position of particle depends on time and velocity of wave is given by Equation 6.The solution of wave can be expressed in the form (by method of variable separation)Ψ(x,t)=F1(x).F2(t)Where F1 and F2 are new functions of x and t, respectively; in such a case x and t are not related and the wave has no velocity. These types of waves are called as standing or stationary waves. Examples of stationary waves are musical drums, violin etc.

Waves can be further (of both kinds) can be classified in to two types depending upon direction of vibration of particles and propagation of the wave.

Transverse waves If the particles in the medium vibrate perpendicular to the direction of propagation of wave then it is called as transverse wave.Examples...

Longitudinal waves If the particles in the medium vibrate parallel to the direction of propagation of wave then it is called as transverse wave. Examples...

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1 CHARACTERISTICS OF WAVE MOTION

1. Wave motion is the disturbance produced in the medium by the repeated periodic motion in the medium by the repeated periodic motion of particle or body

2. Only wave travel forward, whereas particles of the medium vibrate about their mean position3. There is a regular phase change between the various particles of medium. The particles ahead

start vibrating little later than a particle preceding it.

4. The velocity of wave (v=λ/T; λ

2П. 2 П

T=w /k ¿ is different from the velocity with which

particles of medium. The wave travel with a uniform velocity while the velocity of the particles is different at different position. It is maximum at the mean position and zero at extreme points of the particle.

5. The wave motion is of two types longitudinal and transverse. (However, waves may be combination of two!!)

Exercise (Regarding point 4 above)Let wave be represented by y=Asin(kx-wt). As per definition dy/dt is the particle velocity and also called as transverse velocity. Draw waveform at t=0 for y, dy/dt and d2y/dt2 and comment on it: Step 1. y|t=0 =Step 2 Calculate dy/dt|t=0 and d2y/dt2|t=0

Step 3 Draw y|t=0 , dy/dt|t=0 and d2y/dt2|t=0 one below the other (keep same scale for x, don’t forget to show labels on Y-axis)Step 4. Comment on waveforms (relative phase difference between x,dy/dt and x,d2y/dt2. Understand when dy/dt is maximum relative to y and similarly d2y/dt2

Step 5. Get back to me, if you could not complete step 4!!!

1.1 TRANSVERSE WAVES ON A STRING

We will derive an expression for the velocity of a transverse wave travelling along a stretched string. Let the string is uniform and perfectly flexible. Also, let us suppose that the displacement of the string is small and the reflections from the ends are ignored. Consider that a transverse wave is travelling along a string with velocity ‘v’. Further, imagine that the whole string is being moved from right to left with the same velocity so that the wave remains fixed and the particles (of the string) go around the curve successively.

Consider small portion (shaded in the figure), having length Δs. Let m be the mass per unit length. Therefore, the mass of the small element of

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length Δs is m.Δs. As this segment is moving along an arc of circle of radius R, centripetal force is m.Δs.v2/R. This centripetal force is provided by tension in the string. The force T acts on each side of the segment. The horizontal components of the tension T cancel each other. However, tension T on each side contributes Tsinθ along the radial direction (along radius of circle, shown as Fr in the figure). Therefore, total force along radial direction is 2Tsinθ.Therefore, equation of motion becomes,

2Tsinθ = m.Δs.v2/R2T.θ = m.Δs.v2/Rby definition of angle, 2θ = Δs./R (from the geometry shown in the figure )

Therefore,2T.Δs./R = m.Δs.v2/R

i.e. v =√ Tm

Thus velocity of a wave along a string depends on the tension and the mass per unit length.

Discuss how velocity changes with change in T and m What happens when tension increases?What if mass per unit length increases/decreases?

Problem A uniform cord has a mass of 0.300 kg and a length of 6.00 m. The cord passes over a pulley and supports a 2.00 kg object. Find the speed of a pulse travelling along this cord.

1.2 LONGITUDINAL WAVES ON ROD

Consider an elastic rod of density ρ. A point of cross-section is displaced through displacement ξ, which is the same for all the points in cross-section. We consider the portion of the rod of length δx lying between x and x + δx when the rod is at rest. Under the influence of disturbance these coordinates become x+ ξ and x+ ξ+ δx+ δξ where δx+ δξ is strained length.Force per unit area = Young’s modulus X strain

and strain = change in length / Original length

Strain = (δξ+ δx- δx)/ δx = δξ/ δxTherefore,

Force per unit area=Y . δξδx

To write the equation of motion for the portion of rod under consideration, we know force per unit

area is Y . δξδx at x and therefore force at x+ δx at

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Y ( δξδx

+ δ2 ξδx 2 . δx)

The net force per unit area=Y ( δ 2ξδx2 . δx)

Let the area of cross-section is S. Therefore the net force is given by

net force=YS δ 2ξδx 2 . δx

Applying Newton’s law of motion,

F=ma=sρ . (S .δx ) δ2 ξδt 2 =YS δ2 ξ

δx2 . δx

δ2 ξδx 2 = ρ

Yδ2 ξδt2

This is the equation of the wave motion on the rod. The velocity of rod is √Y / ρ. This is clear from the expression that velocity depends only on Young’s modulus and density, not on applied force or the cross sectional area.

1.3 PRESSURE WAVES IN GAS – ACOUSTICS WAVES

The longitudinal waves may propagate through all phases of matter, solid, liquid, gases, plasma in the form of condensation and rarefaction. As a result there is a continuous variation of pressure all along the direction of propagation. Let us consider a fixed mass of a gas of volume Vo, pressure Po and density ρo under condition of equilibrium. When a longitudinal wave passes through it, these values are disturbed from the equilibrium and(let it) becomes Po+p, Vo+v, and ρo+ ρd

The fractional change in volume = v/Vo

Fractional change in density = ρd/ ρo

Consider and infinitesimal thickness δx of gas containing in tube of unit cross-section, extending between the planes at x and x+δx. The planes get displaced to new positions x+ξ and x+δx+ξ+δ ξ.

The volume of the gases gets disturbed because the pressure on both sides does not balance. The net force acting on the gas is

Px- Px+dx = [Px - (Px + ∂ Pxdx

dx)]

= −∂ (P¿¿0+P)∂ x

¿ dx

¿−∂ (P¿¿0+P)∂ x

¿ dx

¿−∂ P∂ x dx -------------- (Eqn 1)

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The equation of motion for the thin slice of gas, according to Newton’s law is

ρ0 dx ∂2 ξ∂ t 2 =−∂P

∂ xdx−−(Eqn2) (equated it with EQN 1)

The bulk modulus of the gas,

B= −dPdV /V

=−V dPdV

The negative sign indicates that as volume increases, pressure decreases.

B=−pv

V o

p=−B vV o

Now,vvo

=Change∈Volumeoriginal volume

=(change∈volume)

Area. thickness=

[ (dξ−δx )−δx ] . Areaδx . Area

= ∂ ξdx (Since Area=1 sq.units)

p=−B ∂ ξdx

Substituting this value of p in equation 2, we get,

ρ0∂2ξ∂ t2 =B ∂2 ξ

∂ x2

∂2 ξ∂ x2 =

ρ0

B∂2 ξ∂ t2 −−−−−−−−−−−−(Eqn 4)

The value of bulk modulus will depend on whether the changes in gas on the passage of wave are isothermal or adiabatic.

[Note: Isothermal consideration (PV=constant, differentiating it PdV +VdP=0hence , p=−dp

dVV

=B)

i.e. negligible temperature variation in layers of rarefaction and compression. This lead to B = p and

hence v=√ pρ

. this resulted in lesser velocity than the experimentally observed values. The Laplace

suggested that the compression and rarefaction are rapid and hence there is no heat exchange with the surrounding. This is also called as adiabatic correction this is used in the following derivation. This is for your information and understanding]

Let us assume that the changes are adiabatic and the gas obeysPVγ = constantWhere, γ is the ratio of the specific heat of the gas at constant pressure to that at constant volume.Taking logarithms and differentiating, we get

dPP

+γ dVV

=0

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dPdV

=−γ PV

Therefore,

B=−V dPdV

=γPThe equation 4 is the differential equation of wave motion and velocity is √ B

ρ0=√ γP

ρ0 Speed of sound depends on bulk modulus and density.

1.3.1 VELOCITIES IN SOLIDS AND LIQUIDS The bulk modulus is reciprocal of compressibility (B=1/β), compressibility is lowest for the solid i.e. bulk modulus is highest; therefore, velocity of sound will be highest in solids. With this logic it may be argued that velocity in gas is lowest, higher in liquids and highest in solids. (Gas: ~343 m/s, Water: 1400 m/s, Aluminum and steel: ~ 6000 m/s). 1.3.2 EFFECT OF TEMPERATURE AND HUMIDITYFor ideal gas for a given mass of gas PV/T is constant. As the density of a given mass is inversely proportional to its volume, the equation may be expressed as P/ρT= const. i.e.P/ρ= const.T Therefore, v=√ γP

ρ0=√γ . const . T . This means that velocity is proportional to square root of temperature. If pressure is changed maintaining the

temperature constant then density will change so that P/ρ is constant and hence will not result in change in velocity.Alternately, PV=RT for a mole of gas. i.e. PM

ρ=RT where M is molar wt.

Therefore, Pρ=RT

M and hence, v =√ γRTM

How velocity will change with change in humidity? (Density of moist air is less than the density of dry air at same pressure)1.4 PROBLEMS

1. On a winter day sound travels with velocity 336 m/s. find the atmospheric temperature. Speed of sound at 0 oC is 332m/s.

2. Suppose you create a pulse by moving the free end of a tring up and down once with your hand. The string is attached at its other end to a distant wall. The pulse reaches the wall in a time t. Which of the following actions, taken by itself, decreases the time it takes the pulse to reach the wall? More than one choice may be correct. (a) Moving your hand more quickly, but still only up and down once by the same amount. (b) Moving your hand more slowly, but still only up and down once by the same amount. (c) Moving your hand a greater distance up and down in the same amount of time. (d) Moving your hand a lesser distance up and down in the same amount of time. (e) Using a heavier string of the same length and under the same

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D

AB

C

tension. (f) Using a lighter string of the same length and under the same tension. (g) Using a string of the same linear mass density but under decreased tension. (h) Using a string of the same linear mass density but under increased tension.

3. When all the strings on a guitar are stretched to the same tension, will the speed of a wave along the more massive bass strings be faster or slower than the speed of a wave on the lighter strings?

4. If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, they do not ascend with constant speed. Explain. (Hint : 1. v will increase, 2. Try after solving next problem)

5. A uniform rope of length 12 m and mass 6 kg. hangs vertically from a rigid support. A block of mass 2 kg is attached at the free end of the rope. A transverse wave of wavelength 0.06 m is produced at lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope? (Ans: 0.12 m. hint: rope is heavy and tension is not same at different point)

6. Find temperature at which the sound travels in H2, with the same speed as in O2 at 1000 oC. Assume γ is same for both the gases. MH2/MO2=1/16 (Ans: -193oC. Hint: start with expression of velocity and use gas equation for one mole)

7. Show that velocity of sound in H2 is 4 times the velocity of sound in O2 (use data provided in previous problems)

8. The speed of sound in oxygen is 470 m/s., what will be the speed in hydrogen at same temperature and pressure?

9. Two blocks each of mass 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB.(refer figure below). The linear mass density of wire AB is 10gm/m. and that of CD is 8 gm/m. Find the speed of transverse pulse produced in AB and in CD (77.9 m/s and 62.6 m/s)

10. Choose correct options (may have multiple correct options). The density of water is greater than air. With this can you say that velocity in air a)is larger than in water b) smaller than water c)is equal to that in water d) Can conclude anything from the given information (insufficient information )

11. Choose correct options. A tuning fork sends sound waves in air. If the temperature of the air increases, which of the parameter will change a) Displacement amplitude b) Frequency c) wavelength d) time period e) velocity

12. The speed of sound depends on a) elastic property but not inertial property b) inertial property but not elastic property c) both elastic and inertial d) neither of them

13. Velocity of sound in air at 14 oC is 340 m/s, what will be its value at 100 oC (γ=1.4 for air)14. Discuss, what will be effect on velocity if pressure is doubled (but temperature is constant) to

the velocity of the sound?15. Starting with general expression for the velocity of sound in air, show that Calculate the

increase in the velocity of sound in air per degree rise in temperature.16. State the wave equation for one-dimensional motion and obtain its solution 17. Verify that y=A.sin(kx-wt) represents solution of one-dimensional wave equation and hence

what is velocity of the wave. [The terms in the equation has usual meaning]18. Verify that y = f1(x-ct) + f2(x+ct) represents general solution of one-dimensional differential

wave equation19. Define particle velocity and acceleration. Derive relation between particle velocity and wave

velocity [Assume y = A.sin(kx-wt)]20. A longitudinal disturbance generated by an earthquake travels 100 km in 3 minutes. If

(average) density of rock is 2700 kg/m3. Calculate bulk modulus. [Ans: 8.3 x 1010 N/m2 ]21. A wave is travelling along a string is given by, y(x,t)=0.000327sin(72.1x-2.72t) a) what is

the amplitude b) What are T, freq, velocity and wavelength of the wave c) what is the

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displacement y at x=22.5 m and t=18.9 s) what are transverse speed and acceleration of the particle at t=8.9 s and x=0.225m

22. A long string has a linear density of 0.01 kg/m and is subjected to a tension of 64 N along x-axis. Find the velocity of the transverse waves on the string.

23. A string is stretched by suspending 5kg. The mass per unit length is 5gm/m. The travelling waves are sent through the string by oscillating one of the ends with freq of 250 Hz and amplitude 5mm. Calculate the velocity and length of the wave. Also write eqn of travelling wave g=10m/s2 (Ans: 100m/s., λ=0.4m)

24. In metallic rod of density 7.5 gm/cc and Young’s modulus 7.5x1011 dynes/cm2 , the frequency of wave is 300Hz, Find wavelength. (Ans: 10.54 m)

25. The freq. Of longitudinal wave produced in a solid rod is 1000Hz. Calculate the wavelength of the waves if the density of the material is 9000kg/m3 Y=9x1011N/m2. (Ans: 10m)

26. Derive expression for velocity of longitudinal wave on rod27. Derive expression for velocity of longitudinal wave in gas28. Derive expression for velocity of transverse wave on string29. What happens to the speed of a wave on a taut string when the frequency is doubled? Assume

that the tension in the string remains the same

2 ULTRASONIC AND ACOUSTICS OF BUILDINGSWe are well aware of the fact that we can hear sound waves in the frequency range from 20 to 20,000 Hz and this is called as audible range; it is not universal range also this range may change with age. Sound waves beyond frequency more than 20kHz are called as ultrasonics. Obviously, these waves travel in air with the velocity of ~330 m/s. There frequency is higher therefore wavelength is small.

2.1 INTRODUCTION

Sounds in the range 20-100 kHz are commonly used for communication and navigation by bats, dolphins, and some other species. They use these waves for navigation and hunting bats can see with the resolution of ~mm; how is answered latter.

As we know, wide variety of medical diagnostic applications use both the echo time and the Doppler shift of the reflected sounds to measure the distance to internal organs and structures and also the speed of movement of these structures. Typical is the 2d-echocardiogram, in which a moving image of the heart's action is produced in video form that gives information about the speed and direction of blood flow and heart valve movements. Ultrasound imaging near the surface of the body is capable of resolutions less than a millimeter. The use of longer wavelengths implies better resolution and hence high frequency (lower wavelength) sound waves are used.

2.2 PROPERTIES OF ULTRASONIC WAVES

1. They have a high energy (higher frequency)2. Just like sound waves, ultrasonic waves get reflected, refracted and absorbed.

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3. They show negligible diffraction because of their small wavelength. Hence they can be transmitted over long distances.

4. They produce intense heating when passed through a substance.

Two important methods for generation of ultrasonic waves namely, piezoelectric oscillator and magnetostriction are discussed below.

2.3 PIEZOELECTRIC EFFECT

If one pair of opposite faces of a quartz crystal is subjected to pressure, the other pair of opposite faces develops equal and opposite charges on them. The sign of the charge is reversed when faces are subjected to tension rather than pressure /compression. The electric charge developed is proportional to the amount of pressure or tension applied. This phenomenon is called as Piezoelectric effects.

The effect is reversible; if the electric field is applied to a pair of opposite faces then it result in contraction or expansion of the other pair of opposite faces.

When the two opposite faces of a quartz crystal, cut appropriately (perpendicular to optic axis!, axis of hexagon, shape of quartz single crystal) subjected to alternating voltage, the other pair of faces experience stresses and strain. i.e. quartz crystal will undergo continuous contraction and expansion; elastic vibrations are set up in the crystal.

When the frequency of alternating voltage is equal to the natural frequency of vibration of the crystal it result in resonance and exhibit large amplitude. These vibrations are longitudinal in nature. The fundamental frequency of crystal of thickness t, exhibit a fundamental frequency of vibration n= 12t √Y

ρ where Y is Young’s modulus and is density.

The inverse piezoelectric effect, when quartz crystal is subjected to an alternating potential it result in elastic vibration in perpendicular direction. (This property is direction dependent and effect is enhanced in certain direction therefore the appropriately cut crystal is used. There is hell lot of incomplete and unnecessary information in text books!!) This phenomenon/principle is therefore used to generate ultrasonic waves. The thickness of the crystal is determined by the desired frequency. A thin wafer element vibrates with a wavelength that is twice its thickness. Therefore, piezoelectric crystals are cut to a thickness that is 1/2 the desired radiated wavelength. The higher the frequency of the transducer, the thinner the active element but there will be limit to it as very thin will be too fragile.

The frequency of vibration is given by,

n=12 t √Y

ρ

Where Y is Youngs modulus and ρ is density and t is thickness

Problem:

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1. A quartz crystal of thickness 0.001 m is vibrating at resonance. Calculate fundamental frequency of vibration. Young’s modulus is 7.9 x1010 N/m2 and density is 2650 kg/m3. (Ans: 2.73x106 Hz)

2. A piezoelectric crystal has a thickness of 1.5 mm. If the velocity of propagation of longitudinal sound wave is 5760 m/s. Calculate fundamental frequency of vibration (Ans: 1.92x106 Hz)

3. Calculate thickness of quartz crystal for generating 500 MHz ultrasonic waves. Young’s modulus is 7.9 x1010 N/m2 and density is 2650 kg/m3.

2.3.1 ADVANTAGES

1. It can generate high frequencies up to 500MHz2. The output power is high. It is not affected by temperature or humidity3. Efficient than magnetostriction oscillator (discussed in next section)4. Produces constant and stable frequency of ultrasonic wave

2.3.2 DISADVANTAGES

1. The cost of quartz crystal is high2. Appropriate cutting and shaping is very complex

2.4 MAGNETOSTRICTION Ultrasonic waves may be produced using the principal of magnetostriction. When a rod of ferromagnetic material like nickel is magnetized, it undergoes a change in length, this is called as magnetostriction.

A nickel rod is placed in a rapidly varying magnetic field, alternately expands and contracts with twice the frequency of the applied magnetic field. By adjusting the frequency of the frequency the alternating field the resonance is achieved and as a result vibrations of large amplitude are produced.

The frequency of vibration of rod is given by,

n=12 l √ Y

ρ

Where l is length of rod, and Y is Young’s modulus and ρ density

2.4.1 ADVANTAGES

1. The construction cost is low2. They are capable of producing large acoustical power with fairly good efficiency (60%)

2.4.2 DISADVANTAGES

1. It can produce frequencies up to 3 MHz2. The frequency of oscillation depends on temperature3. It is difficult to get a single frequency; breadth of resonance curves us large.

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2.5 DETECTION OF ULTRASONIC WAVES

1. Quartz crystal method: This method is based on the principal of piezoelectric effect. A pair of opposite faces of a quartz crystal is exposed to the ultrasonic waves; the opposite pair of faces develops the charges. These charge produced is amplified and detected using an electronic circuit.

2. Sensitive flame method: A narrow sensitive flame is moved along the medium. At the position of antinode the flame is steady (because antinode corresponds to constant pressure). The flame flickers at the position of node because there is a change in pressure (because of compression and rarefaction alternately). In this way position of nodes and antinodes can be found out in the medium and accordingly the wavelength can be found out. (The average distance between two antinodes is equal to half the wavelength)

3. Thermal detectors: A platinum wire is placed in the region to be rested for ultrasonic waves. At nodes, due to alternate compressions and rarefactions alternate cooling and heating is produced. Change in temperature leads to change in resistance and that is detected using appropriate electronic circuit and hence the ultrasonic waves are detected.

4. Kundt’s tube method: If the wavelength of the ultrasonic wave is greater than a few millimeters, Kundt’s tube can be used to form a stationary pattern with a well defined nodes and antinodes. In liquid medium, powdered will fly from apart due to violent pressure variation at antinodes whereas they remain undisturbed because particles there do not vibrate.

2.6 APPLICATIONS OF ULTRASONIC WAVES

1. SONAR (Sound Navigation and Ranging) this is used in submarines, it is equipped with production, transmission and detection of ultrasonic waves. When a pulse of ultrasonic wave is sent through water it is reflected by the obstacles or other submarines present in the water and the reflected waves are detected by the sender. By knowing time interval between the interval between transmission and reception, the distance from the other objects/obstacles may be calculated. (Why radio waves are not used in this application?)

2. Depth of Sea: (Using the same principle mentioned above). The waves are directed to sea-bed from transmitter. They are reflected back from the sea-bed to the receiver. The time interval between the transmission and reception (say t) gives the depth of the sea (vt/2)

3. Ultrasonic soldering: High-frequency sound waves are transmitted through molten solder to remove undesirable surface oxide films from a metal, thereby promoting wetting of the base metal with solder. High frequency waves (i.e. violent pressure vibrations) helps to remove the thin layer of oxide over the metal. Using ultrasonic waves, metals such as Aluminium, susceptible to formation of oxide layer may be soldered and also this may b accomplished without the use of flux.

4. Investing of structure of MatterWe can determine the velocity of ultrasonics in materials and its variation with temperature. As the velocity depends on the elastic and inertial properties (bulk/young’s modulus and density) of material that are in turn depend on atoms and their interactions with each other. Therefore, studying velocity gives information about properties such as specific heat, arrangements of atoms that depends on atoms and their interaction.

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5. Ultrasonic cleaning: This is used if one need quality cleaning of delicate objects such as ornaments (which are difficult to clean with conventional ways as all parts are not accessible) and any materials/parts used in research. The standing wave is produced in the water or some other fluid and objects to be cleaned is immersed in it. Due to high frequency vibrations during rarefaction leads to formation of bubbles, (cavitation). These bubbles formed bursts during compression on the surface of objects and helps to clean the object. The distance between these bubbles will depend on the wavelength (that is ~mm in ultrasonics) and hence will affect the quality of cleaning. This process is widely used in research labs where cleanliness of objects is related with quality of results.

6. Crack in metals: It is used to detect cracks in metal structures. Ultrasonics are made to fall on the metal sheet in which any fault/crack is to be detected and reflected signal is detected by detector. The reflected intensity should be same everywhere, any change in intensity due to crack may be detected and hence the fault/crack in metal sheet may be located.

7. Medical Applications: In general ultrasonics can be used as a diagnostics tool to detect tumors and some defect in the body such as stage of cancer, brain tumor, cardiograph, growth of foetus. In 2d-echocardiograph, it is used to measure the distance to internal organs and structures and also the speed of movement of these structures. A moving image (video) of the heart's action (expansion and contraction) is investigated. Also, this gives information about the speed and direction of blood flow and heart valve movements. Another example is gastric sonography. Surgery: Kidney stone and brain tumors can be removed with the use of ultrasonic waves and serve as painless surgery.Sterilization: Ultrasonics waves can kill bacteria. Therefore they are used for sterilizing milk or food products.

3 ACOUSTICS OF BUILDING

3.1 REVERBERATION

When a source produces sound wave inside a closed building, the waves are reflected by walls, ceilings and other objects in the room. The intensity of the sound wave decreases at every reflection and finally sound becomes inaudible. Therefore, listener receives a) direct waves 2) reflected waves due to multiple reflections. Also, there is some gap between direct and successive/multiple reflected waves received by the listener. Due to this effect, the sound persists for sometime even after source has stopped. This persistence of sound is called as reverberation. The time for which the sound persists is called as reverberation time. The reverberation time is defined as the time taken by sound (due to multiple reflections) to fall below to one millionth of the original value. This time is a measure of minimum audibility after the source of the sound has stopped.

The reverberation time will depend on the size of the room or the auditorium, the nature of reflecting material on wall and ceiling and area of reflecting surfaces.

3.2 SABINE’S FORMULA

Assumptions

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1. The average energy per unit volume is uniform2. The energy is not lost in the auditorium. The energy lost is only due to absorption of

materials of walls and ceiling. Also, due to the escape through windows and ventilators. Starting with these postulates, Sabine derived a formula for reverberation time given by,

T=0.158 V∑ Aα

Where V is volume of the auditorium, A is area of the ceiling and wall etc. and α is absorption.

The reverberation time is proportional to

1. Volume of the auditorium2. Inversely proportional to area of ceiling and walls3. Inversely proportional to total absorption plus transmission through open window

If the reverberation time is too small, the sound dies away almost instantaneously and gives the hall a dead effect. If the reverberation time is too long, each syllable continues to be heard even after the next syllable has been uttered. (This is mixing of direct sound with the multiple reflected previous sound and hence will be difficult to make any sense out of it this is called as unintelligible sound in the field of acoustics.) This makes the sound (speech) unintelligible. Therefore, the value of reverberation time needs to be maintained at optimum value. The rooms of different sizes have been studied the significance of Sabine’s formula and it is established that the reverberation time is 1 to 2 seconds for speech. For music, it is between 2 to 2.5 seconds.

3.3 ABSORPTION

If sound propagating in a medium is incident on the surface, a part of it is absorbed by the surface and partly it is reflected. The amount of sound reflected and absorbed depends on nature of surface on which sound is incident. The absorption coefficient (α) is defined as the sound energy absorbed to the total energy incident on it.

α= energy absorbedtotal energy incident

Sabine chose one square meter of an open window as a standard unit of absorption. All sound waves falling on it (open window) pass through it (transmitted completely) and can be said to be completely absorbed (not reflected). This unit is called as Sabine and open window unit. (O.W.U.)

The absorption coefficient of a material is defined as the ratio of sound energy absorbed by it to that absorbed by an equal area of an open window.

3.3.1 DETERMINATION OF ABSORPTION COEFFICIENT Sabine’s method to calculate reverberation time: An organ pipe (of 512 Hz frequency) is sounded. The reverberation time is determined using a chronograph (crudely we can say that a stop watch is started when an organ pipe is vibrated and the reverberation time i.e. the time for the sound to become inaudible is measured.)

1. The reverberation time (T1) without the absorbing materials in the hall

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According to Sabine’s formula

T 1=0.158V∑ Aα Where ∑ Aα is the total absorbing coefficient, in the hall without material and

V is volume of the hall2. The absorbing material is put inside the room. Let α1 is the coefficient of the absorbing

material and s is surface area

T 2=0.158 V

∑ Aα+α1 s

1T1

− 1T 2

=α 1 s

0.158 VTherefore,

α 1=0.158 V

s ( 1T 1

−1

T 2 )Knowing values of T1, T2, s, and V, value of absorption coefficient of material may be calculated.

3.4 ACOUSTICS OF BUILDING

The branch of Physics that deals with the design and construction of buildings with good acoustic properties is called as Acoustics. It deals with manipulation of 1) reverberation time 2) noise insulation and reduction 3) sound distribution and absorption.

The few requirements of good auditorium are provided below.

1. The quality of sound must be unaltered i.e. relative intensity of the components of the sound must be preserved.

2. The sound should be sufficiently loud and intelligible in every part of hall.3. Sound of each syllable should soon decay so that succeeding syllable may be heard distinctly

(there should not be overlap between the direct sound with multiple reflected of earlier syllable/word/sound). In brief, reverberation time should be optimum.

4. No echoes should be present.5. There should not be undesirable focusing of sound in any part of hall. There should not be

any silence zone or regions of poor audibility in the hall.6. All extraneous noises must be shut out as much as possible.

3.4.1 FACTORS AFFECTING THE ACOUSTICS OF BUILDING

The factors affecting acoustics of building are provided below

1. Reverberation time2. Presence of echoes3. Adequate loudness4. Focusing and interference effects5. Extraneous and inside noise

3.4.1.1 REVERBERATION TIME

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This is the important factor that affects the acoustics of sound. As discussed earlier it should be optimum, smaller lead to dead effect and larger leads to unintelligible sound. After various trials and experiments it is observed that reverberation time should be 1-2 s for speech and 2 to 2.3 for music. To adjust the reverberation time to the best optimum value, the absorption of sound may be increased in following ways

1. Provide windows and openings2. Cover ceilings and walls with sound absorbing materials (ex. Asbestos)3. Cover floor with carpets

3.4.1.2 ECHOES

Echoes can be avoided by covering the long distance walls (responsible for echo) with sound absorbing materials to prevent reflection of sound.

3.4.1.3 ADEQUATE LOUDNESS

Sufficient loudness in every portion of room is important factor for satisfactory hearing. If the loudness of the sound is inadequate then it may be increased by, low ceiling for the reflection of sound towards the audience and/or by providing loud speakers at adequate distances.

3.4.1.4 FOCUSSING (SOUND DISTRIBUTION)If there is any concave surface in the hall, sound is concentrated at its focus region. (See figure below). On the other hand, there might be dead places where intensity of sound is very low. Therefore, these kinds of surfaces should be avoided or it should be covered with absorbent materials.

Consider the following figure. Listener L, will receive direct sound as well as reflected from the curved surface. However there will be certain position where the intensity of sound will be less than at L. The area where sound intensity is low is called as silent zone. In order to avoid this either such shape should be avoided or good absorbing materials should be quoted on the surface.

3.4.1.5 EXTRANEOUS NOISES

The noise which reaches the hall from outside the auditorium/hall through open doors/windows, ventilators. To avoid this auditorium need to be insulated from the outside sound i.e. making the hall air conditioned where it is completely closed,

3.4.1.6 INSIDE NOISE The noise that is produced inside the hall is called as inside noise. It may be produced by machines or air conditioners in the hall. The noise may be minimized by placing machines and air conditioner on sound absorbent pads.

3.4.1.7 ECHELON EFFECT

If regular structure similar to stairs is present in the hall. The sound produced in front of such a structure may produce a musical note due to successive echoes/reflection from the periodic structure. Hence the structure should be avoided or remedy is to cover such a surface with absorbing materials like carpet.

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3.5 QUESTIONS AND PROBLEMS

1. What is reverberation time? What measures may be taken to achieve an optimum value? 2. Discuss reverberation.3. Describe the factors affecting acoustics in the hall.4. Discuss piezoelectric and inverse piezoelectric effect. 5. Calculate the coefficient of absorption of an auditorium of 90 m long, 15m wide and 8 m

high. The reverberation time is 1.2 s. Assume that the absorption of sound waves takes place from the absorbing area of 3500 m2 (can you solve this problem if the area is not specified!)

6. An auditorium has a volume of 1200 m3, has a floor area 200 m2 and ceiling area 200 m2 and wall area 180 m2. The absorption coefficient of wall surface, ceiling and floor surface is 0.032, 0.78 and 0.058 respectively. Calculate the average sound absorption coefficient and reverberation time.

7. A auditorium of 12 X10X5m dimensions has average absorption coefficient of 0.1. Calculate reverberation time.

8. An auditorium of volume 70000 cubic feet has a reverberation time of 0.9 s when empty. Find the reverberation time when 250 persons are present in the auditorium. Absorption coefficient of a person is 0.4. Average absorbing area of a person is 6 sq. feet.

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