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WAVES
Wave is the mode of transfer of energy (and momentum) in which energy propagates without any actual bulk motion of particles of medium with the help of disturbance in the presence or absence of medium.
Wave Motion
The vehicle, which is responsible for the transmission of energy from one place to another through medium without any translation of the medium in the direction of energy flow, is called a wave.
Classically, in general, waves are of two types. They are as follows:
(1) Mechanical Waves :
The wave produced due to vibration of material particle and require a material medium, to propagate through are mechanical waves. e.g. sound wave, water wave and wave in solid.
(2) Electro Magnetic Waves :
Waves which are produced due to the periodic vibration of two mutually perpendicular electric and magnetic fields are Electro magnetic waves. It propagates in a direction perpendicular to both electric and magnetic field. e.g. light waves, X-ray, g-ray etc.
(a) Transverse Wave in a Stretched String :
Consider a transverse pulse produced in a taut string of linear mass density m. Consider a small segment of the pulse, of length Dl , forming an arc of a circle of radius R. A force equal in magnitude to the tension T pulls tangentially on this segment at each end.
Let us set an observer at the centre of the pulse which moves along with the pulse towards rights. For the observer any small length dl of the string as shown will appear to move backward with a velocity v.
Now the small mass of the string is in a circular path of radius R moving with speed v. Therefore the required centripetal force is provided by the only force acting, (neglecting gravity) is the component of tension along the radius.
The net restoring force on the element is
F = 2Tsin(Dq) » T(2Dq) = \(T\frac{\Delta l}{R}\)
The mass of the segment is Dm = mDl
The acceleration of this element toward the centre of the circle is a = \(\frac{{{v}^{2}}}{R}\) , where v is the velocity of the pulse.
Using second law of motion,
\(T\frac{\Delta l}{R}\) = (mDl) \(\left( \frac{{{v}^{2}}}{R} \right)\) or v = \(\sqrt{\frac{T}{\mu }}\)
(b) Longitudinal Wave in Fluids :
Sound wave in air is a longitudinal wave. As a sound wave passes through air, potential energy is associated with periodic compressions and expansions of small volume elements of the air. The property that determines the extent to which an element of the medium changes its volume as the pressure applied to it is increased or decreased is the bulk modulus B.
B = \(\frac{-\Delta p}{\Delta V/V}\), Where \(\frac{\Delta V}{V}\) is the fractional change in volume produced by a change in pressure Dp.
Suppose air of density r is filled inside a tube of cross-sectional area A under a pressure p. Initially the air is at rest.
At t = 0, the piston at the left end of the tube(as shown in the figure) is set in motion toward the right with a speed u. After a time interval Dt, all portions of the air at the left of section 1 are moving with speed u whereas all portions at the right of the section are still at rest. The boundary between the moving and the stationary portions travels to the right with v, the speed of the elastic wave (or sound wave). In the time interval Dt, the piston has moved uDt and the elastic disturbance has traveled a distance vDt.
The mass of air that has attained a velocity u in a time Dt is r(vDt)A. Therefore, the momentum imparted is [rvDt A]u. And, the net impulse acting is (Dp A)Dt
Thus, impulse = change in momentum
(Dp A)Dt = [rv(Dt) A]u or Dp = rvu (1)
\ Dp = B \(\left( \frac{\Delta V}{V} \right)\)
Where V = Av Dt and DV = Au Dt
\ \(\frac{\Delta V}{V}=\frac{Au\Delta t}{Av\Delta t}=\frac{u}{v}\)
Thus, Dp = B \(\frac{u}{v}\) (2)
From (1) and (2)
v = \(\sqrt{\frac{B}{\rho }}\)
(c) Sound Waves :
From practical standpoint it is easier to measure pressure variations in a sound wave than the displacements, so it is worthwhile to develop a relation between the two. Let p be the instantaneous pressure fluctuation at any point, that is, the amount by which the pressure differs from normal atmosphere pressure. If the displacements of two neighboring points x and x + Dx are the same, the gas between these points is neither compressed nor rarefied, there is no volume change, and consequently p =0. Only when y varies from one point to a neighboring point there is a change of volume and therefore of pressure.
The fractional volume change DV/V in an element near point x turns out to be given simple by \(\left( \frac{\partial y}{\partial x} \right)\), which is the rate of change of y and x as we go from one point to the neighboring point. To see why this is so, we note that DV/V is proportional to change in length of an element which has length Dx when no wave disturbance is present, divided by Dx. The change in length is the value of y at the point x + Dx, minus the value at the point x. If Dx is very small, this is approximately multiplied by the derivative of y with respect to x. thus y(x + Dx, t) – y (x, t) = \(\frac{\partial y}{\partial x}\Delta \,x\)
\(\frac{\Delta V}{V}\,\,=\,\,\frac{y\left( x+\Delta x,\,t \right)\,-\,y\,\left( x,\,t \right)}{\Delta x}\,\,=\,\,\frac{\partial y}{\partial x}\) (I)
Now from the definition of the bulk modulus B,
p = – B \(\frac{\Delta V}{V}\) and we find p = – B \(\frac{\partial y}{\partial x}\) = BkA cos(wt - kx)
Maximum amount by which the pressure differs from atmospheric, that is, the maximum value of p, is called the pressure amplitude, denoted P.
Þ P = BkA
Intensity in terms of pressure variation
I = \(\frac{1}{2}\,\omega Bk{{A}^{2}}\) = \(\frac{1}{2}\,\frac{{{P}^{2}}}{\rho v}\) ; Where v is the velocity of the sound.
(1) Transverse Mechanical Waves :
(i) Those in which particles or physical property vibrates perpendicular to the direction of propagation of wave
(ii) During wave propagation, crest and trough are formed.
(iii) Medium should be rigid for the transmission of transverse mechanical wave.
(iv) These waves are generated in media having shearing property (to change shape).
(v) Transverse waves are produced in all wire based musical instrument.
(2) Longitudinal Waves :
(i) Those in which particles or physical property vibrates parallel to the direction of propagation of wave
(ii) Compression and rarefaction are formed. Longitudinal waves are also known as pressure waves or P. waves as pressure change is are associated with compression and rarefaction.
(iii) Air based musical instruments like flute and mouth organ produce this type of wave.
(iv) They do not show polarization phenomenon.
1. Equation of a transverse wave travelling in a rope is given by Y = 5 sin( 4.0t – 0.02 x)
Where y and x are expressed in cm and time in seconds. Calculate the amplitude, frequency, velocity and wavelength of the wave.
Solution :
Comparing this with the standard equation of wave motion y = A sin \(\left( 2\pi ft\,-\,\frac{2\pi }{\lambda }\,x \right)\)
Where a, f and l are amplitude frequency and wavelength respectively.
Thus amplitude A = 5cm
2pf = 4
Þ Frequency f = \(\frac{4}{2\pi }\,\,=\,\,0.673\,\,\,cycle/s\)
Again \(\frac{2\pi }{\lambda }\,\,=\,\,0.02\,\,\,\,\,\,or\,\,\,\,Wavelength\,\,\,\,\lambda \,\,=\,\,\frac{2\pi }{0.02}\,\,=\,\,3.14\,m\)
Velocity of the wave v = \(f\lambda \,\,\,=\,\,\,\frac{4}{2\pi }\,\,\frac{2\pi }{0.02}\,\,\,=\,\,2\,m/s\)
2. Equation of a transverse wave travelling in a rope is given by Y = 5 sin( 4.0t – 0.02 x)
Where y and x are expressed in cm and time in seconds. Calculate The maximum transverse speed and acceleration of a particle in the rope .
Solution :
Transverse velocity of the particle u = \(\frac{dy}{dt}\) = 5 x 4 cos(4.0 t – 0.02x)
= 20 cos(4.0 t – 0.02x)
Maximum velocity of the particle = 20 cm/s
Particle acceleration a = \(\frac{{{d}^{2}}y}{d{{t}^{2}}}\,\,=\,\,-20\,\,x\,4\,\sin \,\left( 4.0\,t\,-\,0.02\,t \right)\)
Maximum particle acceleration = 80 cm/s2
1. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750. What is the velocity of wave in wire?
Solution :
L = 100 cm, f1 = 750 Hz.
v1 = 2Lf1 = 2 ´ 100 ´ 750 = 150000 cms-1 = 1500 ms-1
2. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750.
If the weight is reduced to 4 kg, what is the velocity of wave ?
What is the wavelength and frequency of wave ?
Solution :
\({{v}_{1}}=\sqrt{\frac{{{T}_{1}}}{m}}and{{v}_{2}}=\sqrt{\frac{{{T}_{2}}}{m}}\)
\(\frac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\Rightarrow \frac{{{v}_{2}}}{1500}=\sqrt{\frac{4}{9}}\)
\ v2 = 1000 m s-1
l2 = wave length = 2L = 200 cm = 2 m
f2 = \(\frac{v}{{{\lambda }_{2}}}=\frac{1000}{2}=500\,\,Hz\)
3. Find the ratio of the fundamental frequencies of two identical strings after one of them is stretched by 2% and the other by 4%?
Solution :
\(n=\frac{1}{2l}\sqrt{\frac{T}{m}}.\) If l0 be the initial length and f be the fractional increase in length, l = l0 + fl0. Since tension is proportional to the increase in length, T = k × fl0, where k is a constant.
\(m=\frac{M}{{{l}_{0}}+f{{l}_{0}}}\) where M is the mass of the string
\ \(n=\frac{1}{2{{l}_{0}}(1+f)}\sqrt{\frac{kf{{l}_{0}}\times {{l}_{0}}(1+f)}{M}}=\frac{1}{2}\sqrt{\frac{kf}{M(1+f)}}\)
Since l0, k and M are constants \(n\propto \sqrt{\frac{f}{1+f}}\)
\ \(\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{f}_{1}}(1+{{f}_{2}})}{{{f}_{2}}(1+{{f}_{1}})}}=\sqrt{\frac{0.02(1+0.04)}{0.04(1+0.02)}}=0.71\)
1. Determine the speed of sound waves in water, and find the wavelength of a wave having a frequency of 242 Hz. Take Bwater = 2 ´ 109 Pa.
Solution :
Speed of sound wave, v = \(\sqrt{\frac{B}{\rho }}=\sqrt{\frac{\left( 2\times {{10}^{9}} \right)}{{{10}^{3}}}}=1414\,\,m/s\)
Wavelength l = \(\frac{v}{f}\) = 5.84 m
The motion of sound wave in air is adiabatic. In the case of an ideal gas, the relation between pressure p and volume V during an adiabatic process is given by pVg = constant. Where g is the ratio of the heat capacity at constant pressure to that at constant volume.
After differentiating , we get \(\frac{dp}{dV}{{V}^{\gamma }}+\gamma p{{V}^{\gamma -1}}=0\)
Since B = \(-V\frac{dp}{dV}=\gamma p\)
\ v = \(\sqrt{\frac{\gamma p}{\rho }}\)
(Laplace correction in contrast to Newton’s formula v = \(\sqrt{\frac{P}{\rho }}\) )
Using the gas equation \(\frac{p}{\rho }=\frac{RT}{M}\) where M is the molar mass.
Thus, v = \(\sqrt{\frac{\gamma RT}{M}}\)
2. The ratio of the density of the oxygen to nitrogen is 16 : 14. At what temperature the speed of sound will be the same which is in nitrogen at 200C?
Solution :
The velocity of sound in a gas is given by \(\sqrt{\frac{\gamma P}{\rho }}\,\,=\,\,\,\sqrt{\frac{\gamma RT}{M}}\)
Where M is the molecular mass of the gas. T is absolute temperature at t0C the velocity of sound in oxygen is \(\sqrt{\frac{\gamma \left( 273\,+\,t \right)R}{{{M}_{O}}}}\)
Velocity of sound in Nitrogen at 270C = \(\sqrt{\frac{\gamma R\,\left( 273\,+\,27 \right)}{{{M}_{N}}}}\)
Now, \(\sqrt{\frac{\gamma \left( 273\,+\,t \right)R}{{{M}_{O}}}}\) = \(\sqrt{\frac{\gamma R\,\left( 273\,+\,27 \right)}{{{M}_{N}}}}\)
Þ \(\frac{{{M}_{O}}}{{{M}_{N}}}\,\,=\,\,\frac{273\,+\,t}{300}\)
t = \(\frac{16}{14}\,\,x\,300-273\) = 1530 C
1. A weight of 10 kg suspended form lower end of a uniform steel wire of area of cross-section 0.0045 sq. cm. Find out the ratio of the fundamental frequencies of the wire in the cases, when it is rubbed with a resigned cloth and when it is plucked in the middle. Young’s modulus of the material is 19.6 x 1011dyne/cm2.
Solution:
The frequency of the longitudinal vibration when it is rubbed with resined cloth is given by n1 = \(\frac{1}{2l}\,\,\sqrt{\frac{Y}{\rho }}\)
Where l and r are the length of the wire and density of the material. When the wire is plucked, frequency is given by n2 = \(\frac{1}{2l}\,\,\sqrt{\frac{T}{\mu }}\)
Where m is the mass per unit length.
From these two equations
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Y}{T}\frac{\mu }{\rho }}\,\,=\,\,\sqrt{\frac{Y}{Mg}\frac{a\rho }{\rho }}\)
Where a is the area of cross-section.
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Ya}{Mg}}\,\,=\,\,\sqrt{\frac{19.6\,x\,{{10}^{11}}\,x\,0.0045}{{{10}^{4}}\,x\,980}}\) = 30
2. The vibrations of a string of length 60 cm fixed at both ends are represented by the equation y = 4 sin \(\left( \frac{\pi x}{15} \right)\,cos\,\,\left( 96\,\pi \,t \right)\) where x and y are in cm and t in seconds.
(i) What is the maximum displacement of a point at x = 5 cm?
(ii) Where are the nodes located along the string ?
(iii) What is the velocity of the particle at x = 7.5 cm and at t = 0.25 s ?
(iv) Write down the equations of the component waves whose superpositions give the above wave.
Solution:
y = 4 sin (p x/15) cos (96 p t ) can be broken up into
\(y=2\left[ \sin \left( \frac{\pi \,x\,+\,96\,\pi \,t}{15} \right)\,+\,\sin \left( \frac{\pi \,x\,-\,96\,\pi \,t}{15} \right) \right]\)
Thus the waves are of the same amplitude and frequency but travelling in opposite directions which thus, superimpose to give a standing wave,
(i) At x = 5 cm the standing wave equation gives
y = 4 sin (5p/15) cos (96 p t)
= 4 sin p/3 cos (96p t) = 4 ´ Ö3/2 cos (96 p t)
\ Maximum displacement = 2Ö3.
(ii) The nodes are the points permanently at rest. Thus they are those points for which
sin (p x/15) = 0
i.e. p x/15 = n p , n = 0, 1,2,3, 4 ......
x = 15 n i.e. at x = 0,15,30,45 and 60 cm
(iii) The particle velocity is equal to
\(\begin{align} & \left( \frac{\partial \,y}{\partial \,t} \right)=4\sin \left( \frac{\,\pi \,x}{15} \right)\,\,(96\,\pi \,)\,(-\sin 96\,\pi \,t)\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\,-384\,\pi \sin \,\left( \frac{\pi \,x}{15} \right)\,\sin \,(96\,\pi \,t) \\ & at\,\,\,\,x\,=\,7.5\,\,\,and\,\,\,\,t\,=\,0.\,25\,\,we\,\,\,get \\ & \left( \frac{\partial \,y}{\partial \,t} \right)\,=\,-384\,\pi \sin \,\left( \frac{\pi \,x}{15} \right)\,\sin \,\left( 96\,\pi \,t \right)=-384\pi \sin \left( \frac{\pi }{2} \right)\sin \left( 24\pi \right)=0 \\ \end{align}\)
(iv) The equations of the component waves are
\(\begin{align} & {{y}_{1}}=2\,\sin \,\left( \frac{\pi \,x}{15}\,+\,96\,\pi \,t \right) \\ & and\,{{y}_{2}}=2\,\sin \,\left( \frac{\pi \,x}{15}\,-\,96\,\pi \,t \right) \\ \end{align}\)
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WAVES
Detailed analysis of wave motion needs a mathematical language for describing waves. Transverse waves on a string will be considered first, but the methods introduced are applicable to any wave motion.
We consider a stretched string which in its equilibrium position lies along the x-axis of a coordinate system. During wave motion each particle, labeled by its equilibrium position x, is displaced some distance y in the transverse direction. The value of y depends on which particle of string is being described (that is, on x) and also on the time t at which we look at it. Thus y is a function of x and t : y = f (x, t). If this function is known for a particular wave motion, it can be used to predict the position of any particle at any time. From this the velocity and acceleration of any point, the shape of the string, its slope at any point, and any other kinematics information, can be obtained.
Thus the function y = f (x, t), once it is known, constitutes a complete description of the motion. Such a function is called a wave function; in this section we study wave functions for sinusoidal waves.
As a sinusoidal wave travels from left to right on a string, each particle undergoes simple harmonic motion, but successive particles have successively larger delays, or phase lags, compared to the motion at the left end (at x = 0) where the motion originates. Let the displacement of a particle at the origin (x = 0) be given by
y = A sin wt = A sin2p ft … (1)
The time required for the wave disturbance to travel from x = 0 to some point x to the right of the origin is given by x/v, where v is the wave speed. The motion of point x at time t is the same as the motion of point x = 0 at the earlier time t – x/v. Thus the displacement of point x at time t is obtained simply by replacing t in Eq. (1) by t – (x/v)), and we find
y (x, t) = A sin w \(\left( t-\frac{x}{v} \right)\) … (2)
= \(A\sin 2\pi f\left( t-\frac{x}{v} \right)\) … (3)
The notation y(x, t) is a reminder that the displacement y is a function of both the location x of the point and the time t.
Equation (2) can be rewritten in several alternative forms, conveying the same information in different ways. In terms of the period t and wavelength l, we find, using Eq. (3)
\(y(x,t)=A\sin 2\pi \left( \frac{t}{\tau }-\frac{x}{\lambda } \right)\) … (4)
Another convenient form is obtained by defining a quantity k, called the propagation constant or the wave number:
\(k=\frac{2\pi }{\lambda }\)
In terms of k and w, the wavelength-frequency relation v = lf becomes w = vk, and Eq. (2) can be written
y(x, t) = A sin (wt - kx). … (5)
Which of these various forms is used is a matter of convenience in a specific problem; fundamentally they all say the same thing. It should be mentioned in passing that some authors define the wave number as 1/l rather than 2p/l. In that case, our k is still called the propagation constant.
At any given time t, Eq. (2), (3), (4), or (5) give the displacement y of a particle from its equilibrium position as a function of the coordinate x of the particle. If the wave is a transverse wave in a string, the equation represents the shape of the string at that instant, as if we had taken a photograph of the string. Thus at time t = 0,
y = A sin (–kx) = – A sinkx = – A sin 2p \(\frac{x}{\lambda }.\)
This curve is plotted in Figure. At any given coordinate x, Eq. (2), (3), (4) or (5) gives the displacement y of the particle at that coordinate, as a function of time. Thus at the position x = 0,
y = A sin wt = A sin 2p \(\frac{t}{\tau }.\) This curve is plotted in Figure.
The above formulas may be used to represent a wave traveling in the negative x-direction by making a simple modification. In this case the displacement of point x at time t is the same as the motion of point x = 0 at the later time (t + (x/c)), and in Eq. (2) to (5) we must replace t by (t + (x/c)). Thus, for a wave traveling in the negative x-direction,
y = \(A\sin 2\pi f\left( t+\frac{x}{v} \right)=A\sin 2\pi \left( \frac{t}{\tau }+\frac{x}{\lambda } \right)\) = A sin (wt + kx) … (6)
It is essential to distinguish carefully between the speed of propagation v of the waveform and the particle speed v of a particle of the medium in which the wave is traveling. The wave speed v is given by \(v=\lambda f=\frac{\omega }{k}\)
The particle speed v for any point in a transverse wave (that is, at a fixed value of x) is obtained by taking the derivative of y with respect to t, holding x constant. Such a derivative is called a partial derivative and is written \(\partial \) y/ \(\partial \) t. Thus for a sinusoidal wave given by y = A sin (wt – kx),
we have v \(=\frac{\partial y}{\partial t}=A\omega \,\,\cos (\omega t-kx)\) … (7) (v : particle speed)
The acceleration of the particle is the second partial derivative:
\(a=\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-{{\omega }^{2}}A\sin (\omega t-kx)\)
One may also compute partial derivatives with respect to x, holding t constant. The first derivative \(\partial \) y/\(\partial \) x is the slope of the string at any point. The second partial derivative with respect to x is computed as below
\(\frac{\partial y}{\partial x}=-KA\cos (\omega t-kx)\)
\(\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=-{{K}^{2}}A\sin (\omega t-kx)\)
\ \(\frac{{{\partial }^{2}}y/\partial {{t}^{2}}}{{{\partial }^{2}}y/\partial {{x}^{2}}}=\frac{{{\omega }^{2}}}{{{k}^{2}}}={{v}^{2}},\)
since, v = w/k. the partial differential equation is
\(\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}={{v}^{2}}\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}\) … (8)
This is one of the most important equations on waves. It is called the wave equation, and whenever it occurs, the conclusion is made immediately that y propagates as a travelling wave along the x-axis with a wave speed v.
Wave Speed :
\(v=\frac{\omega }{k}=\frac{2\pi \nu }{2\pi /\lambda }=\nu \lambda \)
Angular Frequency :
\(\omega =2\pi \nu =\frac{2\pi }{T}\)
= frequency, T = time period
Intensity (I) :
It is energy passing perpendicular per unit area per unit time.
\(I=2{{\pi }^{2}}{{\nu }^{2}}{{a}^{2}}\rho v\)
Where p is density of medium and a is amplitude of the wave.
Wave Number (K) :
It is given by \(K=\frac{2\pi }{\lambda }\)
Particle velocity: \({{v}_{p}}\,\,=\,\,\frac{dy}{dt}\,=\,a\omega \,\,\cos \,\,2\pi \,\,\,\left( \frac{t}{T}\mp \,\,\frac{x}{\lambda } \right)\) hence maximum particle velocity is equal to aw
Equation of Progressive Wave :
\(y=a\sin \left( \omega t\mp kx \right)\)
or \(y=a\sin \frac{2\pi }{\lambda }\left( vt\mp x \right)\)
or \(y=a\sin \omega \left( t\mp \frac{x}{v} \right)\)
or \(y=a\sin 2\pi \left( \frac{t}{T}\mp \frac{x}{\lambda } \right)\)