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Chapter I Simple and Damped Simple Harmonic Motion 1.1. Periodic Motion Any motion which is repeated regularly after a fixed interval of time is known as periodic motion. This fixed interval is known as the time period of motion. In a peri...

Chapter I Simple and Damped Simple Harmonic Motion 1.1. Periodic Motion Any motion which is repeated regularly after a fixed interval of time is known as periodic motion. This fixed interval is known as the time period of motion. In a periodic motion, a particle describes the same path again and again in equal intervals of time. The rotation of earth about its own axis is a periodic motion because it is repeated regularly after an interval of one day. Similarly, the revolution of earth around the sun is a periodic motion because it is repeated regularly after an interval of one year. Revolution of moon around the earth, to and fro motion of a pendulum bob, motion of piston in the cylinder of an engine, vibrations of a tuning fork and motion of atoms in a lattice are some other examples of periodic motion. The motion of a particle is said to be vibratory or oscillatory when it moves, back and forth over the same path, in periodic motion. 1.2. Simple Harmonic Motion It is a special case of periodic motion and is very common in nature. It is the most fundamental vibration of one-dimensional system. A particle is said to execute simple harmonic motion when it moves to and fro, about a fixed point, along a straight-line path and repeats this motion periodically. Simple harmonic motion may also be defined as the projection of uniform circular motion on the diameter of the circle of reference. (1)  The term ‘simple harmonic motion’ is generally abbreviated to S.H.M. The inclusion of the word ‘harmonic’ in the term ‘simple harmonic motion’ can be explained based on following two factors. (i) It is this type of motion which is involved in music i.e., harmony. (ii) The displacement of a particle executing simple harmonic motion is expressed in terms of sines and cosines. The expressions containing these functions are generally known as ‘harmonics. When a particle is executing simple harmonic motion, the restoring force acting on it, at any instant, is proportional to its displacement from a fixed point in its path and is always directed towards that point. 1.3. Simple Harmonic Oscillators A particle or a system executing simple harmonic motion is called a simple harmonic oscillator. Examples (i) A metallic bob swinging at the end of a thread or a light rigid rod. (ii) A metallic disc supported by a rigid wire through its centre and oscillating through small angles. (iii) A cylindrical or globular metallic vessel (known as Helm-holtz resonator) containing air when sound passes across its narrow neck (iv) A liquid contained in a U-tube and oscillating about its equilibrium position of equal levels in the two limbs. (2) (v) A hydrometer of constant cross-sectional area floating in a liquid. It executes small vertical oscillation when depressed slightly from its equilibrium position. (vi) An electrical circuit containing an inductance connected across a capacitance carrying a charge. This constitutes an electrical oscillator. (vii) A magnet placed in the stirrup of a vibration magnetometer and vibrating under the action of the earth’s magnetic field. MECHANICAL OSCILLATORS 1.4. Differential Equation of Simple Harmonic Motion Consider a particle of mass m executing S.H.M. Let its displacement from the rest position be x at any instant. Also, let the restoring force be F at that instant. In simple harmonic motion, the restoring force is always proportional to the displacement. Fx or F = -s x (1) where s is a constant of proportionality. This constant is known as force constant or spring constant or stiffiness. The negative sign used in equation (1) signifies that the vectors F x are oppositely directed. In other words, the direction of restoring force is opposite to that of the displacement. (3)  Instantaneous acceleration of the particle can be obtained by differentiating twice the displacement with respect to time. Its value is d2x... It is denoted by x. dt 2 Now, by Newton’s second law of motion, Force = Mass × Acceleration.. or F =m x But F =- sx..  - sx =m x.. Here, m x is the driving force and -s x is the restoring force...  m x +s x =o Dividing both sides by mass m (which is also known as inertia constant), we get.. s x+ x=0 m It may be noted here that s and m are not significant when taken s individually. However, their ratio, i.e., , is a very important factor in m simple harmonic motion. s Therefore, we substitute 2 for. Here  is the angular velocity m of S.H.M such that (4)  2  = 2 f = T where f is the frequency and T is the time period. Sometimes  is also referred to as the angular frequency. s Putting 2 = , we get m x + 2 x = 0 (2) In words, we can say that x and its second order derivative with respect to time give zero when linearly combined. Equation (2) is the differential equation of simple harmonic motion. 1.5. Dimensions and Units of Stiffness Restoring Force Stiffness , s = Displaceme nt Therefore, stiffness is defined as the restoring force per unit displacement. It is assumed that the displacement is small. Dimensional formula of force = MLT-2 Dimensional formula of displacement = L MLT −2  Dimensional formula of stiffness = = ML0T-2 L Therefore, we conclude that stiffness has one dimension in mass, zero dimension in length and -2 dimension in time. In C.G.S. system, the stiffness is expressed in gm./sec 2. or dynes/cm. (5)  In S.I. system, the stiffness is expressed in kgm. /sec.2 or newton/metre. 1.6. Expression for Displacement in Simple Harmonic Motion The distance of the particle, executing S.H.M., at any instant from its mean or equilibrium position is called displacement which is a vector quantity. In order to find the expression for displacement in simple harmonic motion we shall have to solve, for x , the differential equation of S.H.M. The technique of solving second order differential equations is to try certain possible solutions. In this case, let us try the solution x = (A cos t) x where x is a unit vector in the direction of displacement x and A is a constant which has the same dimensions as those of x. This constant helps to make the above equation dimensionally correct because cos t is dimensionless. In magnitude, x = A cos t (3) Differentiating both the sides of equation (3) with respect to time, we get dx = x = -A sin t dt where x is the magnitude of velocity x. Again, differentiating the above equation w.r.t. time, we get d2x 2 = x = −A2 cos t dt (6)  = -2 (A cos t) = -2x or x + 2 x = 0 where x is the magnitude of acceleration x. In vector notation, x + 2 x = 0 Thus, we can build the differential equation of S.H.M from the solution x = A cos t. Therefore, this solution is correct. Let us now try another solution x = (B sin t) x̂ where B is another constant which plays exactly the same role as that of A. In magnitude, x = B sin t (4) Differentiating equation (4) w.r.t. time, we get x = B  cos t Differentiating again the above equation w.r.t. ‘t’, we get x =-B2 sin t = -2 (B sin t) = -2 x  x + 2 x = 0 Thus, the solution x = B sin t is also correct. (7)  The general solution of the second order differential equation is obtained by combining the two solutions given in (3) and (4). Thus, x = A cos t + B sin t Now, put A = a sin  and B = a cos  where ‘a’ is the maximum possible value of displacement and is known as the ‘amplitude of displacement’. Also,  is a constant angle known as “phase constant” or “initial phase” or “epoch”. It is measured in radians. It can haveany value from zero to 2.  x = a sin  cos t + a cos  sin t or x = a sin t cos  + a cos t sin  or x = a sin (t + ) (5) In vector notation, x = {a sin (t + )}. x̂ where the term (t + ) is called the phase of vibration. Here, a = A 2 + B2 A and tan  = B A   = tan −1   B Equation (5) gives the required expression for displacement, in terms of the phase constant , at any instant t. (8) Displacement – Time Graph If it is assumed that the initial phase  is zero, then equation (5) becomes x = a sin (t + 0) = a sin t 2 But = T  2   x = a sin   t T  By assigning different values to t, we get corresponding different magnitudes of x which are given below in a tabular form. Time 0 T/4 T/2 3T/4 T Displacement 0 A 0 -a 0 (in magnitude) Now, if displacement is plotted against time, then graph will be a sine curve (Fig. 1.1). Fig. 1.1. (Displacement-time graph) In this displacement-time graph, time is plotted against X-axis because it is an independent variable and displacement is plotted against Y-axis because it is a dependent variable. 1.7. Expression for Velocity in S.H.M The velocity of a particle executing simple harmonic motion is defined as the rate of change of its displacement with time. We know that x = {a sin (t + )} x̂ (9)  In magnitude, x = a sin (t + ) Differentiating both sides w.r.t. time, we get dx = x = a  cos (t + ) (6) dt In vector notation, x = {a  cos (t + )} x̂ The above equation gives the required expressionfor velocity x̂ in S.H.M. The value of x will be maximum when the value of cos (t + ) is maximum i.e. 1.  ( x )max. = a  In words, the magnitude of maximum velocity is a. This is known as the velocity amplitude. It is defined as the maximum value of the velocity attained by a particle executing simple harmonic motion. 1.8. Velocity – Time Graph Let us assume that the initial phase  is zero. Then the equation (6) becomes x = a  cos (t + 0) = a  cos t 2 But = T  2   x = a  cos   t T By assigning different values to t, we get corresponding different values of velocity x which are given in the following table. (10)  Time 0 T/4 T/2 3T/4 T Velocity a 0 -a 0 a (in magnitude) With the help of the above data, we can plot velocity against time. The velocity time graph is shown in fig. 1.2. This graph is a cosine curve. Fig. 1.2. (Velocity-time graph) Time is plotted against X-axis because it is an independent variable and velocity is plotted against Y-axis because it is a dependent variable. 1.9. Phase Difference between Velocity and Displacement The magnitudes of velocity x and displacement x are respectively given by   x = a cos (t + ) = a sin (t + ) +   2 and x = a sin (t + ) It is clear from the above two equations that the velocity is ahead  of the displacement by a phase angle of radians. In other words, 2 velocity is zero at maximum displacement and is maximum at zero displacement. 1.10. Expression for Acceleration in S.H.M. The acceleration of a particle executing simple harmonic motion is defined as the rate of change of its velocity with time. We know that the magnitude of velocity x is given by (11)  x = a  cos (t + ) Differentiating both sides w.r.t. time t, we get x = -a2 sin (t + ) (7) But x = a sin (t + )  x = -2 {a sin (t + )} = -2 x In vector notations, x = - 2 x The negative sign shows that the acceleration and the displacement are oppositely directed. If the displacement is positive, then acceleration is negative and vice versa. In other words, acceleration is always directed towards the mean or equilibrium position. This is an important characteristic of simple harmonic motion. Again x  x ( 2 is constant) This shows that numerically the acceleration of a particle executing simple harmonic motion is directly proportional to its displacement from the mean position. The direct proportionality between acceleration and displacement is another important characteristic of simple harmonic motion. The above two important characteristics help us to define S.H.M in the following way. The motion of a particle is said to be simple harmonic when it moves wider an acceleration (and hence under a force) which is directly proportional to its displacement from the mean position and is always directed towards the mean position. (12)  Let us now investigate the values of acceleration at the mean position and at the extreme position. At the mean position displacement x = 0  acceleration x = 0 So, acceleration is zero at the mean position. At the extreme position, the displacement x assumes the maximum value ‘a’ (amplitude of displacement). Thus, at the extreme position, the acceleration attains the maximum value a2. This maximum value of acceleration is known as acceleration amplitude. Acceleration amplitude is defined as the maximum value of acceleration attained by a particle executing simple harmonic motion. 1.11. Acceleration – Time Graph If phase constant is taken as zero, then equation (7) becomes x = -a2 sin (t + 0) = -a2 sin t 2 But = T  2   x = - a2 sin   t T By assigning different values to t, we get corresponding values of acceleration x which are given in the following table. Time 0 T/4 T/2 3T/4 T Acceleration 0 -a2 0 a2 0 (13) (in magnitude) Using the above data, acceleration can be plotted against time. Fig. 1.3. (Acceleration-time graph) The acceleration-time graph is a negative sine curve as shown in figure 1.3. Time is plotted agains X-axis and acceleration against Y-axis. 1.12. Phase Difference between Acceleration and Displacement The magnitudes of displacement x and acceleration x are respectively given by x = a sin (t + ) and, x = -a2 sin (t + ) = a2 {-sin(t+)} = a2 sin {(t + ) + } Comparing the expressions for displacement and acceleration, we conclude that the displacement and acceleration are  radians out of phase with each other. 1.13. Energy of a Simple Harmonic Oscillator A simple harmonic oscillator generally possesses the following two types of energy. 1. Potential Energy. This energy is due to displacement from the mean position of a particle executing S.H.M. 2. Kinetic Energy. This energy is due to the velocity of a particle executing S.H.M. (14)  At any instant, the total mechanical energy of an oscillator is the sum of its potential energy and kinetic energy. Here, it is assumed that no dissipative or frictional forces are acting on the system. In other words, the system is conservative. So, the total energy of the simple harmonic oscillator is conserved. 1.14. Equation of S.H.M (from energy considerations) Consider a particle executing S.H.M. Let us calculate the work done on the particle in order to displace it by an amount dx from the mean position. When the particle suffers a displacement away from the mean position, a restoring force F = - s x opposes the motion. Therefore, the applied force F should be equal and opposite to the restoring force.  F=s x Work is given by the scalar or dot product of the applied force and the displacement.  Work dW done for displacement xd = F.dx Total work W done to move the body from mean position 0 to displacement x can be found out by integrating dW. x Thus W =  F.dx 0 x =  sx.dx 0 x = s  x.dx 0 In magnitude, x. dx = x.dx [ angle between x and dx is zero] (15)  x  W = s  xdx 1 or W = sx 2 2 This work done on the particle is stored as potential energy. 1 2  Potential energy = sx 2 1 Kinetic energy of the particle = m x 2 2 where m is the mass of the particle and x is its velocity. The total mechanical energy E of the particle is the sum of the two energies-kinetic and potential. 1 1  E = sx 2 + mx 2 (8) 2 2 Differentiating both the sides of equation (8) w.r.t. ‘t’, we get dE 1 1 = s. 2x. x + m.2x.x dt 2 2 According to the law of conservation of energy, E is a constant quantity. dE  =0 dt or sx x + m x x = 0 or x (sx + m x) = 0 But x  0 , except at the extreme position. (16)   m x + sx = 0 s or x + x=0 m s Put = 2 m  x + 2 x = 0 In vector form, x + 2 x = 0 This is the equation of S.H.M, derived from energy considerations. Total Energy at any Instant We have already derived that 1 1 E= m x 2 + sx 2 (8) 2 2 where x = a sin (t + ) and x = acos (t+) Putting these values in equation (8), we get 1 1 E = m a 22 cos2 (t + ) + s a 2 sin 2 (t + ) 2 2 s But 2 = or m2 = s m 1 1  E = s a 2 cos2 (t + ) + sa 2 sin 2 (t + ) 2 2 1 = s a 2 cos2 (t + ) + sin 2 (t + ) 2 (17)  1 2 or E= sa (9) 2 It is clear from the above equation that total energy E at any instant is a constant quantity and is independent of time. Maximum Kinetic Energy If m is the mass of a particle and x its velocity, then 1 kinetic energy = m x 2 2 where x =a cos (t + ) (From eq. (6)) 1  Kinetic energy = m a 2 2 cos2 (t + ) 2 Kinetic energy will be maximum when the value of cos 2 (t+) is maximum i.e. 1. 1  Maximum kinetic energy = m a 22 2 1 2 = sa [ m2 = s] (10) 2 Maximum Potential Energy We have already derived that 1 2 potential energy = sx 2 where x = a sin (t + ) (From eq. (5)) (18)  The magnitude of potential energy will be maximum when the displacement x assumes the maximum value ‘a’ i.e. amplitude of displacement. Amplitude ‘a’ can be both positive and negative. So, the system has maximum potential energy at x =  a. When the displacement is maximum, velocity is zero. So, we may say that the system possesses maximum potential energy at the point where velocity is zero. 1 2  Maximum potential energy = sa (11) 2 It is clear from equations (11), (10) and (9) that the values of maximum potential energy, maximum kinetic energy and energy at any instant are all equal. Note: The results obtained for linear S.H.M are equally valid for angular S.H.M if mass m is replaced by moment of inertia I and displacement x by an infinitesimally small angular displacement 0. Average Kinetic Energy 1 Kinetic energy at any instant = m x 2 2 where x = a  cos (t + ) Average kinetic energy for time T (i.e. one complete time period) is given by T 1  2 ma 2 cos2 (t + )dt 2 0 T (19)  1 1 T =. ma 22 T 2 0 cos2 (t + )dt m2 a 2 T 1 = 2T 0 2 {(1 + cos 2( t + )}dt = m2 a 2 4T  1.dt + 0 T T 0 cos 2t + )dt  m2 a 2 1 =.T = m2 a 2 4T 4 But m2 = s 1 2  Average K.E. = sa (12) 4 Average Potential Energy 1 2 Potential energy at any instant = sx 2 where x = a sin (t + )  Instantaneous potential energy 1 2 = sa sin 2 (t + ) 2 Average potential energy for time T (i.e., one complete time period) is given by T 1 2 2 0 2 sa sin (t + )dt T 1 1 T =. sa 2 0 sin 2 (t + )dt T 2 (20)  sa 2 1 = T 0 1 − cos 2 (t + )dt 2T 2 = sa 2 4T  1.dt −  T 0 0 T cos 2(t + )dt  sa 2 1 =.T = s a 2 4T 4 1 2  Average potential energy = sa (13) 4 It is clear from equations (12) and (13) that the average kinetic energy of a harmonic oscillator is equal to the average potential energy. And each is equal to one half the total energy of the oscillator. 1.15 Graphical Representation of Energy Let us first study the variation of potential energy with displacement. 1 2 We know that P.E = sx 2 It is clear from this equation that if we plot a graph between potential energy and displacement, it will be parabolic in shape. It is further clear from the above equation that graph will be symmetric about x = 0 (i.e. equilibrium position) As is shown in fig. 1.4, the vertex of the parabola is at x = 0. Its physical significance is that energy is stored both when x is positive and negative. We also know that the potential energy attains the maximum value 1 2 s a at x = a. Here x =  a, means that P.E is stored in a mechanical 2 oscillator, e.g. in a spring, at the time of compression as well as (21) extension. Again, the potential energy is zero at the mean position (i.e. at x = 0). Fig. 1.4. (P.E – displacement curve) All the above facts have been depicted in potential energy- displacement graph shown in fig. 1.4. Now, let us study the variation of kinetic energy with displacement. we know that 1 K.E. = mx 2 2 It is clear from the above equation that the kinetic energy- displacement curve will also parabolic. Even if the kinetic energy is plotted against velocity, the graph will be parabolic in shape. To sum up, the kinetic energy-displacement curve is parabolic with respect to both displacement and velocity. Since velocity is maximum at the mean position, (at x = 0), therefore the kinetic energy will also be maximum at the mean position 1 2 i.e. s a. At the extreme position, the particle will momentarily come to 2 rest and velocity will be zero. Therefore at the extreme position, the magnitude of kinetic energy will be zero. On the basis of the above considerations, the kinetic energy- displacement curve will be as shown in fig. 1.5. Fig. 1.5. (K.E-displacement curve) (22)  Let us now graphically represent the continuous exchange of kinetic and potential energies in simple harmonic motion. For this purpose, we will plot three curves at the same place. These three curves will be (i) K.E.-displacement curve,(ii) P.E.-displacement curve and (iii) Total energy-displacement curve. Such an energy distribution-displacement curve is shown in figure 1.6. The curve for the total energy E is a horizontal line. This is because of the fact that E is constant. It may be noted that the particle cannot go beyond the points where the horizontal line is intersecting the potential energy curve. It is due to the fact that potential energy cannot be greater than total energy. For any value of x, the sum of the ordinates of the P.E. and K.E. curves is equal to E, where E is the total constant energy of the particle. It is clear from the graph that as the particle moves away from the mean position, the potential energy increases at the expense of kinetic energy. When the particle moves away from the extreme position, the kinetic energy increases at the expenses at the expense of potential energy. Fig. 1.6. (Energy-displacement curve) The point of intersection of the K.E. curve and P.E. curve E a represents. This means that at the position x =  between the mean 2 2 and the extreme position, the energy is half kinetic and half potential. The inversion of one curve over the other shows that there is a phase (23)   difference of between the velocity as related to K.E. and the 2 displacement as related to P.E. 1.16 Simple Harmonic Oscillations in an Electrical System The basic oscillating circuit, also known as tank circuit, consists of an inductance coil connected across a condenser (Fig. 1.7). This inductance should be a pure inductance. The condenser is first charged by a high tension battery. The electrostatic energy of the condenser resides in the dielectric medium. When the condenser begins to discharge itself through the inductance, a current begins to flow through the coil. This sets up a magnetic field around the coil As the condenser discharges, the electrostatic energy of the condenser is converted into the magnetic energy. There is a continuous exchange of energy between the inductance and the capacitance. It may be noted here that the tank circuit is a resistance-free circuit. Therefore, there is no dissipation of energy in the form of heat. Hence the total energy is conserved. dI Let L be the self-inductance of the coil and be the rate of dt growth of current. Fig. 1.7. (Oscillatory circuit) According to Faraday’s law of electromagnetic induction. dI V=-L dt (24) where V is the voltage developed across the inductance at any instant. But dq I= ( Current is the rate of flow of charge) dt where q is the charge on the capacitor at any time t. d 2q  V = - L 2 = - L q dt The negative sign shows that the voltage V opposes the growth of current. q Voltage across the capacitor = C where C is the capacitance of the condenser. Applying Kirchoff’s 2nd law to the mesh containing inductance and capacitance, we get q Lq + =0 C Dividing both sides by L, we get 1 q + q = 0. LC But 1 = 2 LC  q + 2 q = 0 (25)  This is the required equation for simple harmonic oscillations in an electrical system. Its solution is q = q0 sin (t + ) where q0 is the maximum value of the charge. It is also known as amplitude of charge. Comparison of Equations of Mechanical and Electrical Oscillators Quantity Mechanical Oscillator Electrical Oscillator = n of S.H.M. x + 2 x = 0 q + 2 q = 0 Force mx + sx = 0 q Lq + = 0 C It is clear from the above comparison that (a) q plays the same role in an electrical oscillator as x plays in a mechanical oscillator. (b) the inductance L plays the same part in an electrical circuit as is played by inertial mass m in the mechanical system. For this reason, the inductance L is sometimes known as magnetic field inertia. 1 (c) the equivalent of , in an electrical circuit, is stiffness s in the C mechanical system. Note: A wide range of frequencies can be produced by combining inductance and capacitance of suitable values. (26) 1.17. Addition of Two Simple Harmonic Motions in one Dimension Case (i) Vibrations of equal angular frequencies but different amplitudes and phases. When two simple harmonic vibrations act on the same system at the same time, then the resulting motion of the system can be found by the vector addition of their individual displacements. Let the displacements of two simple harmonic motions be represented in magnitude by x1 = a sin (t + 1) (1) and x2 = b sin (t + 2) (2) where a and b represent the amplitudes, 1 and 2 the phases and  the angular frequency of two simple harmonic vibrations respectively. Thus the magnitude of displacement x of the resultant S.H.M is given by x = x1 + x2 = a sin (t + 1) + b sin (t + 2) = a sin t cos 1 +a cos  sin 1 + b sin t cos 2 + b cos t sin 2 = cos t (a sin 1 + b sin 2)+ sin t (a cos 1 + b cos 2) (3) Put a sin 1 + b sin 2 = R sin 0 (4) and a cos 1 + b cos 2 = R cos 0 (5) where R is given by squaring, adding and simplifying (4) and (5). i.e. R2 = a2 + b2 + 2ab (sin 1 sin 2 + cos 1 cos 2) (27)  = a2 + b2 + 2ab cos (2 - 1) = a2 + b2 + 2ab cos  where phase difference,  = 2 - 1 = constant. Also,  is given by dividing (4) by (5). a sin 1 + b sin 2  tan  = a cos 1 + b cos  2 Hence equation (3) becomes x = R cos t sin  + R sin t cos  = R sin (t + ) Here R represents the amplitude,  the phase constant and  the frequency of the resultant S.H.M. Case (ii) Vibrations of different frequencies but same amplitudes and phases. In this case, the magnitude of displacements of two simple harmonic vibrations can be represented as x1 = a sin 1t and x2 = a sin 2t where a represents the amplitude of two vibrations and 1, 2 their respective frequencies such that 2>1. The magnitude of displacement x of the resultant S.H.M. is then given by x = x1 + x2 = a (sin 1t+ sin 2t) (28)  2 − 1  + 2 = 2a cos t sin 1 t 2 2   + 2  = A sin  1 t  2  2 − 1 where A = 2a cos.t 2 = amplitude of the resultant displacement which grows and decays between the maximum values of 2a. When 1 and 2 are nearly equal, this growth and decay of amplitude is recorded as beats of frequency (2-1). Thus, the resultant S.H.M. represents sinusoidal vibration of 1 + 2 average frequency. 2 Fig. 1.8. 1.19. Complex Number Notation The mathematical discussion of an oscillatory motion is made easy by the use of complex numbers. A complex number z may be written as z + a + ib where a and b are ordinary real numbers and i is the square root of -1 i.e. i = -1 Squaring both the sides, we get i2 = -1 Addition and Subtraction of two Complex Numbers Let z1 = a1 + ib1 (29) and z2 = a2 + ib2  z1+z2 = (a1+ib1) + (a2+ib2 = (a1+a2) + i(b1+b2) and z1-z2 = (a1+ib1) – (a2+ib2) = = (a1-a2) + i(b1-b2) Multiplication of two Complex Number z1z2 = (a1+ib1)(a2+ib2) = a1a2+ia1b2 + ia2b1 + i2b1b2 But i2 = -1  z1z2 = (a1a2 – b1b2) + i(a1b2+a2b1) Division of two Complex Numbers z1 a1 + ib 1 = z 2 a 2 + ib 2 We will try make the denominator real, z1 (a1 + ib 1 )(a 2 − ib 2 ) = a 2 (a 2 + ib 2 )(a 2 − ib 2 ) (a1a 2 + b1b 2 ) + i(a 2 b1 − a1b 2 ) = a 22 + b 22 Complex conjugate z*: Complex conjugate z* of any complex number is obtained by changing the sign of i wherever it appears. If z = a + ib then, z* = a – ib (30) Product of a Complex number z and its Conjugate z* zz* = (a+ib)(a-ib) = a2 + b2 Thus, the product of a complex number and its conjugate is a real number. Modulus of Complex number It is denoted by | z |. z = zz * = a 2 + b 2 1.20. Exponential Series By the definition of exponential series, we have (ix ) 2 (ix )3 (ix ) 4 (ix )5 e = 1 + ix + ix + + + +... 2! 3! 4! 5! i 2 x 2 i3x 3 i 4 x 4 i5 x 5 = 1 + ix + + + + +... (1) 2! 3! 4! 5! But i=i i2 = -1 i3 = -i i4 = +1 i5 = i Putting these values in (1), we get x 2 ix 3 x 4 ix 5 e = 1 + ix + ix + + + +... 2! 3! 4! 5! (31)   x2 x4   ix 3 ix 5   = 1 − +   +... +  ix − + +...  2! 4i   3! 5!  or eix = cos x + i sin x Similarly, e-ix = cos x - i sin x 1.21 Damped Simple Harmonic Motion In an ideal simple harmonic motion, the displacement continues to follow a sine curve for an infinite time. This is due to the fact that the total energy of the simple harmonic oscillator does not change. But, in actual practice, the simple harmonic oscillator always experiences frictional or resistive forces. Due to these opposing forces, some energy is lost, and the oscillations get damped. In a mechanical oscillator, the amplitude of a freely swinging pendulum decreases continuously with time because energy is being continuously dissipated to overcome the resistance offered by air. Finally, the oscillations die out. In the case of an electrical circuit, where inductance, capacitance and resistance are connected in series, energy is dissipated in resistance. In other words, the resistance of the LCR circuit produces damping. Thus electrical oscillations are damped. Now consider a body of mass m attached to a spring as shown in fig. 1.9. Fig. 1.9. When the body is put into motion, it will experience a damped simple harmonic motion. The whole treatment of the damping process (32) assumes that the dissipation is due to a resistive force proportional to the velocity of the vibrating system. This assumption is valid only so long as the damping force is small. Therefore, the resistive force acting on the body is a function of the velocity of the body.  Resistive force  x = r x where r is a constant of proportionality known as ‘damping factor’. This constant has the dimensions of force per unit velocity. It may be defined as the damping force per unit mass when the system is moving with unit velocity. The resistive force will act in a direction opposite to that of the displacement. Therefore, the equation of motion of such a system is given by mx = − sx − r x where s is stiffness constant. or mx = + sx − r x = 0 or mx = + rx + sx = 0 (A) Dividing both sides by m, we get r s x + x + x = 0 (1) m m x +  x + 2 x = 0 (2) (33)  r where = m s and = m The quantity ‘’ has the dimensions of frequency and  represents the angular frequency of the system, if the damping were absent. here  and  are quantities of the same dimensions. Equation (2) describes a damped simple harmonic motion. Let us now solve equation (A). To do so, let us try the solution, x = C et Here et is non-dimensional. It is because has the dimensions of inverse time and C has the dimensions of length. Differentiating, we can write dx = x = Cet dt d2x and 2 = x =  2Cet dt  From = n (A), we get m2 Cet + rCet + sCet = 0 or Cet(m3 + r + s) = 0 The above equation will be satisfied for all values of t, if m2 + r + s = 0 (34)  It is a quadratic equation in . Solving for , we get − r  r 2 − 4ms = 2m r r 2 − 4ms =−  2m 4m 2 r r2 s or =−  − 2m 4m 2 m Possible Cases Case I. r2 s When 2  4m m 2 i.e. when  2 4 or 2> 42 i.e. > 2 The above inequality implies that the damping factor r is greater than the stiffness ‘s’. In this case, the system is heavily damped. r Now put =p 2m 1/ 2  r2 s and  −  =q  4m 2 m    (35)   =-pq and so, x = Ce-t = Ce(pq)t = Ce-pte.qt = e.-pt.Ceqt = e-pt(C1eqt+C2e-qt) where C1 and C2 are arbitrary constants. Each of these two constants has the same dimensions as those of C. Put C1 + C2 = F and C1 – C2 = G where F and G are arbitrary constants whose values are determined by boundary conditions. Adding, we get, 2C1 = F + G F+G or C1 = 2 F−G Subtracting, we get 2C 2 = F − G or C 2 = 2  F + G qt F − G −qt  x = e −pt  e + e   2 2  − pt   e qt + e −qt   e qt − e −qt  =e F  + G       2   2  = e-pt [F cosh qt + G sinh qt] It is clear from the above equation that the displacement x is not a function of sin t or cos t (periodic functions of time). So, the motion of (36) the system is periodic. When the system is disturbed, it will take a long time to come back to the equilibrium position. This is shown in fig. 1.10. Case II. r2 s When = 4m 2 m i.e.  = 2 This is the condition for critical damping. When a system satisfies this condition, it quickly comes back to its normal position. This property is made use of in the construction of pointer type instruments. In these instruments, the pointer takes very short time to come back to the normal position. These instruments are called dead beat. In this case, the system is critically damped. 1/ 2  r2 s Here q =  2 −  =0  4m m  x = Cet = Ce(-p q)t = Ce-pt In this particular case, the constant Can be written as C + A + Bt where A has the dimensions of length and B that of velocity. x = (A + Bt)e-pt In vector notation, x = (A+Bt)e-pt x̂ It is clear from the expression for displacement that the behavior of the system is non-oscillatory. The decay of displacement with time is exponential (Fig. 1.10). (37)  Fig. 1.10. Case III. r2 s When  4m 2 m i.e. < 2 In this case, the system is lightly damped. 1/ 2 rt  r 2 s  −  − 2 m  4 m 2 m  Now x = Ce t 1/ 2 rt  r 2 s  −  − 2 m  4 m 2 m  = Ce t 1/ 2 − rt  2 m.e  −1  2 r2  = Ce   m − 4m 2  t   1/ 2 s r2  Put  −  = `  m 4m 2    rt − `  x = Ce 2 m e.i t (Ce ) rt −  i`t =e 2m (C e ) rt − m it =e 2 1 + C2e −i`t (1) Both the constants C1 and C2 have the dimensions of C. It may be noted here that we are allowed the use of two constants because we are dealing with a second order differential equation. (38)  A i Put C1 = e 2i A i and C2 = e 2i Both A and  are constants. Putting these values of C1 and C2 in equation (1), we get rt −  A i i`t A -i -i`t  x=e 2m  2i e e − 2i e e  − rt  ei (`t +) e −i (`t + +   = Ae 2m  −   2i 2i  − rt  ei (`t +) − e −i (`t +)  = Ae 2m    2i  rt − or x = Ae 2m sin (`t + ) (1)  rt  − 2m   In vector notation, x = Ae sin (`t + )x̂     The constant A is the maximum amplitude when the damping is absent. It is clear from the above equation that the displacement x varies sinusoidally with time. So, the behavior of the system is oscillatory in nature. The sinusoidal variation of displacement with time is shown in the graph (Fig. 1.11). It is clear from the graph that the amplitude of damped oscillatory motion is decaying with time. It may be pointed out here that the decay of amplitude is a direct consequence of the presence of damping term r x in the equation of motion of S.H.M. (39)  Fig. 1.11. (Oscillatory damped S.H.M.) It is further clear from equation (1) that the lightly damped system oscillates with a new frequency ` which is different from . 1/ 2 s r2  `=  −  2  m 4 m  1/ 2  2 r2  =   −  2  4 m  > and tan  .  r (79)  s When  is very low, is much greater than  m. In this case,   will come out to be negative and the force will lag behind the velocity in phase.  s  When the driving frequency  tends towards zero, tan   Z m    s  will tend towards minus infinity and  will approach -90o. When  = 0, then tan  = 0 and so  will be zero. Therefore there will be no phase lag between force and velocity. Then, the velocity resonance will occur. It is for this reason that 0 is referred to as the frequency of velocity resonance. (80) Problem 1 (a) Obtain general expression for the acceleration of damped mechanical oscillator driven by force F0 cos t. (b) Deduce an expression for the frequency of maximum acceleration. (c) Prove that if r = sm then acceleration amplitude at the frequency of velocity resonance is equal to the limit of acceleration amplitude at high frequency. Solution (a) The equation of force for forced oscillator is mx + rx + sx = F0 cos t Solving it, we get F0 x= sin (t - ) (1) Z m where Zm = mechanical impedance 2  s = r +  m −  2   s m - and tan  =  r Differentiating (1) w.r.t. time, we get F0 velocity, v = x = cos (t - ) Zm (81)  - F0  acceleration, a = x = sin (t - ) (2) Zm (b) Amplitude of acceleration from equation (2) is given by F0 F = 0 Zm Zm  Zm Amplitude will be maximum if is minimum  d  Zm  and so,   should be zero. d    1/ 2 d 1 2  s  2 i.e. r +  m −   =0 d       1/ 2 d  r2 s 2 2s m   +m + 4 − 2  =0 2 or d  2    or 2  1 2 −2 r  + m 2 + s 2−4 − 2sm−2 1/ 2  (− 2r  2 −3 ) − 4s 2 −5 + 4sm−3 = 0 This will be possible if, 2r 2 4s 2 4sm − + 3 =0 3 5  4s 2 or − 2r 2 − + 4sm = 0 2 4s 2 or = 4sm − 2r 2  2 (82)  4s 2 s2 s or gw 2 = = = 4sm − 2r 2 r2 r2 sm − m− 2 2s s angular frequency at which = = r2 amplitude of acc. will be max. m− 2s s (c) We know that frequency of velocity resonance, 0 = m 2  s At this value of , Z m = r +  m −  = r 2    Amplitude of acceleration at velocity resonance s F0 F0  m = F0 s = = Zm r r m When r = sm, then amplitude of acceleration F0 s F0 = = (1) sm m m And at very high frequency i.e. when →, 2  s Z m = r +  m −  → m 2    s    m → 0 and r 2 can be neglected as compared to    Amplitude of acceleration at high frequency (83)  F0  F0  F0 = = = (2) Z m m m Hence (1) = (2) Problem 2 Show that in a forced oscillator, (a) displacement amplitude at low frequency (→0),  s  (b) velocity amplitude at velocity resonance  = 0 =    m  (c) acceleration amplitude at high frequency, are all frequency- independent. Solution (a) In the case of forced oscillator, displacement is given by F0 x= sin (t - ) (1) Z m F0 where amplitude of displacement = Z m 2 2 s s When →0, then Z m = r  m −  →    2 2 s In Z m = r  m −  ,   when →0 then, m = 0 and r2 is neglected as compared to .  Displacement amplitude at low frequency (84)  F0 F = = 0 which is independent of . s s   (b) From equation (1) we can get F0 velocity, v = x = cos (t - ) (2) Zm F0  Velocity amplitude = Zm 2 2s Now, Z m = r  m −    s  At  = , m 2 s2 m Z m = r  m−s  =r   m s  s F  At  = , velocity amplitude = 0 m Zm F0 = which is independent of . r (c) From equation (2), we can get − F0 acceleration, x = sin (t - ) Zm 2 s s2  Acceleration amplitude = Z m = r  m −  →    (85)  2 2 s Now, Z m = r  m −     when →, then Zm = r 2 + 2 m 2 →  m [r2 neglected as compared to ] F0  Acceleration amplitude at high frequency = Zm F0 F0 = which is independent of . m m (86)  Chapter 3 Coupled Oscillations 3.1. Stiffness-coupled Pendulums Consider two identical pendulums, each having mass m and suspended from light rigid rods, each of length l. The masses of the pendulums are connected by a light spring in such a manner that the distance between the centres of masses is exactly equal to the natural length of the connecting spring when the system is in the undisturbed position. Let the system of two coupled pendulums be made to oscillate such that the oscillations have small amplitude (Fig. 3.1). Fig. 3.1. We shall have two different equations of motion for such a system. Let s be the stiffness of the connecting spring. If x and y represent the displacement of the two masses, then the coupling term will be ( s x−y ) When x is greater than y , the spring will suffer an extension in its natural length. In such a situation, the spring will favour mass at the same time. It is for this reason that the coupling term will be negative in one equation of motion and positive in the other. The two equations of motion, one for each pendulum, will be m x = −mg x l ( −s x − y ) (1) (87) and m y = −mg y l +s x −y( ) (2) Dividing equations (1) and (2) by m, we get x = −g x s − x−y l m ( ) and y = −g y s − x−y l m ( ) g Putting = 02 and rearranging the above equations, we get l x + o2 x = − s m x−y ( ) (3) and y + o2 y = − s m y−x ( ) (4) Here 0 is the frequency with which each pendulum will vibrate when free to do so. In other words, it is the natural frequency of each pendulum. Adding equations (3) and (4), we get ( + x + y + o2 x + u = 0 ) (5) Subtracting equation (4) from(3), we get x - y + o2 ( x - y) = − 2s m x−y ( ) (6) It will be quite difficult to solve these equations directly for x and y. However, the problem will be considerably solved by introducing two normal co-ordinates X and Y so that, (88)  X=x+y (a) and Y=x−y (b) Differentiating these two equations twice w.r.t. ‘t’, we get x = x + y and y = x − y Rewriting equations (5) and (6), we get  + 02 X = 0 X (7)  + 02 Y = − 2s Y Y m or  +  02 + 2s Y = 0 Y (8)  m The motion of the coupled system is described completely by equations (7) and (8). Special Cases Case I. If Y =0, then from equation (b), we get x=y In this case, the motion of the system will be completely described by equation (7). Moreover, the coupling term will be rendered completely meaningless. Fig. 3.2. (89)  Both the pendulums will be oscillating in phase (3.2). The distances between the centres of masses will be equal to the natural length of the light spring. In fact, the system will behave as if there is no spring in it. Case II. If X =0, then from equation (a), we get x=−y In this case, the motion of the system will be completely described by equation (8). The coupling term will be effective. The pendulums will always be out of phase (Fig. 3.3). The connecting spring will not retain its natural length. It will either be extended or compressed. (90)  Chapter 4 Wave Motion 4.1. Types of Waves A wave is the transmission of disturbance from point to point in a medium. It is one of the most important means for the transference of energy. A wave-motion is a form of disturbance which travels in a medium and is due to the repeated periodic motion of the particles of the medium about their mean positions, the disturbance being handed over from one particle to the next particle of the medium. It may be noted here that there is no bodily transfer of the medium through which wave propagates. It is only the state of motion which travels forward. Wave-motion can be demonstrated by moving the loose end A of a string quickly up and down keeping the other end B of the string fixed. Transverse waves are seen to travel from left to right in the form of crests C, C and troughs T, T. (A crest is plane of maximum +ve amplitude and trough is a plane of maximum –ve amplitude). Transverse waves are formed when the particles of the medium get displaced in a direction perpendicular to the direction of wave propagation. If the string were infinitely long, then the waves formed would be called progressive waves. These waves travel in an unbounded medium and are free from possible reflections. In a progressive wave, the disturbance is continuously transmitted along the same direction. In a (91) progressive wave-motion, there is always a transmission of energy along the direction of propagation of the wave. If, however, the string is limited in extent by fixing it at both ends A and B, then the waves travelling along the string would be reflected at both ends and on superimposition give rise to standing or stationary waves. In a stationary wave, the energy is not transferred from one place to another. If we consider a plane perpendicular to the direction in which the wave is propagated, then all oscillators lying in that plane of the medium have the same phase. A plane of common phase progressing through the medium can be regarded as a plane wave. A wave front is defined as a surface at every point of which particles of the medium are in the same phase of vibration. The waves in which the surfaces of the same phase are spheres are called spherical waves. These waves are formed when a stone is dropped on the surface of water in a pond or when an explosion takes place. After travelling some distance, spherical waves become plane waves. A small section of a large spherical wave may also be regarded as a plane wave. 4.2. Derivation of Wave Equation Consider a uniform string of linear density  [linear density is mass per unit length]. A string is a cord whose length is large compared to its diameter and which is perfectly flexible and uniform. Suppose a transverse wave propagates in the string. The string will vibrate at right angles to its length. A constant large tension T will act throughout the length of the string. We will ignore the effect of gravity on the string. (92)  Consider a short, curved element of length ds of a very short section of the string executing vertical simple harmonic motions (Fig. 4.2). Fig. 4.2. 1/ 2   y  2  Then ds = 1 +    dx [From partial differentiation]   x   Since we are considering a very short section of the string and also y very small oscillations, therefore is very small and so x ds  dx. Since the string is perfectly flexible therefore the same tension will act at each point of the string. The tension will act along the tangent at every point. At the upper end of the element of the string, the tension T acts at an acute angle (+d) with the horizontal. It may be noted here that angle  itself is very small. Resolving this force into two components, we get (i) T cos (+d) as the horizontal component and (ii) T sin (+d) as the vertical component. At the lower end of the string, tension T acts at an acute angle  with the horizontal. Resolving this force also into two components, we get (93)  (i) T cos  as the horizontal component and (ii) T sin  as the vertical component. We will ignore the effect of horizontal components T cos  and T cos (+d), because we are considering vertical oscillations. The resultant force in the vertically upward direction is T sin (+d)- T sin . It may be noted here that T sin (+d) is greater than T sin  because it involves sine of a greater angle. Mass of the string element =  dx According to newton’s second law of motion, Force = Mass x Acceleration 2y  T sin (+d) - T sin  = dx x 2 (1) t 2y Since is the double rate of change of vertical displacement t 2 with time, so it is the acceleration of the string in y-direction. Now, let us try to express the net force [T sin (+d)-T sin ] in terms of partial differentials. Since  is a small angle, therefore sine of the angles can be equated with the tangents of the angles. y Since tangent of an angle gives gradient , x  y   y   2y  T[sin(  + d) - sin ] = T   −    = T 2 dx  x  z + dz  x  x  x (94)  [From partial differentiation] Rewriting equation (1), we get 2 y 2 y T 2 dx = dx x 2 x t 2 y 2 y or T 2 dx =  2 x t 2 y  2 y or = (A) x 2 T t 2  Put = c −2 T where c is the velocity of wave propagation. 2 y 1 2 y  = x 2 c 2 t 2 This is a one-dimensional differential equation of wave motion which relates the displacement y of the small section of the string acting as a simple oscillator to distance x and time t. The important feature of this wave equation is that the coefficient of the derivative on the right- hand side represents the reciprocal of the square of wave velocity. So, we need not solve this equation to obtain the wave velocity. Note The common feature of all waves is the transmission of disturbance with a velocity depending on the medium through which wave is propagating. In the case of an electromagnetic wave, y corresponds to E (intensity of electric field) or to H (intensity of magnetic field0. In the (95) case of a sound wave, y corresponds to pressure at a point. In general, the disturbance is represented by wave function  of the wave. Therefore, the general form of wave equation is 2  1 2  = x 2 c 2 t 2 In order to determine the form of a wave and its wavelength etc., the wave function  is determined for that particular situation. Velocity of Transverse Waves Equation (A) is the differential equation of a vibrating string. If m is the mass per unit length of the string, then equation (A) maybe rewritten as 2 y m 2 y = x 2 T t 2 Comparing the above equation with the differential wave equation 2 y 1 2 y = , we get x 2 c 2 t 2 1 m = c2 T T or c= m This is the required equation for the velocity of transverse waves moving along a string. This velocity depends on (i) applied tension T(ii) mass m per unit length. (96)  Again, if ‘n’ is the frequency and ‘’ is the wavelength, then T 1 T 1 T c = n = or n = = m  m 2 m where l is length of the string producing fundamental note. Solution of Wave Equation The wave equation is 2 y 1 2 y = x 2 c 2 t 2 It is clear from the structure of the wave equation that its solution should be a function of two variables x and t. Let us first try the solution y = f1 (ct – x) y Now, = −f `1 (ct − x ) (1) x where f`1(ct-x) is the differentiation of f1(ct-x) w.r.t. x. 2y Again, = −f `1 (ct − x ) (2) x 2 where –f``1(ct-x) is the differentiation of f`1(ct-x) w.r.t. x y Similarly, = cf `1 (ct − x ) (3) x 2y and = c12f1`` (ct − x ) (4) x 2 From equation (2) and (4), we get (97)  2 y 1 2 y = x 2 c 2 t 2 Hence y = f1(ct-x) is a solution of the wave equation. Proceeding in the same manner, we can show that y = f 2(ct+x) is also a solution of the wave equation. Relation between Particle velocity and Wave velocity y From equation (1), f``1 (ct-x) =- x Putting this value of f`1(ct-x) in equation (3), we get y y = −c t x y Here is the particle velocity at a time t and at a point distant x t y from the origin. is the slope of the y-x curve at the same instant and at x the same point.  Particle velocity – Wave velocity x slope of y-x curve. The above relation is for those waves which travel along the positive direction of X-axis. 4.3. Characteristic Impedance of a String When a transverse wave propagates in a string, the string offers certain opposition to the propagation of wave. This opposition is known as impedance. It is known as transverse impedance in the case of propagation of transverse waves in a string. Transverse impedance Z is defined as the ratio of transverse force F to transverse velocity v. (98)  F Thus, Z= v Consider a uniform string and at its one end apply an alternating force F0 eit where F0 is the amplitude of the force and  is the angular frequency. When the direction of this applied force is perpendicular to the length of the string, a transverse wave will begin to propagate in the string. A constant tension T will act throughout the length of the string. Let us study the various forces which act at one end of the string. These forces are shown in figure 4.3. Fig. 4.3. Tension T is resolved into two components (i) horizontal component T cos and (ii) vertical component T sin. We will concern ourselves only with the vertical component T sin . Now, F0 eit = -T sin  (1) Negative sign is introduced here because the applied force and T sin  are equal and opposite at all times. If  is small, sin  can be replaced by tan . But tan  is equal to y the gradient. x  Rewriting equation (1), we get  y  F0 e it = −T   x  (99)  The expression for displacement of a wave may be written as y = A ei(t-kx) (2) Here amplitude A maybe simple or complex. It consists of many components. Some of the components will be in phase with the driving force while others will be out of phase. Differentiating partially both sides of equation (2) w.r.t. x, we get y = −ikA ei( t -kx) x At the end where the alternating force is applied, x=0. Therefore y gradient at that end will be given by x y = −ik Ae it x  y  it   = −ik Ae  x  x =0  F0 eit = −T(−ik Ae it ) or F0 e it = −ik TAe it or F0 = ikTA F0  A= ikT  But k= c F0  c   A=   iT    (100)  F0  c  i ( t −kx ) or y = Ae i ( t −kx ) =  e it    F0  c  i ( t −kx ) or y=  e i  T  Differentiating both sides w.r.t ‘t’, we get y = y = v t F0  c  i( - kx) =  (i) e i  T  c  v = F0  ei ( t − kx ) T This is the expression for transverse velocity. Maximum value of velocity i.e. velocity amplitude c = F0   T Maximum value of force = F0 Force  Characteristic impedance Z = Velocity F0 or Specific impedance = c F0   T T or Z= c But T = c2 (101)  c 2  Z= c = c It may be noted that impedance will depend on two factors- elasticity and inertia. 4.4. Reflection and Transmission of Waves on a String at a Boundary Consider two sections of a string joined smoothly at a point (Fig. 4.4). By smooth joining, we mean that the same tension T will act along both the sections. Le the linear densities of the two sections by 1 and 2. Since T is same and linear densities are different, therefore the wave velocities in the two sections will be different. Let c 1 and c2 be the wave velocities in the two sections. The characteristic impedances of the two sections will be 1c1 and 2c2 respectively. i.e. Z1 = 1 c1 and Z2 = 2 c 2 A wave travelling in one section of the string will encounter a different impedance when it reaches the second section. A part of the wave will be reflected, and a part of the wave will be transmitted. Let yi , y r and y t denote the displacements of the incident wave, reflected wave and transmitted wave respectively. The magnitudes of these displacements will be given by the equations, (102)  y i = A1e i ( t − k1x ) , y r = B1ei ( t − k1x ) and y t = A 2 e i ( t − k 2 x ) Fig. 4.4. Here, A1, B1 and A2 represent the amplitudes of the incident wave, reflected wave and transmitted wave respectively. In the expressions for yi and yr, it may be noted that the signs of k1x (in the power of e) are different. We will apply two boundary conditions in the present problem. There is a discontinuity in impedance at the boundary x = 0. But there is no discontinuity in displacement. At all times, during propagation of the wave, the displacements to the right and left of the boundary will be the same. Applying this boundary condition, we see that the sum of the magnitudes of the displacements of incident and reflected waves must be equal to the displacement of the transmitted wae.  yi + yr = yt or A1e i ( t − k1x ) + B1e i ( t + kz ) = A 2 ei ( t − k 2 z ) At the boundary, x = 0.  A1ei ( t −ok1 ) + B1ei ( t + ok1 ) = A 2 e i ( t −o.k 2 ) or A1e it + B1eit = A 2 eit (103) or e it (A1 + B1 ) = A 2 eit A1 + B1 = A2 (1) y The second boundary condition demands that the gradient is x continuous at the boundary. This would mean that even the transverse force will be continuous, because, as already proved, the magnitude of  y  transverse force is T  . It will be interesting to discuss a situation  x  where this boundary condition is not satisfied. Let us assume that there is a difference in the magnitude of forces immediately to the right and left of boundary x = 0. This net force will be acting on an infinitesimally small mass of the string. According to the kinematic relation, F F = ma or a = , the net force will produce an infinitely great m acceleration in that section of the string. This is absurd. Thus, it is of paramount importance that the second boundary condition, viz., continuity of transverse force, be satisfied. applying this boundary condition, we get  T (y i + y r ) = T  y t x x Here, T is not to be cancelled. As we proceed further in the problem, it will prove useful. y i y y Now T +T r =T t x x x Putting the respective values, we get (104)  T  x  A1ei (t −k1x ) + T  x  B1ei ( t + k1x )  =T  x  A1ei ( t − k 2 x )    or T A1ei ( t −k1x ) − i(t − k1x ) + TB1 e(it + k1x ) x x   = T A1ei ( t −k 2 x ) + TB1 i(t − k 2 x ) x x  i ( t −k1x )  or − ik 1T A1 e + ik 1TB1 e(it + k1x ) x x = − kiT A 2 ei ( t − k 2 x ) But at the boundary, x = 0  it  − ik 1T A1 e = −ik 2TA 2eit x This equation is valid for all values of t. Cancelling out exponential eit and operator i , we get -k1TA1 + k1 TB1 = -k2 TA2    But, k= , k1 = , k 2 = c c1 c2 T T T  − A1 +  B1 = − A 2 c1 c1 c2 But it can be proved that the magnitude of specific impedance is given by T Z = c = c (105)  T T  = Z1 and = Z2 c1 c2 So − Z1A1 + Z1B1 = −Z 2 A 2 or Z1A1 − Z1B1 ) = Z 2 A 2 or Z1(A1-B1)=Z2A2 In vector notation, Z1 (A1 − B1 ) = Z2 A 2 (2) Let us now find an expression for the reflection co-efficient of amplitude. Reflection co-efficient of amplitude is measured by the ratio of the amplitude of reflected wave to the amplitude of incident wave. B1  Reflection co-efficient of amplitude = A1 B1 The ration can be found from equations (1) and (2) by A1 eliminating A2. Putting the value of A2 from equation (1) in equation (2), we get Z1 (A1 − B1 ) = Z2 (A1 + B1 ) or ( ) A1 Z1 − Z2 = B1 (Z1 + Z2 ) B1 Z1 − Z2  = A1 Z1 + Z2 = Reflection co-efficient of amplitude. (106)  Let us now calculate the transmission co-efficient of amplitude. It is measured by the ratio of the amplitude of transmitted wave to the amplitude of incident wave. A2  Transmission co-efficient of amplitude = A1 A2 The ratio can be found from equations (1) and (2) by A1 eliminating B1. From equation (1), B1 = A2-A1 Putting this value of B1 in equation (2), we get Z1A1 − (A 2 − A1 ) = Z2 A 2 or Z1 (2A1 − A2 ) = Z2A 2 or ( ) 2A1 Z1 = Z1 + Z2 A 2 A2 2Z1  = A1 Z1 + Z 2 = Transmission co-efficient of amplitude. Following points are worth mentioning about these two co-efficient of amplitude. (a) These two co-efficients are independent of the angular frequency  of the wave. Hence they are valid for waves of all frequencies. (107)  (b) These are valid for all types of waves, namely transverse waves, acoustic waves, voltage waves, current waves, and electromagnetic waves. (c) These co-efficients are real. Thus, they are free from all phase changes except the phase change of  radians on which the sign of a term depends. (d) These co-efficients completely depend on the ratios of the impedances. Special Cases Case (a). When Z2 = 0. B1 Z1 − 0 = =1 A1 Z1 + 0  B1 = A1 i.e., amplitude of reflected wave is equal to the amplitude of incident wave. A2 2Z1 Also, = =2 A1 Z1 + 0 A2 = 2A1 i.e., amplitude of transmitted wave is equal to twice the amplitude of incident wave. Case (b). When Z2 = . (108)  B1 Z1 − Z 2 = A1 Z1 + Z 2 Dividing the numerator and denominator of R.H.S. by Z2 , we get Z1 Z1 −1 −1 B2 Z2  = = A1 Z1 Z1