13.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A. v = ω A (13.8) The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the Fig. 13.12 The acceleration, a(t), of the particle P′ isgeometry of Fig. 13.11, it is clear that the velocity the projection of the acceleration a of theof the projection particle P′ at time t is reference particle P. v(t) = –ωA sin (ωt + φ ) (13.9) Eq. (13.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity v(t) given by Eq. (13.9) with respect to time: d a (t ) = v (t ) (13.12) d t We note from Eq. (13.11) the important Fig. 13.11 The velocity, v (t), of the particle P′ is property that acceleration of a particle in SHM the projection of the velocity v of the is proportional to displacement. For x(t) > 0, reference particle, P. a(t) < 0 and for x(t) < 0, a(t) > 0. Thus, whatever Reprint 2025-26 OSCILLATIONS 267 the value of x between –A and A, the acceleration (b) Using Eq. (13.9), the speed of the body a(t) is always directed towards the centre. = – (5.0 m)(2π s–1) sin [(2π s–1) ×1.5 s For simplicity, let us put φ = 0 and write the + π/4] expression for x (t), v (t) and a(t) = – (5.0 m)(2π s–1) sin [(3π + π/4)] x(t) = A cos ωt, v(t) = – ω Asin ωt, a(t)=–ω2 A cos ωt = 10π × 0.707 m s–1 The corresponding plots are shown in Fig. 13.13. = 22 m s–1 All quantities vary sinusoidally with time; only (c) Using Eq.(13.10), the acceleration of the their maxima differ and the different plots differ body in phase. x varies between –A to A; v(t) varies = –(2π s–1)2 × displacement from –ωA to ωA and a(t) from –ω2A to ω2A. With = – (2π s–1)2 × (–3.535 m) respect to displacement plot, velocity plot has a = 140 m s–2 ⊳ phase difference of π/2 and acceleration plot has a phase difference of π. 13.6 FORCE LAW FOR SIMPLE HARMONIC MOTION Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 13.11), the force acting on a particle of mass m in SHM is F (t ) = ma = –mω2 x (t) i.e., F (t) = –k x (t) (13.13) where k = mω2 (13.14a) k or ω = (13.14b) m Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (13.4) for displacement or by Eq. (13.13) that gives Fig. 13.13 Displacement, velocity and acceleration of its force law. Going from Eq. (13.4) to Eq. (13.13) a particle in simple harmonic motion have required us to differentiate two times. Likewise, the same period T, but they differ in phase by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4). u Example 13.5 A body oscillates with SHM Note that the force in Eq. (13.13) is linearly according to the equation (in SI units), proportional to x(t). A particle oscillating under x = 5 cos [2π t + π/4]. such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force At t = 1.5 s, calculate the (a) displacement, may contain small additional terms proportional (b) speed and (c) acceleration of the body. to x2, x3, etc. These then are called non-linear oscillators. Answer The angular frequency ω of the body = 2π s–1 and its time period T = 1 s. u Example 13.6 Two identical springs of At t = 1.5 s spring constant k are attached to a block (a) displacement = (5.0 m) cos [(2π s–1)× of mass m and to fixed supports as shown 1.5 s + π/4] in Fig. 13.14. Show that when the mass is = (5.0 m) cos [(3π + π/4)] displaced from its equilibrium position on = –5.0 × 0.707 m either side, it executes a simple harmonic = –3.535 m motion. Find the period of oscillations. Reprint 2025-26 268 PHYSICS 13.7 ENERGY IN SIMPLE HARMONIC MOTION Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values. In section 13.5 we have seen that the velocity of a particle executing SHM, is a periodic Fig. 13.14 function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) Answer Let the mass be displaced by a small of such a particle, which is defined as distance x to the right side of the equilibrium position, as shown in Fig. 13.15. Under this 1 2 K = mv situation the spring on the left side gets 2 1 2 2 2 = m ω A sin (ωt + φ) 2 1 2 2 = k A sin (ωt + φ) (13.15) 2 is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean Fig. 13.15 position. Note, since the sign of v is immaterial in K, the period of K is T/2. elongated by a length equal to x and that on What is the potential energy (U) of a particle the right side gets compressed by the same executing simple harmonic motion? In length. The forces acting on the mass are Chapter 6, we have seen that the concept of then, potential energy is possible only for conservative forces. The spring force F = –kx is a conservative F1 = –k x (force exerted by the spring on force, with associated potential energy the left side, trying to pull the mass towards the mean 1 2 U = k x position) (13.16) 2 F2 = –k x (force exerted by the spring on Hence the potential energy of a particle the right side, trying to push the executing simple harmonic motion is, mass towards the mean position) 1 2 The net force, F, acting on the mass is then U(x) = k x given by, 2 F = –2kx 1 2 2 Hence the force acting on the mass is = k A cos (ωt + φ) (13.17) 2proportional to the displacement and is directed towards the mean position; therefore, the motion Thus, the potential energy of a particle executed by the mass is simple harmonic. The executing simple harmonic motion is also time period of oscillations is, periodic, with period T/2, being zero at the mean m position and maximum at the extreme T = 2 π displacements. 2k ⊳ Reprint 2025-26 OSCILLATIONS 269 It follows from Eqs. (13.15) and (13.17) that Observe that both kinetic energy and the total energy, E, of the system is, potential energy in SHM are seen to be always positive in Fig. 13.16. Kinetic energy can, of E = U + K course, be never negative, since it is proportional to the square of speed. Potential 1 1 = k A 2 cos 2 (ωt + φ) + k A 2 sin 2 (ωt + φ) energy is positive by choice of the undermined 2 2 constant in potential energy. Both kinetic energy and potential energy peak twice during each period of SHM. For x = 0, the energy is 1 2 2 2 = k A cos (ωt + φ) + sin (ωt + φ)  kinetic; at the extremes x = ±A, it is all   2 potential energy. In the course of motion Using the familiar trigonometric identity, the between these limits, kinetic energy increases value of the expression in the brackets is unity. at the expense of potential energy or Thus, vice-versa. 1 2 E = k A (13.18) u Example 13.7 A block whose mass is 1 kg 2 is fastened to a spring. The spring has a The total mechanical energy of a harmonic spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from itsoscillator is thus independent of time as equilibrium position at x = 0 on a frictionlessexpected for motion under any conservative surface from rest at t = 0. Calculate the force. The time and displacement dependence kinetic, potential and total energies of the of the potential and kinetic energies of a block when it is 5 cm away from the mean linear simple harmonic oscillator are shown position. in Fig. 13.16. Answer The block executes SHM, its angular frequency, as given by Eq. (13.14b), is k ω = m 1 50 N m– = 1kg = 7.07 rad s–1 Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Fig. 13.16 Kinetic energy, potential energy and total energy as a function of time [shown in (a)] Therefore, when the particle is 5 cm away from and displacement [shown in (b)] of a particle in SHM. The kinetic energy and potential the mean position, we have energy both repeat after a period T/2. The total energy remains constant at all t or x. 0.05 = 0.1 cos (7.07t) Reprint 2025-26 270 PHYSICS let it go. The stone executes a to and fro motion,Or cos (7.07t) = 0.5 and hence it is periodic with a period of about two seconds. 3 We shall show that this periodic motion is sin (7.07t) = = 0.866 2 simple harmonic for small displacements from Then, the velocity of the block at x = 5 cm is = 0.1 × 7.07 × 0.866 m s–1 = 0.61 m s–1 Hence the K.E. of the block, 1 2 = m v 2 = ½[1kg × (0.6123 m s–1 )2 ] = 0.19 J (a) The P.E. of the block, 1 2 = k x 2 = ½(50 N m–1 × 0.05 m × 0.05 m) = 0.0625 J The total energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J we also know that at maximum displacement, K.E. is zero and hence the total energy of the (b) system is equal to the P.E. Therefore, the total Fig. 13.17 (a) A bob oscillating about its mean position. (b) The radial force T-mg cosθ energy of the system, provides centripetal force but no torque = ½(50 N m–1 × 0.1 m × 0.1 m ) about the support. The tangential force mg sinθ provides the restoring torque. = 0.25 J which is same as the sum of the two energies at the mean position. Consider simple pendulum a displacement of 5 cm. This is in conformity — a small bob of mass m tied to an inextensible with the principle of conservation of energy. ⊳ massless string of length L. The other end of the string is fixed to a rigid support. The bob13.8 The Simple Pendulum oscillates in a plane about the vertical lineIt is said that Galileo measured the periods of a through the support. Fig. 13.17(a) shows thisswinging chandelier in a church by his pulse system. Fig. 13.17(b) is a kind of ‘free-body’beats. He observed that the motion of the chandelier was periodic. The system is a kind diagram of the simple pendulum showing the of pendulum. You can also make your own forces acting on the bob. pendulum by tying a piece of stone to a long Let θ be the angle made by the string with unstretchable thread, approximately 100 cm the vertical. When the bob is at the mean long. Suspend your pendulum from a suitable position, θ = 0 support so that it is free to oscillate. Displace There are only two forces acting on the bob; the stone to one side by a small distance and the tension T along the string and the vertical Reprint 2025-26 OSCILLATIONS 271 force due to gravity (=mg). The force mg can be Table 13.1 sin θ as ma function of angle θ resolved into the component mg cosθ along the string and mg sinθ perpendicular to it. Since (degrees) (radians) sin the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration (ω2L) and also a tangental acceleration; the latter arises since motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force T –mg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to Equation (13.24) is mathematically, identical towork with torque about the support since the radial force gives zero torque. Torque τ about Eq. (13.11) except that the variable is angular displacement. Hence we have proved that forthe support is entirely provided by the tangental small q, the motion of the bob is simple harmonic.component of force From Eqs. (13.24) and (13.11), τ = –L (mg sinθ ) (13.19) This is the restoring torque that tends to reduce mgL ω =angular displacement — hence the negative Isign. By Newton’s law of rotational motion, τ = I α (13.20) and where I is the moment of inertia of the system about the support and α is the angular I T = 2π (13.25)acceleration. Thus, mgL I α = –m g sin θ L (13.21) Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL2. Eq. (13.25) then gives the well-known Or, formula for time period of a simple pendulum. m g L α = − sin θ (13.22) L I T = 2π (13.26) g We can simplify Eq. (13.22) if we assume that the displacement θ is small. We know that sin θ u Example 13.8 What is the length of a can be expressed as, simple pendulum, which ticks seconds ? θ3 θ5 Answer From Eq. (13.26), the time period of a sin θ = θ− + ± ... (13.23) 3! 5! simple pendulum is given by, L where θ is in radians. T = 2π Now if θ is small, sin θ can be approximated g by θ and Eq. (13.22) can then be written as, From this relation one gets, gT 2 mgL L = 2 θ α = − (13.24) 4π I The time period of a simple pendulum, which In Table 13.1, we have listed the angle θ in ticks seconds, is 2 s. Therefore, for g = 9.8 m s–2 degrees, its equivalent in radians, and the value of the function sin θ . From this table it and T = 2 s, L is can be seen that for θ as large as 20 degrees, 9.8(m s –2 ) × 4(s 2 ) = sin θ is nearly the same as θ expressed 2 4π in radians. = 1 m ⊳ Reprint 2025-26 272 PHYSICS SUMMARY 1. The motion that repeats itself is called periodlic motion. 2. The period T is the time reequired for one complete oscillation, or cycle. It is related to the frequency v by, 1 T = v The frequency ν of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz : 1 hertz = 1 Hz = 1 oscillation per second = 1s–1 3. In simple harmonic motion (SHM), the displacement x (t) of a particle from its equilibrium position is given by, x (t) = A cos (ωt + φ ) (displacement), in which A is the amplitude of the displacement, the quantity (ωt + φ ) is the phase of the motion, and φ is the phase constant. The angular frequency ω is related to the period and frequency of the motion by, 2π ω= = 2πν (angular frequency). T 4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs. 5. The particle velocity and acceleration during SHM as functions of time are given by, v (t) = –ωA sin (ωt + φ ) (velocity), a (t) = –ω2A cos (ωt + φ ) = –ω2x (t) (acceleration), Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude vm=ω A and acceleration amplitude am =ω 2A, respectively. 6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion. 7. A particle executing simple harmonic motion has, at any time, kinetic energy K = ½ mv2 and potential energy U = ½ kx2. If no friction is present the mechanical energy of the system, E = K + U always remains constant even though K and U change with time. 8. A particle of mass m oscillating under the influence of Hooke’s law restoring force given by F = – k x exhibits simple harmonic motion with k ω = (angular frequency) m m T = 2π (period) k Such a system is also called a linear oscillator. 9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by, L T = 2π g Reprint 2025-26 OSCILLATIONS 273 POINTS TO PONDER 1. The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer. 2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = – k x is simple harmonic. 3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: –mω2r. In the latter case, the phases of motion, in two perpendicular directions (x and y) must differ by π/2. Thus, for example, a particle subject to a force –mω2r with initial position (0, A) and velocity (ωA, 0) will move uniformly in a circle of radius A. 4. For linear simple harmonic motion with a given ω, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase. 5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity. 6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes. 7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law). 8. The motion of a simple pendulum is simple harmonic for small angular displacement. 9. For motion of a particle to be simple harmonic, its displacement x must be expressible in either of the following forms : x = A cos ωt + B sin ωt x = A cos (ωt + α ), x = B sin (ωt + β ) The three forms are completely equivalent (any one can be expressed in terms of any other two forms). Thus, damped simple harmonic motion is not strictly simple harmonic. It is approximately so only for time intervals much less than 2m/b where b is the damping constant. Reprint 2025-26 274 PHYSICS Exercises 13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position. 13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? Fig. 18.18 Reprint 2025-26 OSCILLATIONS 275 13.4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt – cos ωt (b) sin3 ωt (c) 3 cos (π/4 – 2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (–ω2t2) (f) 1 + ωt + ω2t2 13.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. 13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = –200x2 (c) a = –10x (d) a = 100x3 13.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions. 13.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ? 13.9 A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Fig. 13.19 Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Reprint 2025-26 276 PHYSICS 13.10 In Exercise 13.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? 13.11 Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 13.20 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case. 13.12 Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s). (a) x = –2 sin (3t + π/3) (b) x = cos (π/6 – t) (c) x = 3 sin (2πt + π/4) (d) x = 2 cos πt 13.13 Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. Fig. 13.21 (a) What is the maximum extension of the spring in the two cases ? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ? Reprint 2025-26 OSCILLATIONS 277 13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ? 13.15 The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2) 13.16 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ? 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period hρ T = 2π ρ1 g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid). 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion. Reprint 2025-26 CHAPTER FOURTEEN WAVES 14.1 INTRODUCTION In the previous Chapter, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects? A material medium provides

14.6 Reflection of waves some cork pieces on the disturbed surface, it is seen that

14.7 Beats the cork pieces move up and down but do not move away Summary from the centre of disturbance. This shows that the water Points to ponder mass does not flow outward with the circles, but rather a Exercises moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this Chapter, we will study such waves. Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of sig- nals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For exam- ple, sound waves may be first converted into an electric cur- rent signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a Reprint 2025-26 WAVES 279 satellite. Detection of the original signal will usu- We shall illustrate this connection through ally involve these steps in reverse order. simple examples. Not all waves require a medium for their Consider a collection of springs connected to propagation. We know that light waves can one another as shown in Fig. 14.1. If the spring travel through vacuum. The light emitted by at one end is pulled suddenly and released, the stars, which are hundreds of light years away, disturbance travels to the other end. What has reaches us through inter-stellar space, which is practically a vacuum. The most familiar type of waves such as waves on a string, water waves, sound waves, seismic waves, etc. is the so-called mechanical waves. Fig. 14.1 A collection of springs connected to each These waves require a medium for propagation, other. The end A is pulled suddenly they cannot propagate through vacuum. They generating a disturbance, which then involve oscillations of constituent particles and propagates to the other end. depend on the elastic properties of the medium. The electromagnetic waves that you will learn happened? The first spring is disturbed from its in Class XII are a different type of wave. equilibrium length. Since the second spring is Electromagnetic waves do not necessarily require connected to the first, it is also stretched or a medium - they can travel through vacuum. compressed, and so on. The disturbance moves Light, radiowaves, X-rays, are all electromagnetic from one end to the other; but each spring only waves. In vacuum, all electromagnetic waves executes small oscillations about its equilibrium have the same speed c, whose value is : position. As a practical example of this situation, consider a stationary train at a railway station. c = 299, 792, 458 ms–1. (14.1) Different bogies of the train are coupled to each A third kind of wave is the so-called Matter other through a spring coupling. When an waves. They are associated with constituents of engine is attached at one end, it gives a push to matter : electrons, protons, neutrons, atoms and the bogie next to it; this push is transmitted from molecules. They arise in quantum mechanical one bogie to another without the entire train description of nature that you will learn in your being bodily displaced. later studies. Though conceptually more abstract Now let us consider the propagation of sound than mechanical or electro-magnetic waves, they waves in air. As the wave passes through air, it have already found applications in several compresses or expands a small region of air. This devices basic to modern technology; matter causes a change in the density of that region, waves associated with electrons are employed say δρ, this change induces a change in pressure, in electron microscopes. δp, in that region. Pressure is force per unit area, In this chapter we will study mechanical so there is a restoring force proportional to waves, which require a material medium for the disturbance, just like in a spring. In this their propagation. case, the quantity similar to extension or The aesthetic influence of waves on art and compression of the spring is the change in literature is seen from very early times; yet the density. If a region is compressed, the molecules first scientific analysis of wave motion dates back in that region are packed together, and they tend to the seventeenth century. Some of the famous to move out to the adjoining region, thereby scientists associated with the physics of wave increasing the density or creating compression motion are Christiaan Huygens (1629-1695), in the adjoining region. Consequently, the air Robert Hooke and Isaac Newton. The in the first region undergoes rarefaction. If a understanding of physics of waves followed the region is comparatively rarefied the surrounding physics of oscillations of masses tied to springs air will rush in making the rarefaction move to and physics of the simple pendulum. Waves in the adjoining region. Thus, the compression or elastic media are intimately connected with rarefaction moves from one region to another, harmonic oscillations. (Stretched strings, coiled making the propagation of a disturbance springs, air, etc., are examples of elastic media). possible in air. Reprint 2025-26 280 PHYSICS In solids, similar arguments can be made. In a crystalline solid, atoms or group of atoms are arranged in a periodic lattice. In these, each atom or group of atoms is in equilibrium, due to forces from the surrounding atoms. Displacing one atom, keeping the others fixed, leads to restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, with springs between pairs of them. In the subsequent sections of this chapter we are going to discuss various characteristic Fig. 14.3 A harmonic (sinusoidal) wave travelling properties of waves. along a stretched string is an example of a transverse wave. An element of the string 14.2 TRANSVERSE AND LONGITUDINAL in the region of the wave oscillates about WAVES its equilibrium position perpendicular to the direction of wave propagation. We have seen that motion of mechanical waves involves oscillations of constituents of the position as the pulse or wave passes through medium. If the constituents of the medium them. The oscillations are normal to the oscillate perpendicular to the direction of wave direction of wave motion along the string, so this propagation, we call the wave a transverse wave. is an example of transverse wave. If they oscillate along the direction of wave We can look at a wave in two ways. We can fix propagation, we call the wave a longitudinal an instant of time and picture the wave in space. wave. This will give us the shape of the wave as a Fig.14.2 shows the propagation of a single whole in space at a given instant. Another way pulse along a string, resulting from a single up is to fix a location i.e. fix our attention on a and down jerk. If the string is very long compared particular element of string and see its oscillatory motion in time. Fig. 14.4 describes the situation for longitudinal waves in the most familiar example of the propagation of sound waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a Fig. 14.2 When a pulse travels along the length of a stretched string (x-direction), the elements of the string oscillate up and down (y- direction) to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig. 14.3 shows a similar situation, but this time the external agent gives a continuous periodic sinusoidal up Fig. 14.4 Longitudinal waves (sound) generated in a and down jerk to one end of the string. The pipe filled with air by moving the piston up resulting disturbance on the string is then a and down. A volume element of air oscillates sinusoidal wave. In either case the elements of in the direction parallel to the direction of the string oscillate about their equilibrium mean wave propagation. Reprint 2025-26 WAVES 281 sinusoidal wave will be generated propagating u Example 14.1 Given below are some in air along the length of the pipe. This is clearly examples of wave motion. State in each case an example of longitudinal waves. if the wave motion is transverse, longitudinal The waves considered above, transverse or or a combination of both: longitudinal, are travelling or progressive waves (a) Motion of a kink in a longitudinal spring since they travel from one part of the medium produced by displacing one end of the to another. The material medium as a whole spring sideways. does not move, as already noted. A stream, for (b) Waves produced in a cylinder example, constitutes motion of water as a whole. containing a liquid by moving its piston back and forth.In a water wave, it is the disturbance that moves, (c) Waves produced by a motorboat sailing not water as a whole. Likewise a wind (motion in water. of air as a whole) should not be confused with a (d) Ultrasonic waves in air produced by a sound wave which is a propagation of vibrating quartz crystal. disturbance (in pressure density) in air, without the motion of air medium as a whole. Answer In transverse waves, the particle motion is (a) Transverse and longitudinal normal to the direction of propagation of the (b) Longitudinal wave. Therefore, as the wave propagates, each (c) Transverse and longitudinal element of the medium undergoes a shearing (d) Longitudinal ⊳ strain. Transverse waves can, therefore, be 14.3 DISPLACEMENT RELATION INpropagated only in those media, which can A PROGRESSIVE WAVE sustain shearing stress, such as solids and not in fluids. Fluids, as well as, solids can sustain For mathematical description of a travelling compressive strain; therefore, longitudinal wave, we need a function of both position x and time t. Such a function at every instant shouldwaves can be propagated in all elastic media. give the shape of the wave at that instant. Also,For example, in medium like steel, both at every given location, it should describe thetransverse and longitudinal waves can motion of the constituent of the medium at thatpropagate, while air can sustain only location. If we wish to describe a sinusoidal longitudinal waves. The waves on the surface travelling wave (such as the one shown in Fig. of water are of two kinds: capillary waves and 14.3) the corresponding function must also be gravity waves. The former are ripples of fairly sinusoidal. For convenience, we shall take the short wavelength—not more than a few wave to be transverse so that if the position of centimetre—and the restoring force that the constituents of the medium is denoted by x, produces them is the surface tension of water. the displacement from the equilibrium position Gravity waves have wavelengths typically may be denoted by y. A sinusoidal travelling ranging from several metres to several hundred wave is then described by: meters. The restoring force that produces these y ( x , t ) = a sin( kx −ωt + φ) (14.2)waves is the pull of gravity, which tends to keep The term φ in the argument of sine functionthe water surface at its lowest level. The means equivalently that we are considering aoscillations of the particles in these waves are linear combination of sine and cosine functions:not confined to the surface only, but extend with diminishing amplitude to the very bottom. The y ( x , t ) = A sin(kx − ωt ) + B cos(kx − ωt ) (14.3) particle motion in water waves involves a From Equations (14.2) and (14.3), complicated motion—they not only move up and  B down but also back and forth. The waves in an a = A 2 + B 2 and φ= tan −1ocean are the combination of both longitudinal  A  and transverse waves. To understand why Equation (14.2) It is found that, generally, transverse and represents a sinusoidal travelling wave, take a longitudinal waves travel with different speed fixed instant, say t = t0. Then, the argument of in the same medium. the sine function in Equation (14.2) is simply Reprint 2025-26 282 PHYSICS kx + constant. Thus, the shape of the wave (at any fixed instant) as a function of x is a sine wave. Similarly, take a fixed location, say x = x0. Then, the argument of the sine function in Equation (14.2) is constant -ωt. The displacement y, at a fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as t increases, x must increase in the positive direction to keep kx – ωt + φ constant. Thus, Eq. (14.2) represents a sinusiodal (harmonic) wave travelling along the positive direction of the x-axis. On the other hand, a function y ( x , t ) = a sin( kx + ω t + φ ) (14.4) represents a wave travelling in the negative direction of x-axis. Fig. (14.5) gives the names of the various physical quantities appearing in Eq. (14.2) that we now interpret. y(x,t) : displacement as a function of position x and time t a : amplitude of a wave ω : angular frequency of the wave Fig. 14.6 A harmonic wave progressing along the k : angular wave number positive direction of x-axis at different times. kx–ωt+φ : initial phase angle (a+x = 0, t = 0) Using the plots of Fig. 14.6, we now define Fig. 14.5 The meaning of standard symbols in the various quantities of Eq. (14.2). Eq. (14.2) 14.3.1 Amplitude and Phase Fig. 14.6 shows the plots of Eq. (14.2) for In Eq. (14.2), since the sine function varies different values of time differing by equal between 1 and –1, the displacement y (x,t) varies intervals of time. In a wave, the crest is the between a and –a. We can take a to be a positive point of maximum positive displacement, the constant, without any loss of generality. Then, trough is the point of maximum negative a represents the maximum displacement of the displacement. To see how a wave travels, we constituents of the medium from their can fix attention on a crest and see how it equilibrium position. Note that the displacement progresses with time. In the figure, this is y may be positive or negative, but a is positive. shown by a cross (×) on the crest. In the same It is called the amplitude of the wave. manner, we can see the motion of a particular The quantity (kx – ωt + φ) appearing as the constituent of the medium at a fixed location, argument of the sine function in Eq. (14.2) issay at the origin of the x-axis. This is shown called the phase of the wave. Given theby a solid dot (•). The plots of Fig. 14.6 show amplitude a, the phase determines thethat with time, the solid dot (•) at the origin displacement of the wave at any position andmoves periodically, i.e., the particle at the origin oscillates about its mean position as at any instant. Clearly φ is the phase at x = 0 the wave progresses. This is true for any other and t = 0. Hence, φ is called the initial phase location also. We also see that during the time angle. By suitable choice of origin on the x-axis the solid dot (•) has completed one full and the intial time, it is possible to have φ = 0. oscillation, the crest has moved further by a Thus there is no loss of generality in dropping certain distance. φ, i.e., in taking Eq. (14.2) with φ = 0. Reprint 2025-26 WAVES 283 14.3.2 Wavelength and Angular Wave Number The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ. For simplicity, we can choose points of the same phase to be crests or troughs. The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking φ = 0 in Eq. (14.2), Fig. 14.7 An element of a string at a fixed location the displacement at t = 0 is given by oscillates in time with amplitude a and period T, as the wave passes over it. y ( x ,0) = a sin kx (14.5) Since the sine function repeats its value after Now, the period of oscillation of the wave is the every 2π change in angle, time it takes for an element to complete one full oscillation. That is − a sin ωt = −a sin ω(t + T) = − a sin(ωt + ωT) That is the displacements at points x and at Since sine function repeats after every 2π, 2nπx + 2π k ωT = 2π or ω = (14.7) T are the same, where n=1,2,3,... The 1east is called the angular frequency of the wave.distance between points with the same ω displacement (at any given instant of time) is Its SI unit is rad s –1. The frequency ν is the obtained by taking n = 1. λ is then given by number of oscillations per second. Therefore, 1 ω 2π 2π ν= = (14.8) λ= or k = (14.6) T 2π k λ ν is usually measured in hertz. k is the angular wave number or propagation In the discussion above, reference has always constant; its SI unit is radian per metre or been made to a wave travelling along a string or rad m−*1 a transverse wave. In a longitudinal wave, the displacement of an element of the medium is 14.3.3 Period, Angular Frequency and parallel to the direction of propagation of the Frequency wave. In Eq. (14.2), the displacement function Fig. 14.7 shows again a sinusoidal plot. It for a longitudinal wave is written as, describes not the shape of the wave at a certain s(x, t) = a sin (kx – ωt + φ) (14.9) instant but the displacement of an element (at any fixed location) of the medium as a function where s(x, t) is the displacement of an element of time. We may for, simplicity, take Eq. (14.2) of the medium in the direction of propagation with φ = 0 and monitor the motion of the element of the wave at position x and time t. In Eq. (14.9), a is the displacement amplitude; othersay at x = 0 . We then get quantities have the same meaning as in case y (0, t ) = a sin( −ωt ) of a transverse wave except that the displacement function y (x, t) is to be replaced = −a sin ωt by the function s (x, t). * Here again, ‘radian’ could be dropped and the units could be written merely as m–1. Thus, k represents 2π times the number of waves (or the total phase difference) that can be accommodated per unit length, with SI units m–1. Reprint 2025-26 284 PHYSICS the shape of the wave at two instants of time, u Example 14.2 A wave travelling along a which differ by a small time internal ∆t. The string is described by, entire wave pattern is seen to shift to the right (positive direction of x-axis) by a distance ∆x. In y(x, t) = 0.005 sin (80.0 x – 3.0 t), particular, the crest shown by a dot (• ) moves a in which the numerical constants are in SI units (0.005 m, 80.0 rad m–1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ? Answer On comparing this displacement equation with Eq. (14.2), Fig. 14.8 Progression of a harmonic wave from time y (x, t) = a sin (kx – ωt), t to t + ∆t. where ∆t is a small interval. The wave pattern as a whole shifts to the we find right. The crest of the wave (or a point with (a) the amplitude of the wave is 0.005 m = 5 mm. any fixed phase) moves right by the distance (b) the angular wave number k and angular ∆x in time ∆t. frequency ω are distance ∆x in time ∆t. The speed of the wave is k = 80.0 m–1 and ω = 3.0 s–1 then ∆x/∆t. We can put the dot (• ) on a point We, then, relate the wavelength λ to k through with any other phase. It will move with the same Eq. (14.6), speed v (otherwise the wave pattern will not λ = 2π/k remain fixed). The motion of a fixed phase point on the wave is given by 2π = −1 kx – ωt = constant (14.10) 80.0 m Thus, as time t changes, the position x of the = 7.85 cm fixed phase point must change so that the phase (c) Now, we relate T to ω by the relation remains constant. Thus, T = 2π/ω kx – ωt = k(x+∆x) – ω(t+∆t) 2π or k ∆x – ω∆t =0 = −1 3.0 s Taking ∆x, ∆t vanishingly small, this gives = 2.09 s dx ω and frequency, v = 1/T = 0.48 Hz = = v (14.11) dt k The displacement y at x = 30.0 cm and Relating ω to T and k to λ, we get time t = 20 s is given by 2πν λ y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20) v = = λν= (14.12) 2π/λ T = (0.005 m) sin (–36 + 12π) = (0.005 m) sin (1.699) Eq. (14.12), a general relation for all progressive = (0.005 m) sin (970) j 5 mm ⊳ waves, shows that in the time required for one full oscillation by any constituent of the medium, the 14.4 THE SPEED OF A TRAVELLING WAVE wave pattern travels a distance equal to the To determine the speed of propagation of a wavelength of the wave. It should be noted that travelling wave, we can fix our attention on any the speed of a mechanical wave is determined by particular point on the wave (characterised by the inertial (linear mass density for strings, mass some value of the phase) and see how that point density in general) and elastic properties (Young’s moves in time. It is convenient to look at the modulus for linear media/ shear modulus, bulk motion of the crest of the wave. Fig. 14.8 gives modulus) of the medium. The medium determines Reprint 2025-26 WAVES 285 the speed; Eq. (14.12) then relates wavelength to arising due to an external force). It does not frequency for the given speed. Of course, as depend on wavelength or frequency of the wave remarked earlier, the medium can support both itself. In higher studies, you will come across transverse and longitudinal waves, which will have waves whose speed is not independent of different speeds in the same medium. Later in this frequency of the wave. Of the two parameters λ chapter, we shall obtain specific expressions for and ν the source of disturbance determines the the speed of mechanical waves in some media. frequency of the wave generated. Given the speed of the wave in the medium and the 14.4.1 Speed of a Transverse Wave on frequency Eq. (14.12) then fixes the wavelength Stretched String v The speed of a mechanical wave is determined λ = (14.15) by the restoring force setup in the medium when ν it is disturbed and the inertial properties (mass u Example 14.3 A steel wire 0.72 m longdensity) of the medium. The speed is expected to has a mass of 5.0 ×10–3 kg. If the wire isbe directly related to the former and inversely to under a tension of 60 N, what is the speedthe latter. For waves on a string, the restoring of transverse waves on the wire ?force is provided by the tension T in the string. The inertial property will in this case be linear Answer Mass per unit length of the wire,mass density µ, which is mass m of the string divided by its length L. Using Newton’s Laws of −3 5.0 × 10 kgMotion, an exact formula for the wave speed on µ = 0. 72 m a string can be derived, but this derivation is outside the scope of this book. We shall, = 6.9 ×10–3 kg m–1 therefore, use dimensional analysis. We already know that dimensional analysis alone can never Tension, T = 60 N yield the exact formula. The overall The speed of wave on the wire is given by dimensionless constant is always left T 60 N −1undetermined by dimensional analysis. = v = = 93 m s ⊳ µ 6.9 × 10− 3 kg m −1 The dimension of µ is [ML–1] and that of T is like force, namely [MLT–2]. We need to combine 14.4.2 Speed of a Longitudinal Wavethese dimensions to get the dimension of speed (Speed of Sound)v [LT–1]. Simple inspection shows that the quantity T/µ has the relevant dimension In a longitudinal wave, the constituents of the − 2 medium oscillate forward and backward in the MLT   2 − 2   =  L T  direction of propagation of the wave. We have [ ML ]   already seen that the sound waves travel in the Thus if T and µ are assumed to be the only form of compressions and rarefactions of small volume elements of air. The elastic property thatrelevant physical quantities, determines the stress under compressional strain is the bulk modulus of the medium defined T v = C (14.13) by (see Chapter 8) µ ∆P where C is the undetermined constant of B = (14.16)dimensional analysis. In the exact formula, it −∆V/V turms out, C=1. The speed of transverse waves Here, the change in pressure ∆P produces a on a stretched string is given by ∆V volumetric strain . B has the same dimension V T v = (14.14) as pressure and given in SI units in terms of µ pascal (Pa). The inertial property relevant for the Note the important point that the speed v propagation of wave is the mass density ρ, with depends only on the properties of the medium T dimensions [ML–3]. Simple inspection reveals and µ (T is a property of the stretched string that quantity B/ρ has the relevant dimension: Reprint 2025-26 286 PHYSICS ML − 2 T − 2   2 − 2 Liquids and solids generally have higher speed    L T =  (14.17) of sound than gases. [Note for solids, the speed − 3    ML    being referred to is the speed of longitudinal Thus, if B and ρ are considered to be the only waves in the solid]. This happens because they are much more difficult to compress than gasesrelevant physical quantities, and so have much higher values of bulk modulus. B Now, see Eq. (14.19). Solids and liquids have v = C (14.18) higher mass densities ( ρ) than gases. But the ρ corresponding increase in both the modulus (B)where, as before, C is the undetermined constant of solids and liquids is much higher. This is thefrom dimensional analysis. The exact derivation reason why the sound waves travel faster inshows that C=1. Thus, the general formula for solids and liquids.longitudinal waves in a medium is: We can estimate the speed of sound in a gas B in the ideal gas approximation. For an ideal gas, v = (14.19) ρ the pressure P, volume V and temperature T are related by (see Chapter 10). For a linear medium, like a solid bar, the lateral expansion of the bar is negligible and we PV = NkBT (14.21) may consider it to be only under longitudinal where N is the number of molecules in volumestrain. In that case, the relevant modulus of V, kB is the Boltzmann constant and T theelasticity is Young’s modulus, which has the temperature of the gas (in Kelvin). Therefore, forsame dimension as the Bulk modulus. an isothermal change it follows from Eq.(14.21) Dimensional analysis for this case is the same that as before and yields a relation like Eq. (14.18), V∆P + P∆V = 0 with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of ∆ P or − = Plongitudinal waves in a solid bar is given by ∆V/V Hence, substituting in Eq. (14.16), we have Y v = (14.20) B = P ρ Therefore, from Eq. (14.19) the speed of a where Y is the Young’s modulus of the material longitudinal wave in an ideal gas is given by, of the bar. Table 14.1 gives the speed of sound in some media. v = P (14.22) ρ Table 14.1 Speed of Sound in some Media This relation was first given by Newton and is known as Newton’s formula. u Example 14.4 Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10–3 kg. Answer We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is: ρo = (mass of one mole of air)/ (volume of one mole of air at STP) −3 29.0 × 10 kg = −3 3 22.4 × 10 m = 1.29 kg m–3 Reprint 2025-26 WAVES 287 According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP, = 280 m s–1 (14.23) ⊳ The result shown in Eq.(14.23) is about 15% smaller as compared to the experimental value of 331 m s–1 as given in Table 14.1. Where did we go wrong ? If we examine the basic assumption made by Newton that the pressure variations in a medium during propagation of sound are isothermal, we find that this is not correct. It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal. For adiabatic processes the ideal Fig. 14.9 Two pulses having equal and opposite gas satisfies the relation (see Section 11.8), γ displacements moving in opposite PV = constant directions. The overlapping pulses add up i.e. ∆(PVγ ) = 0 to zero displacement in curve (c). or P γ V γ –1 ∆V + V γ ∆P = 0 pulses. Figure 14.9 shows the situation when where γ is the ratio of two specific heats, two pulses of equal and opposite shapes move Cp/Cv. towards each other. When the pulses overlap, Thus, for an ideal gas the adiabatic bulk the resultant displacement is the algebraic summodulus is given by, of the displacement due to each pulse. This is ∆P known as the principle of superposition of waves. Bad = − ∆V/V According to this principle, each pulse moves = γP as if others are not present. The constituents of The speed of sound is, therefore, from Eq. the medium, therefore, suffer displacments due (14.19), given by, to both and since the displacements can be positive and negative, the net displacement is γ P an algebraic sum of the two. Fig. 14.9 gives v = (14.24) ρ graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the This modification of Newton’s formula is referred displacements due to the two pulses have exactly to as the Laplace correction. For air cancelled each other and there is zero γ = 7/5. Now using Eq. (14.24) to estimate the speed displacement throughout. of sound in air at STP, we get a value 331.3 m s–1, To put the principle of superposition which agrees with the measured speed. mathematically, let y1 (x,t) and y2 (x,t) be the displacements due to two wave disturbances in