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

10.7 POLARISATION Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.17). Such a wave could be described by the following equation FIGURE 10.17 (a) The curves represent the displacement of a string at t = 0 and at t = Dt, respectively when a sinusoidal wave is propagating in the +x-direction. (b) The curve represents the time variation of the displacement at x = 0 when a sinusoidal wave is propagating in the +x-direction. At x = Dx, the time variation of the displacement will be slightly displaced to the right. y (x,t) = a sin (kx – wt) (10.15) where a and w(= 2pn) represent the amplitude and the angular frequency of the wave, respectively; further, 2 π λ = (10.16) k represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 14 of Class XI textbook. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised 269 Reprint 2025-26 Physics wave. Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by z (x,t) = a sin (kx – wt) (10.17) It should be mentioned that the linearly polarised waves [described by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation. Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid. Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1, it is observed that its intensity is reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P2 be placed before P1. As expected, the light from the lamp is reduced in intensity on passing through P2 alone. But now rotating P1 has a dramatic effect on the light coming from P2. In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2 (Fig. 10.18). The experiment at figure 10.18 can be easily understood by assuming that light passing through the polaroid P2 gets polarised along the pass- axis of P2. If the pass-axis of P2 makes an angle q with the pass-axis of P1, then when the polarised beam passes through the polaroid P2, the component E cos q (along the pass-axis of P2) will pass through P2. Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as: I = I0 cos2q (10.18) where I0 is the intensity of the polarized light after passing through P1. This is known as Malus’ law. The above discussion shows that the Reprint 2025-26 Wave Optics FIGURE 10.18 (a) Passage of light through two polaroids P2 and P1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0° to 90°. Notice that the light seen through a single polaroid P1 does not vary with angle. (b) Behaviour of the electric vector when light passes through two polaroids. The transmitted polarisation is the component parallel to the polaroid axis. The double arrows show the oscillations of the electric vector. intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids. Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras. Example 10.2 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? Solution Let I0 be the intensity of polarised light after passing through the first polariser P1. Then the intensity of light after passing through second polariser P2 will be I = I 0cos 2θ, where q is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (p/2–q). Hence the intensity of light emerging from P3 will be  π  – θ I = I 0 cos 2θ cos 2  2  EXAMPLE = I0 cos2q sin2q =(I0/4) sin22q Therefore, the transmitted intensity will be maximum when q = p/4. 10.2 271 Reprint 2025-26 Physics SUMMARY 1. Huygens’ principle tells us that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time. 2. Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. When the speed of light is independent of direction, the secondary waves are spherical. The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. This principle leads to the well known laws of reflection and refraction. 3. The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. When we consider the intensity of light due to these sources at the given point, there is an interference term in addition to the sum of the individual intensities. But this term is important only if it has a non-zero average, which occurs only if the sources have the same frequency and a stable phase difference. 4. Young’s double slit of separation d gives equally spaced interference fringes. 5. A single slit of width a gives a diffraction pattern with a central λ 2λ maximum. The intensity falls to zero at angles of ± , ± , etc., a a with successively weaker secondary maxima in between. 6. Natural light, e.g., from the sun is unpolarised. This means the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polaroid transmits only one component (parallel to a special axis). The resulting light is called linearly polarised or plane polarised. When this kind of light is viewed through a second polaroid whose axis turns through 2p, two maxima and minima of intensity are seen. POINTS TO PONDER 1. Waves from a point source spread out in all directions, while light was seen to travel along narrow rays. It required the insight and experiment of Huygens, Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light. 2. The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive, as shown in Young’s experiment. 3. Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light. 4. Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarisation phenomena are special to transverse waves like light waves. Reprint 2025-26 Wave Optics EXERCISES

14.5 THE PRINCIPLE OF SUPERPOSITION the medium. If the waves arrive in a region OF WAVES simultaneously, and therefore, overlap, the net displacement y (x,t) is given by What happens when two wave pulses travelling in opposite directions cross each other y (x, t) = y1(x, t) + y2(x, t) (14.25) (Fig. 14.9)? It turns out that wave pulses If we have two or more waves moving in the continue to retain their identities after they have medium the resultant waveform is the sum of crossed. However, during the time they overlap, wave functions of individual waves. That is, if the wave pattern is different from either of the the wave functions of the moving waves are Reprint 2025-26 288 PHYSICS y1 = f1(x–vt), y2 = f2(x–vt), .......... .......... yn = fn (x–vt) then the wave function describing the disturbance in the medium is y = f1(x – vt)+ f2(x – vt)+ ...+ fn(x – vt) n = ∑ f ( x − vt ) (14.26) i i =1 The principle of superposition is basic to the phenomenon of interference. For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequency) and k (wave number), and, therefore, the same wavelength Fig. 14.10 The resultant of two harmonic waves of λ. Their wave speed will be identical. Let us equal amplitude and wavelength further assume that their amplitudes are equal according to the principle of superposition. and they are both travelling in the positive The amplitude of the resultant wave depends on the phase difference φ, whichdirection of x-axis. The waves only differ in their is zero for (a) and π for (b)initial phase. According to Eq. (14.2), the two waves are described by the functions: φ between the constituent two waves: y1(x, t) = a sin (kx – ωt) (14.27) A(φ) = 2a cos ½φ (14.32) For φ = 0, when the waves are in phase, and y2(x, t) = a sin (kx – ωt + φ ) (14.28) y ( x , t ) = 2a sin ( kx − ωt ) (14.33) The net displacement is then, by the principle i.e., the resultant wave has amplitude 2a, theof superposition, given by largest possible value for A. For φ = π , the y (x, t) = a sin (kx – ωt) + a sin (kx – ωt + φ) waves are completely, out of phase and the (14.29) resultant wave has zero displacement   ( kx − ωt ) + ( kx − ωt + φ)  φ everywhere at all times = a  2sin   cos  y (x, t) = 0 (14.34)   2  2  Eq. (14.33) refers to the so-called constructive (14.30) interference of the two waves where the where we have used the familiar trignometric amplitudes add up in the resultant wave. Eq. identity for sin A + sin B . We then have (14.34) is the case of destructive intereference where the amplitudes subtract out in the φ  φ resultant wave. Fig. 14.10 shows these two cases y ( x , t ) = 2 a cos sin  kx − ωt +  (14.31) 2  2  of interference of waves arising from the principle of superposition.Eq. (14.31) is also a harmonic travelling wave in the positive direction of x-axis, with the same 14.6 REFLECTION OF WAVES frequency and wavelength. However, its initial So far we considered waves propagating in an φ unbounded medium. What happens if a pulse phase angle is . The significant thing is that 2 or a wave meets a boundary? If the boundary is its amplitude is a function of the phase difference rigid, the pulse or wave gets reflected. The Reprint 2025-26 WAVES 289 phenomenon of echo is an example of reflection If on the other hand, the boundary point is by a rigid boundary. If the boundary is not not rigid but completely free to move (such as in completely rigid or is an interface between two the case of a string tied to a freely moving ring different elastic media, the situation is some on a rod), the reflected pulse has the same phase what complicated. A part of the incident wave is and amplitude (assuming no energy dissipation) reflected and a part is transmitted into the as the incident pulse. The net maximum second medium. If a wave is incident obliquely displacement at the boundary is then twice the on the boundary between two different media amplitude of each pulse. An example of non- rigid the transmitted wave is called the refracted boundary is the open end of an organ pipe. wave. The incident and refracted waves obey To summarise, a travelling wave or pulse Snell’s law of refraction, and the incident and suffers a phase change of π on reflection at a reflected waves obey the usual laws of rigid boundary and no phase change on reflection. reflection at an open boundary. To put this Fig. 14.11 shows a pulse travelling along a mathematically, let the incident travelling stretched string and being reflected by the wave be boundary. Assuming there is no absorption of y 2 ( x , t ) = a sin ( kx − ωt )energy by the boundary, the reflected wave has the same shape as the incident pulse but it At a rigid boundary, the reflected wave is given suffers a phase change of π or 1800 on reflection. by This is because the boundary is rigid and the yr(x, t) = a sin (kx – ωt + π). disturbance must have zero displacement at all = – a sin (kx – ωt) (14.35) times at the boundary. By the principle of At an open boundary, the reflected wave is given superposition, this is possible only if the reflected by and incident waves differ by a phase of π, so that yr(x, t) = a sin (kx – ωt + 0). the resultant displacement is zero. This = a sin (kx – ωt) (14.36) reasoning is based on boundary condition on a Clearly, at the rigid boundary, y = y 2 + y r = 0rigid wall. We can arrive at the same conclusion dynamically also. As the pulse arrives at the wall, at all times. it exerts a force on the wall. By Newton’s Third 14.6.1 Standing Waves and Normal Modes Law, the wall exerts an equal and opposite force We considered above reflection at one boundary. on the string generating a reflected pulse that But there are familiar situations (a string fixed differs by a phase of π. at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries. In a string, for example, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Eqs. (14.2) and (14.4), with φ = 0, we get: y1(x, t) = a sin (kx – ωt) y2(x, t) = a sin (kx + ωt) The resultant wave on the string is, according to the principle of superposition: Fig. 14.11 Reflection of a pulse meeting a rigid boundary. y (x, t) = y1(x, t) + y2(x, t) Reprint 2025-26 290 PHYSICS = a [sin (kx – ωt) + sin (kx + ωt)] nodes; the points at which the amplitude is the largest are called antinodes. Fig. 14.12 showsUsing the familiar trignometric identity a stationary wave pattern resulting fromSin (A+B) + Sin (A–B) = 2 sin A cosB we get, superposition of two travelling waves in y (x, t) = 2a sin kx cos ωt (14.37) opposite directions. Note the important difference in the wave The most significant feature of stationary pattern described by Eq. (14.37) from that waves is that the boundary conditions constrain described by Eq. (14.2) or Eq. (14.4). The terms the possible wavelengths or frequencies of kx and ωt appear separately, not in the vibration of the system. The system cannot combination kx - ωt. The amplitude of this wave oscillate with any arbitrary frequency (contrast is 2a sin kx. Thus, in this wave pattern, the this with a harmonic travelling wave), but is amplitude varies from point-to-point, but each characterised by a set of natural frequencies or element of the string oscillates with the same normal modes of oscillation. Let us determine angular frequency ω or time period. There is no these normal modes for a stretched string fixed phase difference between oscillations of different at both ends. elements of the wave. The string as a whole First, from Eq. (14.37), the positions of nodes vibrates in phase with differing amplitudes at (where the amplitude is zero) are given by different points. The wave pattern is neither sin kx = 0 . moving to the right nor to the left. Hence, they which implies are called standing or stationary waves. The kx = nπ; n = 0, 1, 2, 3, ... amplitude is fixed at a given location but, as Since, k = 2π/λ , we get remarked earlier, it is different at different locations. The points at which the amplitude is nλ zero (i.e., where there is no motion at all) are x = ; n = 0, 1, 2, 3, ... (14.38) 2 Fig. 14.12 Stationary waves arising from superposition of two harmonic waves travelling in opposite directions. Note that the positions of zero displacement (nodes) remain fixed at all times. Reprint 2025-26 WAVES 291 Clearly, the distance between any two speed of wave determined by the properties of λ the medium. The n = 2 frequency is called the successive nodes is . In the same way, the second harmonic; n = 3 is the third harmonic 2 and so on. We can label the various harmonics bypositions of antinodes (where the amplitude is the symbol νn ( n = 1, 2, ...).the largest) are given by the largest value of sin Fig. 14.13 shows the first six harmonics of akx : sin k x = 1 stretched string fixed at either end. A string need not vibrate in one of these modes only.which implies Generally, the vibration of a string will be a kx = (n + ½) π ; n = 0, 1, 2, 3, ... superposition of different modes; some modes With k = 2π/λ, we get may be more strongly excited and some less. Musical instruments like sitar or violin are λ based on this principle. Where the string is x = (n + ½) ; n = 0, 1, 2, 3, ... (14.39) 2 plucked or bowed, determines which modes are Again the distance between any two consecutive more prominent than others. Let us next consider normal modes of λ antinodes is . Eq. (14.38) can be applied to oscillation of an air column with one end closed 2 the case of a stretched string of length L fixed at both ends. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by λ L = n ; n = 1, 2, 3, ... (14.40) 2 Thus, the possible wavelengths of stationary waves are constrained by the relation 2L λ = ; n = 1, 2, 3, … (14.41) n with corresponding frequencies nv v = , for n = 1, 2, 3, (14.42) 2L We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed at either end v it is given by v = , corresponding 2 L Fig. 14.13 The first six harmonics of vibrations of a stretched to n = 1 of Eq. (14.42). Here v is the string fixed at both ends. Reprint 2025-26 292 PHYSICS and the other open. A glass tube partially filled modes of this system is more complex. This with water illustrates this system. The end in problem involves wave propagation in two contact with water is a node, while the open end dimensions. However, the underlying physics is is an antinode. At the node the pressure the same. changes are the largest, while the displacement is minimum (zero). At the open end - the u Example 14.5 A pipe, 30.0 cm long, is open antinode, it is just the other way - least pressure at both ends. Which harmonic mode of the change and maximum amplitude of pipe resonates a 1.1 kHz source? Will displacement. Taking the end in contact with resonance with the same source be water to be x = 0, the node condition (Eq. 14.38) observed if one end of the pipe is closed ? is already satisfied. If the other end x = L is an Take the speed of sound in air as antinode, Eq. (14.39) gives 330 m s–1.  1  λ n +L = , for n = 0, 1, 2, 3, …  2  2 Answer The first harmonic frequency is given by The possible wavelengths are then restricted by v v the relation : ν1 = λ1 = 2 L (open pipe) where L is the length of the pipe. The frequency 2 L λ = , for n = 0, 1, 2, 3,... (14.43) of its nth harmonic is: ( n + 1 / 2 ) nv νn = 2L , for n = 1, 2, 3, ... (open pipe) The normal modes – the natural frequencies – of the system are First few modes of an open pipe are shown in Fig. 14.15.  1  v For L = 30.0 cm, v = 330 m s–1, ; n = 0, 1, 2, 3, ... (14.44) ν =  n + 2  2 L n 330 (m s − 1 ) νn = = 550 n s–1 The fundamental frequency corresponds to n = 0, 0.6 (m) v Clearly, a source of frequency 1.1 kHz will and is given by . The higher frequencies resonate at v2, i.e. the second harmonic. 4 L are odd harmonics, i.e., odd multiples of the v v fundamental frequency : 3 , 5 , etc. 4 L 4 L Fig. 14.14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open air column at both ends generates all harmonics (See Fig. 14.15). The systems above, strings and air columns, can also undergo forced oscillations (Chapter 13). If the external frequency is close to one of the natural frequencies, the system shows resonance. Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no Fundamental point on the circumference of the membrane or third fifth vibrates. Estimation of the frequencies of normal first harmonic harmonic harmonic Reprint 2025-26 WAVES 293 Fig. 14.15 Standing waves in an open pipe, first four harmonics are depicted. while tuning their instruments with each other. They go on tuning until their sensitive ears do seventh ninth eleventh not detect any beats. harmonic harmonic harmonic To see this mathematically, let us consider two harmonic sound waves of nearly equal Fig. 14.14 Normal modes of an air column open at angular frequency ω1 and ω2 and fix the location one end and closed at the other end. Only to be x = 0 for convenience. Eq. (14.2) with a the odd harmonics are seen to be possible. suitable choice of phase (φ = π/2 for each) and, assuming equal amplitudes, gives Now if one end of the pipe is closed (Fig. 14.15), it follows from Eq. (14.15) that the fundamental s1 = a cos ω1t and s2 = a cos ω2t (14.45) frequency is Here we have replaced the symbol y by s, v v since we are referring to longitudinal not transverse displacement. Let ω1 be the (slightly) ν1 = λ1 = 4 L (pipe closed at one end) greater of the two frequencies. The resultant and only the odd numbered harmonics are displacement is, by the principle of present : superposition, s = s1 + s2 = a (cos ω1 t + cos ω2 t) 3v 5v ν3 = , ν5 = , and so on. Using the familiar trignometric identity for 4 L 4 L cos A + cosB, we get For L = 30 cm and v = 330 m s–1, the (ω1 - ω2 ) t (ω1 + ω2 ) t fundamental frequency of the pipe closed at one = 2 a cos cos (14.46) end is 275 Hz and the source frequency 2 2 corresponds to its fourth harmonic. Since this which may be written as : harmonic is not a possible mode, no resonance s = [2 a cos ωb t ] cos ωat (14.47) will be observed with the source, the moment If |ω1 – ω2| <<ω1, ω2, ωa >> ωb, th one end is closed. ⊳ where 14.7 BEATS (ω1 − ω2 ) (ω1 + ω2 ) ωb = and ωa = ‘Beats’ is an interesting phenomenon arising 2 2 from interference of waves. When two harmonic Now if we assume |ω1 – ω2| <<ω1, which means sound waves of close (but not equal) frequencies ωa >> ωb, we can interpret Eq. (14.47) as follows. are heard at the same time, we hear a sound of The resultant wave is oscillating with the average similar frequency (the average of two close angular frequency ωa; however its amplitude is frequencies), but we hear something else also. not constant in time, unlike a pure harmonic We hear audibly distinct waxing and waning of wave. The amplitude is the largest when the the intensity of the sound, with a frequency term cos ωb t takes its limit +1 or –1. In other equal to the difference in the two close words, the intensity of the resultant wave waxes frequencies. Artists use this phenomenon often and wanes with a frequency which is 2ωb = ω1 – Reprint 2025-26 294 PHYSICS ω2. Since ω = 2πν, the beat frequency νbeat, is given by νbeat = ν1 – ν2 (14.48) Fig. 14.16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. Musical Pillars Temples often have some pillars portraying human figures playing musical instru- ments, but seldom do these pillars themselves produce music. At the Nellaiappar temple in Tamil Nadu, gentle taps on a cluster of pillars carved out of a single piece of rock produce the basic notes of Indian classical music, viz. Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa. Vibrations of these pillars depend on elasticity of the stone used, Fig. 14.16 Superposition of two harmonic waves, one its density and shape. of frequency 11 Hz (a), and the other of Musical pillars are categorised into three frequency 9Hz (b), giving rise to beats of frequency 2 Hz, as shown in (c). types: The first is called the Shruti Pillar, as it can produce the basic notes — the “swaras”. The second type is the Gana u Example 14.6 Two sitar strings A and B Thoongal, which generates the basic tunes playing the note ‘Dha’ are slightly out of that make up the “ragas”. The third variety tune and produce beats of frequency 5 Hz. is the Laya Thoongal pillars that produce The tension of the string B is slightly “taal” (beats) when tapped. The pillars at the increased and the beat frequency is found Nellaiappar temple are a combination of the to decrease to 3 Hz. What is the original Shruti and Laya types. frequency of B if the frequency of A is Archaeologists date the Nelliappar 427 Hz ? temple to the 7th century and claim it was built by successive rulers of the Pandyan Answer Increase in the tension of a string dynasty. increases its frequency. If the original frequency The musical pillars of Nelliappar and of B (νB) were greater than that of A (νA ), further several other temples in southern India like increase in νB should have resulted in an those at Hampi (picture), Kanyakumari, and increase in the beat frequency. But the beat Thiruvananthapuram are unique to the frequency is found to decrease. This shows that country and have no parallel in any other νB < νA. Since νA – νB = 5 Hz, and νA = 427 Hz, we part of the world. get νB = 422 Hz. ⊳ Reprint 2025-26 WAVES 295 SUMMARY 1. Mechanical waves can exist in material media and are governed by Newton’s Laws. 2. Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation. 3. Longitudinal waves are waves in which the particles of the medium oscillate along the direction of wave propagation. 4. Progressive wave is a wave that moves from one point of medium to another. 5. The displacement in a sinusoidal wave propagating in the positive x direction is given by y (x, t) = a sin (kx – ωt + φ) where a is the amplitude of the wave, k is the angular wave number, ω is the angular frequency, (kx – ωt + φ) is the phase, and φ is the phase constant or phase angle. 6. Wavelength λ of a progressive wave is the distance between two consecutive points of the same phase at a given time. In a stationary wave, it is twice the distance between two consecutive nodes or antinodes. 7. Period T of oscillation of a wave is defined as the time any element of the medium takes to move through one complete oscillation. It is related to the angular frequency ω through the relation 2π T = ω 8. Frequency v of a wave is defined as 1/T and is related to angular frequency by ω ν = 2 π ω λ 9. Speed of a progressive wave is given by v = = = λν k T 10. The speed of a transverse wave on a stretched string is set by the properties of the string. The speed on a string with tension T and linear mass density µ is T v = µ 11. Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases. The speed v of sound wave in a fluid having bulk modulus B and density ρ is B v = ρ The speed of longitudinal waves in a metallic bar is Y v = ρ For gases, since B = γP, the speed of sound is γP v = ρ Reprint 2025-26 296 PHYSICS 12. When two or more waves traverse simultaneously in the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is known as the principle of superposition of waves n f i ( x − vt ) y = ∑ i =1 13. Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φ, the result is a single wave with the same frequency ω :  1   1  y (x, t) =  2a cos 2 φ sin  kx − ωt + 2 φ If φ = 0 or an integral multiple of 2π, the waves are exactly in phase and the interference is constructive; if φ= π, they are exactly out of phase and the interference is destructive. 14. A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change. For an incident wave yi (x, t) = a sin (kx – ωt ) the reflected wave at a rigid boundary is yr (x, t) = – a sin (kx + ωt ) For reflection at an open boundary yr (x,t ) = a sin (kx + ωt) 15. The interference of two identical waves moving in opposite directions produces standing waves. For a string with fixed ends, the standing wave is given by y (x, t) = [2a sin kx ] cos ωt Standing waves are characterised by fixed locations of zero displacement called nodes and fixed locations of maximum displacements called antinodes. The separation between two consecutive nodes or antinodes is λ/2. A stretched string of length L fixed at both the ends vibrates with frequencies given by n v v = , n = 1, 2, 3, ... 2 L The set of frequencies given by the above relation are called the normal modes of oscillation of the system. The oscillation mode with lowest frequency is called the fundamental mode or the first harmonic. The second harmonic is the oscillation mode with n = 2 and so on. A pipe of length L with one end closed and other end open (such as air columns) vibrates with frequencies given by v v = ( n + ½) , n = 0, 1, 2, 3, ... 2L The set of frequencies represented by the above relation are the normal modes of oscillation of such a system. The lowest frequency given by v/4L is the fundamental mode or the first harmonic. 16. A string of length L fixed at both ends or an air column closed at one end and open at the other end or open at both the ends, vibrates with certain frequencies called their normal modes. Each of these frequencies is a resonant frequency of the system. 17. Beats arise when two waves having slightly different frequencies, ν1 and ν2 and comparable amplitudes, are superposed. The beat frequency is νbeat = ν1 ~ ν2 Reprint 2025-26 WAVES 297 POINTS TO PONDER 1. A wave is not motion of matter as a whole in a medium. A wind is different from the sound wave in air. The former involves motion of air from one place to the other. The latter involves compressions and rarefactions of layers of air. 2. In a wave, energy and not the matter is transferred from one point to the other. 3. In a mechanical wave, energy transfer takes place because of the coupling through elastic forces between neighbouring oscillating parts of the medium. 4. Transverse waves can propagate only in medium with shear modulus of elasticity, Longitudinal waves need bulk modulus of elasticity and are therefore, possible in all media, solids, liquids and gases. 5. In a harmonic progressive wave of a given frequency, all particles have the same amplitude but different phases at a given instant of time. In a stationary wave, all particles between two nodes have the same phase at a given instant but have different amplitudes. 6. Relative to an observer at rest in a medium the speed of a mechanical wave in that medium (v) depends only on elastic and other properties (such as mass density) of the medium. It does not depend on the velocity of the source. EXERCISES 14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? 14.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1 ? (g = 9.8 m s–2) 14.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1. γP 14.4 Use the formula v = to explain why the speed of sound in air ρ (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. Reprint 2025-26 298 PHYSICS 14.5 You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave : (a) (x – vt )2 (b) log [(x + vt)/x0] (c) 1/(x + vt) 14.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s –1 and in water 1486 m s–1. 14.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1 ? The operating frequency of the scanner is 4.2 MHz. 14.8 A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave ? If it is travelling, what are the speed and direction of its propagation ? (b) What are its amplitude and frequency ? (c) What is the initial phase at the origin ? (d) What is the least distance between two successive crests in the wave ? 14.9 For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ? 14.10 For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4 14.11 The transverse displacement of a string (clamped at its both ends) is given by  2π  y(x, t) = 0.06 sin  3 x  cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg. Answer the following : (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave ? Reprint 2025-26 WAVES 299 (c) Determine the tension in the string. 14.12 (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end? 14.13 Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos (3x) sin (10t) (b) y = 2 x − vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t 14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string? 14.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected. 14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel? 14.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1). 14.18 Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B? 14.19 Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium. Reprint 2025-26