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

3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. R = A + B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos θ SN = B sin θ OS2 = (A + B cos θ)2 + (B sin θ)2 or, R2 = A2 + B2 + 2AB cos θ R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 The average acceleration for three time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and the acceleration of an object are always along -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  direction or in the opposite direction). ⊳ However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (b) The time taken to return to the same level is Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r′ be the position vectors and v and r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 3.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 3.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 3.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? Reprint 2025-26 48 PHYSICS 3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 3.16 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 3.17 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 2.0 j m s-2. (a) At what time is ( ) the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t1 to t2) 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is 2 2 2 2 −1 −1 R = v r + v w = 35 + 12 m s = 37 m s The direction θ that R makes with the vertical is given by v w 12 tan θ = = = 0.343 Fig. 3.8 (a) Two non-colinear vectors a and b. v r 35 (b) Resolving a vector A in terms of vectors Or, θ = tan-1 ( 0.343 ) = 19° a and b. Therefore, the boy should hold his umbrella We say that A has been resolved into two in the vertical plane at an angle of about 19o component vectors λ a and µ b along a and b with the vertical towards the east. ⊳ Reprint 2025-26 32 PHYSICS respectively. Using this method one can resolve and A2 is parallel to ɵj, we have :a given vector into two component vectors along a set of two vectors – all the three lie in the same A1= Ax ɵi , A2 = Ay ɵj (3.11) plane. It is convenient to resolve a general vector where Ax and Ay are real numbers.along the axes of a rectangular coordinate system using vectors of unit magnitude. These Thus, A = Ax ɵi + Ay ɵj (3.12) are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and This is represented in Fig. 3.9(c). The quantities points in a particular direction. It has no Ax and Ay are called x-, and y- components of the dimension and unit. It is used to specify a vector A. Note that Ax is itself not a vector, but direction only. Unit vectors along the x-, y- and ɵi is a vector, and so is Ay ɵj. Using simplez-axes of a rectangular coordinate system are Ax trigonometry, we can express Ax and Ay in terms denoted by ɵi , ɵj and ˆk , respectively, as shown of the magnitude of A and the angle θ it makes in Fig. 3.9(a). with the x-axis : Since these are unit vectors, we have Ax = A cos θ Ay = A sin θ (3.13) ˆi  = ˆj  = ˆk =1 (3.9) As is clear from Eq. (3.13), a component of a These unit vectors are perpendicular to each vector can be positive, negative or zero other. In this text, they are printed in bold face depending on the value of θ. with a cap (^) to distinguish them from other Now, we have two ways to specify a vector A vectors. Since we are dealing with motion in two in a plane. It can be specified by : dimensions in this chapter, we require use of (i) its magnitude A and the direction θ it makes only two unit vectors. If we multiply a unit vector, with the x-axis; or say ˆn by a scalar, the result is a vector (ii) its components Ax and Ay λ = λ ˆn. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained using Eq. (3.13). If Ax and Ay are given, A and θ A = |A| ˆn (3.10) can be obtained as follows : where ˆn is a unit vector along A. 2 2 2 2 2 2 A x + A y = A cos θ + A sin θ We can now resolve a vector A in terms = A2 of component vectors that lie along unit vectors iˆ and ɵj. Consider a vector A that lies in x-y Or, A = A 2x + Ay2 (3.14) plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate Ay A y tan θ = , θ = tan− 1 axes as in Fig. 3.9(b), and get vectors A1 and A2 And A x A x (3.15) such that A1 + A2 = A. Since A1 is parallel to ɵi Fig. 3.9 (a) Unit vectors ɵi , ɵj and ɵk lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of ɵi and ɵj . Reprint 2025-26 MOTION IN A PLANE 33 So far we have considered a vector lying in ɵ ɵ B = B x i + B y jan x-y plane. The same procedure can be used Let R be their sum. We haveto resolve a general vector A into three components along x-, y-, and z-axes in three R = A + B dimensions. If α, β, and γ are the angles* ɵ ɵ ɵ ɵ = + (3.19a) ( A x i + A y j ) ( B x i + B y j )between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : ɵ ɵ j (3.19b) R = ( A x + B x ) i + ( A y + B y ) ɵ ɵ SinceR = R x i + R y j (3.20) we have, R x = A x + B x , R y = A y + B y (3.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have ɵ ɵ ɵ A = A x i + Ay j + A z k ɵ ɵ ɵ B = B x i + B y j + B z k (d) ɵ ɵ ɵ R = A + B = R x i + R y j + R z kFig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes with R x = A x + B x y = A y + B yA x = A cos α, A y = A cos β, A z = A cos γ (3.16a) R In general, we have R z = A z + B z (3.22) A = Ax ˆi + Ay ˆj + Az kˆ (3.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = A x2 + Ay2 + A z2 (3.16c) example, if vectors a, b and c are given as ɵ ɵ ɵ A position vector r can be expressed as a = a x i + a y j + a z k ɵ ɵ ɵ r = x i + y j + z k (3.17) ɵ ɵ ɵ b = b x i + b y j + b z k where x, y, and z are the components of r along ɵ ɵ ɵ x-, y-, z-axes, respectively. c = c x i + c y j + c z k (3.23a)