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

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

9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.