4.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 4.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 4.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 4.10 Circular motion agency is needed to provide force to move a body from rest. 4.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion. Summary In these examples, the external agency of force (hands, Points to ponder wind, stream, etc) is in contact with the object. This is not Exercises always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? Reprint 2025-26 50 PHYSICS 4.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion isThe question posed above appears to be simple. possible with no frictional forces opposing. ThisHowever, it took ages to answer it. Indeed, the is what Galileo did.correct answer to this question given by Galileo in the seventeenth century was the foundation 4.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an interme- something external is required to keep it moving. diate situation. Galileo concluded that an object According to this view, for example, an arrow moving on a frictionless horizontal plane must shot from a bow keeps flying since the air behind neither have acceleration nor retardation, i.e. it the arrow keeps pushing it. The view was part of should move with constant velocity (Fig. 4.1(a)).an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force (i) (ii) (iii)is required to keep a body in motion. Fig. 4.1(a) Aristotelian law of motion is flawed, as we shall Another experiment by Galileo leading to thesee. However, it is a natural view that anyone same conclusion involves a double inclined plane.would hold from common experience. Even a A ball released from rest on one of the planes rollssmall child playing with a simple (non-electric) down and climbs up the other. If the planes are toy-car on a floor knows intuitively that it needs smooth, the final height of the ball is nearly the to constantly drag the string attached to the toy- same as the initial height (a little less but never car with some force to keep it going. If it releases greater). In the ideal situation, when friction is the string, it comes to rest. This experience is absent, the final height of the ball is the same common to most terrestrial motion. External as its initial height. forces seem to be needed to keep bodies in If the slope of the second plane is decreased motion. Left to themselves, all bodies eventually and the experiment repeated, the ball will still come to rest. reach the same height, but in doing so, it will What is the flaw in Aristotle’s argument? The travel a longer distance. In the limiting case, when answer is: a moving toy car comes to rest because the slope of the second plane is zero (i.e. is a the external force of friction on the car by the floor horizontal) the ball travels an infinite distance. opposes its motion. To counter this force, the child In other words, its motion never ceases. This is, has to apply an external force on the car in the of course, an idealised situation (Fig. 4.1(b)). direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle Fig. 4.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. Reprint 2025-26 LAWS OF MOTION 51 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he formu- Galileo thus, arrived at a new insight on lated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, Two kinds of situations are encountered in thea body at rest continues to remain at rest and a application of this law in practice. In somebody in motion continues to move with a uniform examples, we know that the net external forcevelocity. This property of the body is called on the object is zero. In that case we caninertia. Inertia means ‘resistance to change’. A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 4.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. Reprint 2025-26 52 PHYSICS More often, however, we do not know all the The acceleration of the car cannot be accounted forces to begin with. In that case, if we know for by any internal force. This might sound that an object is unaccelerated (i.e. it is either surprising, but it is true. The only conceivable at rest or in uniform linear motion), we can infer external force along the road is the force of from the first law that the net external force on friction. It is the frictional force that accelerates the object must be zero. Gravity is everywhere. the car as a whole. (You will learn about friction For terrestrial phenomena, in particular, every in section 4.9). When the car moves with object experiences gravitational force due to the constant velocity, there is no net external force. earth. Also objects in motion generally experience The property of inertia contained in the First friction, viscous drag, etc. If then, on earth, an law is evident in many situations. Suppose we object is at rest or in uniform linear motion, it is are standing in a stationary bus and the driver not because there are no forces acting on it, but starts the bus suddenly. We get thrown because the various external forces cancel out backward with a jerk. Why ? Our feet are in touch i.e. add up to zero net external force. with the floor. If there were no friction, we would Consider a book at rest on a horizontal surface remain where we were, while the floor of the bus Fig. (4.2(a)). It is subject to two external forces : would simply slip forward under our feet and the the force due to gravity (i.e. its weight W) acting back of the bus would hit us. However, downward and the upward force on the book by fortunately, there is some friction between the the table, the normal force R . R is a self-adjusting feet and the floor. If the start is not too sudden, force. This is an example of the kind of situation i.e. if the acceleration is moderate, the frictional mentioned above. The forces are not quite known force would be enough to accelerate our feet fully but the state of motion is known. We observe along with the bus. But our body is not strictly the book to be at rest. Therefore, we conclude a rigid body. It is deformable, i.e. it allows some from the first law that the magnitude of R equals relative displacement between different parts. that of W. A statement often encountered is : What this means is that while our feet go with “Since W = R, forces cancel and, therefore, the book the bus, the rest of the body remains where it is is at rest”. This is incorrect reasoning. The correct due to inertia. Relative to the bus, therefore, we statement is : “Since the book is observed to be at are thrown backward. As soon as that happens, rest, the net external force on it must be zero, however, the muscular forces on the rest of the according to the first law. This implies that the body (by the feet) come into play to move the body normal force R must be equal and opposite to the along with the bus. A similar thing happens weight W ”. when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest. ⊳ Example 4.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a Fig. 4.2 (a) a book at rest on the table, and (b) a car constant rate of 100 m s–2. What is the moving with uniform velocity. The net force acceleration of the astronaut the instant after is zero in each case. he is outside the spaceship ? (Assume that Consider the motion of a car starting from there are no nearby stars to exert rest, picking up speed and then moving on a gravitational force on him.) smooth straight road with uniform speed (Fig. Answer Since there are no nearby stars to exert(4.2(b)). When the car is stationary, there is no gravitational force on him and the smallnet force acting on it. During pick-up, it spaceship exerts negligible gravitationalaccelerates. This must happen due to a net attraction on him, the net force acting on theexternal force. Note, it has to be an external force. Reprint 2025-26 LAWS OF MOTION 53 astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. ⊳ you may have noticed, he draws in the hands backward in the act of catching the ball4.5 NEWTON’S SECOND LAW OF MOTION (Fig. 4.3). The novice, on the other hand, The first law refers to the simple case when the keeps his hands fixed and tries to catch the net external force on a body is zero. The second ball almost instantly. He needs to provide a law of motion refers to the general situation when much greater force to stop the ball instantly, there is a net external force acting on the body. and this hurts. The conclusion is clear: force It relates the net external force to the not only depends on the change in momentum, acceleration of the body. but also on how fast the change is brought Momentum about. The same change in momentum Momentum of a body is defined to be the product brought about in a shorter time needs a of its mass m and velocity v, and is denoted greater applied force. In short, the greater the by p: rate of change of momentum, the greater is p = m v (4.1) the force. Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. • If two stones, one light and the other heavy, are dropped from the top of a building, a Fig. 4.3 Force not only depends on the change in person on the ground will find it easier to catch momentum but also on how fast the change is brought about. A seasoned cricketer draws the light stone than the heavy stone. The in his hands during a catch, allowing greater mass of a body is thus an important time for the ball to stop and hence requires a parameter that determines the effect of force smaller force. on its motion. • Speed is another important parameter to consider. A bullet fired by a gun can easily • Observations confirm that the product of pierce human tissue before it stops, resulting mass and velocity (i.e. momentum) is basic to in casualty. The same bullet fired with the effect of force on motion. Suppose a fixed moderate speed will not cause much damage. force is applied for a certain interval of time Thus for a given mass, the greater the speed, on two bodies of different masses, initially at the greater is the opposing force needed to stop rest, the lighter body picks up a greater speed the body in a certain time. Taken together, than the heavier body. However, at the end of the product of mass and velocity, that is the time interval, observations show that each momentum, is evidently a relevant variable body acquires the same momentum. Thus of motion. The greater the change in the the same force for the same time causes momentum in a given time, the greater is the the same change in momentum for force that needs to be applied. • A seasoned cricketer catches a cricket ball different bodies. This is a crucial clue to the second law of motion. coming in with great speed far more easily • In the preceding observations, the vector than a novice, who can hurt his hands in the Reprint 2025-26 54 PHYSICS character of momentum has not been evident. ∆p ∆p In the examples so far, momentum and change F ∝ or F = k ∆t ∆ t in momentum both have the same direction. But this is not always the case. Suppose a where k is a constant of proportionality. Taking stone is rotated with uniform speed in a ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 4.4). A force is needed d p to cause this change in momentum vector. respect to t, denoted by . Thus dt This force is provided by our hand through the string. Experience suggests that our hand d p F = k (4.2) needs to exert a greater force if the stone is d t rotated at greater speed or in a circle of For a body of fixed mass m, smaller radius, or both. This corresponds to greater acceleration or equivalently a greater d p d d v = (m v ) = m = m a (4.3) rate of change in momentum vector. This d t d t d t suggests that the greater the rate of change i.e the Second Law can also be written as in momentum vector the greater is the force F = k m a (4.4) applied. which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (4.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp F = = m a (4.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Let us note at this stage some important points Fig. 4.4 Force is necessary for changing the direction about the second law : of momentum, even if its magnitude is constant. We can feel this while rotating a 1. In the second law, F = 0 implies a = 0. The second stone in a horizontal circle with uniform speed law is obviously consistent with the first law. by means of a string. 2. The second law of motion is a vector law. It is These qualitative observations lead to the equivalent to three equations, one for each second law of motion expressed by Newton as component of the vectors : follows : d p x F x = = ma xThe rate of change of momentum of a body is d t directly proportional to the applied force and d p ytakes place in the direction in which the force F y = = ma y acts. d t dp zThus, if under the action of a force F for time F z = =m a z (4.6) dtinterval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum This means that if a force is not parallel to the velocity of the body, but makes some anglep = m v changes by ∆ p = m ∆v . According to the with it, it changes only the component of Second Law, velocity along the direction of force. The Reprint 2025-26 LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical – 90 × 90 – u 2 gravitational force, the horizontal component m s −2 = – 6750 m s −2 a = = of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force = 0.04 kg × 6750 m s-2 = 270 N on the particle and a stands for acceleration of the particle. It turns out, however, that the The actual resistive force, and therefore, law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The answer therefore, only indicates the average even more generally, to a system of particles. resistive force. ⊳ In that case, F refers to the total external force ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is 1 2 the acceleration of the centre of mass of the mass m is described by y = ut + gt . Find 2 system about which we shall study in detail in the force acting on the particle. Chapter 6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y = ut + gt 2 Now, d y v = = u + gt d t dv acceleration, a = = g d t Fig. 4.5 Acceleration at an instant is determined by Then the force is given by Eq. (4.5) the force at that instant. The moment after a F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, if of a particle under acceleration due to gravity air resistance is neglected. The stone carries no memory of its acceleration with the train and y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces determined by the force here and now, not by back, the force on the ball by the wall acts for a any history of the motion of the particle very short time when the two are in contact, yet the force is large enough to reverse the momentum (See Fig. 4.5). of the ball. Often, in these situations, the force ⊳ and the time duration are difficult to ascertain Example 4.2 A bullet of mass 0.04 kg separately. However, the product of force and time, moving with a speed of 90 m s–1 enters a which is the change in momentum of the body heavy wooden block and is stopped after a remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2025-26 56 PHYSICS A large force acting for a short time to produce a Thus, according to Newtonian mechanics, finite change in momentum is called an impulsive force never occurs singly in nature. Force is the force. In the history of science, impulsive forces were mutual interaction between two bodies. Forces put in a conceptually different category from always occur in pairs. Further, the mutual forces ordinary forces. Newtonian mechanics has no such between two bodies are always equal and distinction. Impulsive force is like any other force – opposite. This idea was expressed by Newton in except that it is large and acts for a short time. the form of the third law of motion. ⊳ To every action, there is always an equal and Example 4.4 A batsman hits back a ball opposite reaction. straight in the direction of the bowler without changing its initial speed of 12 m s–1. Newton’s wording of the third law is so crisp and If the mass of the ball is 0.15 kg, determine beautiful that it has become a part of common the impulse imparted to the ball. (Assume language. For the same reason perhaps, linear motion of the ball) misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of theAnswer Change in momentum terms : action and reaction. = 0.15 × 12–(–0.15×12) 1. The terms action and reaction in the third law = 3.6 N s, mean nothing else but ‘force’. Using different Impulse = 3.6 N s, terms for the same physical concept in the direction from the batsman to the bowler. can sometimes be confusing. A simple and clear way of stating the third law is as This is an example where the force on the ball follows :by the batsman and the time of contact of the ball and the bat are difficult to know, but the Forces always occur in pairs. Force on a impulse is readily calculated. ⊳ body A by B is equal and opposite to the force on the body B by A.

5.9 THE POTENTIAL ENERGY OF A SPRING The spring force is an example of a variable force which is conservative. Fig. 5.7 shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force Fs is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 5.7(b)] or negative [Fig. 5.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as Fs = − kx The constant k is called the spring constant. Its unit is N m-1. The spring is said to be stiff if k is large and soft if k is small. Fig. 5.7 Illustration of the spring force with a block Suppose that we pull the block outwards as in attached to the free end of the spring. Fig. 5.7(b). If the extension is xm, the work done by (a) The spring force Fs is zero when the the spring force is displacement x from the equilibrium position is zero. (b) For the stretched spring x > 0 xm xm and Fs < 0 (c) For the compressed spring d x x < 0 and Fs > 0.(d) The plot of Fs versus x. Fs d x = −∫kx W s = ∫ 0 0 The area of the shaded triangle represents the work done by the spring force. Due to the k x m2 opposing signs of Fs and x, this work done is = − (5.15) 2 2 negative, W s = −kx m / 2 . This expression may also be obtained by considering the area of the triangle as in The same is true when the spring is Fig. 5.7(d). Note that the work done by the compressed with a displacement xc (< 0). Theexternal pulling force F is positive since it overcomes the spring force. spring force does work Ws = − kx c2 / 2 while the Reprint 2025-26 WORK, ENERGY AND POWER 81 2 and vice versa, however, the total mechanical external force F does work + kxc / 2. If the block energy remains constant. This is graphically is moved from an initial displacement xi to a depicted in Fig. 5.8. final displacement xf , the work done by the spring force Ws is xf k x 2f k x i2 (5.17) k x d x = − Ws = − ∫ 2 2 x i Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to xi ; x i k x i2 k x i2 k x dx = − Ws = − ∫ 2 2 x i = 0 (5.18) Fig. 5.8 Parabolic plots of the potential energy V and The work done by the spring force in a cyclic kinetic energy K of a block attached to a spring obeying Hooke’s law. The two plotsprocess is zero. We have explicitly demonstrated are complementary, one decreasing as the that the spring force (i) is position dependent other increases. The total mechanical only as first stated by Hooke, (Fs = − kx); (ii) energy E = K + V remains constant. does work which only depends on the initial and final positions, e.g. Eq. (5.17). Thus, the spring ⊳ Example 5.8 To simulate car accidents, autoforce is a conservative force. manufacturers study the collisions of moving We define the potential energy V(x) of the spring cars with mounted springs of different springto be zero when block and spring system is in the constants. Consider a typical simulation withequilibrium position. For an extension (or a car of mass 1000 kg moving with a speedcompression) x the above analysis suggests that 18.0 km/h on a smooth road and colliding kx 2 with a horizontally mounted spring of spring V(x) = (5.19) constant 5.25 × 103 N m–1. What is the 2 maximum compression of the spring ?You may easily verify that − dV/dx = −k x, the spring force. If the block of mass m in Fig. 5.7 is extended to xm and released from rest, then its Answer At maximum compression the kinetic total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the where x lies between – xm and + xm, will be given by potential energy of the spring. The kinetic energy of the moving car is 1 2 1 2 1 2 k x m = k x + m v 1 2 2 2 K = mv2 2where we have invoked the conservation of mechanical energy. This suggests that the speed 1 3 and the kinetic energy will be maximum at the = × 10 × 5 × 5 2 equilibrium position, x = 0, i.e., K = 1.25 × 104 J 1 2 1 2 m v m = k x m where we have converted 18 km h–1 to 5 m s–1 [It is 2 2 useful to remember that 36 km h–1 = 10 m s–1]. where vm is the maximum speed. At maximum compression xm, the potential energy V of the spring is equal to the kinetic k or v m = x m energy K of the moving car from the principle of m conservation of mechanical energy. Note that k/m has the dimensions of [T-2] and our equation is dimensionally correct. The 1 2 V = k x m kinetic energy gets converted to potential energy 2 Reprint 2025-26 82 PHYSICS = 1.25 × 104 J We obtain xm = 2.00 m We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction. ⊳ We conclude this section by making a few Fig. 5.9 The forces acting on the car. remarks on conservative forces. (i) Information on time is absent from the above 1 2 ∆K = Kf − Ki = 0 − m v discussions. In the example considered 2 above, we can calculate the compression, but The work done by the net force is not the time over which the compression 1 2 occurs. A solution of Newton’s Second Law W = − kx m −µm g x m 2 for this system is required for temporal information. Equating we have (ii) Not all forces are conservative. Friction, for 1 2 1 2 example, is a non-conservative force. The m v = k x m + µm g x m 2 2 principle of conservation of energy will have Now µmg = 0.5 × 103 × 10 = 5 × 103 N (taking to be modified in this case. This is illustrated g =10.0 m s-2). After rearranging the above in Example 5.9. equation we obtain the following quadratic(iii) The zero of the potential energy is arbitrary. equation in the unknown xm. It is set according to convenience. For the spring force we took V(x) = 0, at x = 0, i.e. the 2 2 k x m + 2µm g x m − m v = 0 unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later chapter we shall see that for the force where we take the positive square root since due to the universal law of gravitation, the zero is best defined at an infinite distance xm is positive. Putting in numerical values we obtain from the gravitational source. However, once the zero of the potential energy is fixed in a xm = 1.35 m given discussion, it must be consistently which, as expected, is less than the result in adhered to throughout the discussion. You Example 5.8. cannot change horses in midstream ! If the two forces on the body consist of a conservative force Fc and a non-conservative⊳ force Fnc , the conservation of mechanical energy Example 5.9 Consider Example 5.8 taking formula will have to be modified. By the WE the coefficient of friction, µ, to be 0.5 and theorem calculate the maximum compression of the spring. (Fc+ Fnc) ∆x = ∆K But Fc ∆x = − ∆V Answer In presence of friction, both the spring Hence, ∆(K + V) = Fnc ∆x force and the frictional force act so as to oppose ∆E = Fnc ∆x the compression of the spring as shown in where E is the total mechanical energy. Over Fig. 5.9. the path this assumes the form We invoke the work-energy theorem, rather Ef − Ei = Wnc than the conservation of mechanical energy. where Wnc is the total work done by the The change in kinetic energy is non-conservative forces over the path. Note that Reprint 2025-26 WORK, ENERGY AND POWER 83 unlike the conservative force, Wnc depends on Our electricity bills carry the energy the particular path i to f. ⊳ consumption in units of kWh. Note that kWh is a unit of energy and not of power.

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