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

13.2 PERIODIC AND OSCILLATORY MOTIONS Very often, the body undergoing periodic motion has an equilibrium position somewhereFig. 13.1 shows some periodic motions. Suppose inside its path. When the body is at this positionan insect climbs up a ramp and falls down, it no net external force acts on it. Therefore, if it iscomes back to the initial point and repeats the left there at rest, it remains there forever. If the process identically. If you draw a graph of its body is given a small displacement from the height above the ground versus time, it would position, a force comes into play which tries to look something like Fig. 13.1 (a). If a child climbs bring the body back to the equilibrium point, up a step, comes down, and repeats the process giving rise to oscillations or vibrations. For identically, its height above the ground would example, a ball placed in a bowl will be in look like that in Fig. 13.1 (b). When you play the equilibrium at the bottom. If displaced a little game of bouncing a ball off the ground, between from the point, it will perform oscillations in the your palm and the ground, its height versus time bowl. Every oscillatory motion is periodic, but graph would look like the one in Fig. 13.1 (c). every periodic motion need not be oscillatory. Note that both the curved parts in Fig. 13.1 (c) Circular motion is a periodic motion, but it is are sections of a parabola given by the Newton’s not oscillatory. equation of motion (see section 2.6), There is no significant difference between 1 2 oscillations and vibrations. It seems that when h = ut + gt for downward motion, and 2 the frequency is small, we call it oscillation (like, the oscillation of a branch of a tree), while when 1 2 h = ut – gt for upward motion, the frequency is high, we call it vibration (like, 2 the vibration of a string of a musical instrument). with different values of u in each case. These Simple harmonic motion is the simplest form are examples of periodic motion. Thus, a motion of oscillatory motion. This motion arises when that repeats itself at regular intervals of time is the force on the oscillating body is directly called periodic motion. proportional to its displacement from the mean position, which is also the equilibrium position. Further, at any point in its oscillation, this force is directed towards the mean position. In practice, oscillating bodies eventually (a) come to rest at their equilibrium positions because of the damping due to friction and other dissipative causes. However, they can be forced to remain oscillating by means of some external periodic agency. We discuss the phenomena of damped and forced oscillations later in the chapter. Any material medium can be pictured as a (b) collection of a large number of coupled oscillators. The collective oscillations of the constituents of a medium manifest themselves as waves. Examples of waves include water waves, seismic waves, electromagnetic waves. We shall study the wave phenomenon in the next chapter. (c) 13.2.1 Period and frequency We have seen that any motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called its Fig. 13.1 Examples of periodic motion. The period T period. Let us denote the period by the symbol is shown in each case. T. Its SI unit is second. For periodic motions, Reprint 2025-26 OSCILLATIONS 261 which are either too fast or too slow on the scale as a displacement variable [see Fig.13.2(b)]. The of seconds, other convenient units of time are term displacement is not always to be referred used. The period of vibrations of a quartz crystal is expressed in units of microseconds (10–6 s) abbreviated as µs. On the other hand, the orbital period of the planet Mercury is 88 earth days. The Halley’s comet appears after every 76 years. The reciprocal of T gives the number of repetitions that occur per unit time. This quantity is called the frequency of the periodic motion. It is represented by the symbol ν. The relation between ν and T is Fig. 13.2(a) A block attached to a spring, the other ν = 1/T (13.1) end of which is fixed to a rigid wall. The block moves on a frictionless surface. The The unit of ν is thus s–1. After the discoverer of motion of the block can be described in radio waves, Heinrich Rudolph Hertz (1857–1894), terms of its distance or displacement x a special name has been given to the unit of from the equilibrium position. frequency. It is called hertz (abbreviated as Hz). Thus, 1 hertz = 1 Hz =1 oscillation per second =1 s–1 (13.2) Note, that the frequency, ν, is not necessarily an integer. u Example 13.1 On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period. Fig.13.2(b) An oscillating simple pendulum; its Answer The beat frequency of heart = 75/(1 min) motion can be described in terms of = 75/(60 s) angular displacement θ from the vertical. = 1.25 s–1 = 1.25 Hz in the context of position only. There can be The time period T = 1/(1.25 s–1) many other kinds of displacement variables. The = 0.8 s ⊳ voltage across a capacitor, changing with time in an AC circuit, is also a displacement variable.13.2.2 Displacement In the same way, pressure variations in time inIn section 3.2, we defined displacement of a the propagation of sound wave, the changingparticle as the change in its position vector. In electric and magnetic fields in a light wave arethis chapter, we use the term displacement examples of displacement in different contexts.in a more general sense. It refers to change The displacement variable may take bothwith time of any physical property under positive and negative values. In experiments onconsideration. For example, in case of rectilinear oscillations, the displacement is measured formotion of a steel ball on a surface, the distance different times.from the starting point as a function of time is The displacement can be represented by a its position displacement. The choice of origin mathematical function of time. In case of periodic is a matter of convenience. Consider a block motion, this function is periodic in time. One of attached to a spring, the other end of the spring the simplest periodic functions is given by is fixed to a rigid wall [see Fig.13.2(a)]. Generally, it is convenient to measure displacement of the f (t) = A cos ωt (13.3a) body from its equilibrium position. For an If the argument of this function, ωt, is oscillating simple pendulum, the angle from the increased by an integral multiple of 2π radians, vertical as a function of time may be regarded the value of the function remains the same. The Reprint 2025-26 262 PHYSICS function f (t) is then periodic and its period, T, (ii) This is an example of a periodic motion. It is given by can be noted that each term represents a 2 π periodic function with a different angular T = (13.3b) frequency. Since period is the least interval ω of time after which a function repeats its Thus, the function f (t) is periodic with period T, value, sin ωt has a period T0= 2π/ω ; cos 2 ωt f (t) = f (t+T ) has a period π/ω =T0/2; and sin 4 ωt has a period 2π/4ω = T0/4. The period of the firstThe same result is obviously correct if we term is a multiple of the periods of the lastconsider a sine function, f (t ) = A sin ωt. Further, two terms. Therefore, the smallest intervala linear combination of sine and cosine functions of time after which the sum of the threelike, terms repeats is T0, and thus, the sum is a f (t) = A sin ωt + B cos ωt (13.3c) periodic function with a period 2π/ω. is also a periodic function with the same period (iii) The function e–ωt is not periodic, itT. Taking, decreases monotonically with increasing A = D cos φ and B = D sin φ time and tends to zero as t → ∞ and thus, Eq. (13.3c) can be written as, never repeats its value. (iv) The function log(ωt) increases f (t) = D sin (ωt + φ ) , (13.3d) monotonically with time t. It, therefore, Here D and φ are constant given by never repeats its value and is a non- periodic function. It may be noted that as  B  t → ∞, log(ωt) diverges to ∞. It, therefore, 2 2 – 1 D = A + B and φ= tan  A  cannot represent any kind of physical displacement. ⊳ The great importance of periodic sine and cosine functions is due to a remarkable result 13.3 SIMPLE HARMONIC MOTION proved by the French mathematician, Jean Consider a particle oscillating back and forth Baptiste Joseph Fourier (1768–1830): Any about the origin of an x-axis between the limits periodic function can be expressed as a +A and –A as shown in Fig. 13.3. This oscillatory superposition of sine and cosine functions motion is said to be simple harmonic if the of different time periods with suitable displacement x of the particle from the origin coefficients. varies with time as : x (t) = A cos (ω t + φ) (13.4) u Example 13.2 Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [ω is any positive constant]. (i) sin ωt + cos ωt Fig. 13.3 A particle vibrating back and forth about the origin of x-axis, between the limits +A (ii) sin ωt + cos 2 ωt + sin 4 ωt and –A. (iii) e–ωt (iv) log (ωt) where A, ω and φ are constants. Thus, simple harmonic motion (SHM) is not Answer any periodic motion but one in which displacement is a sinusoidal function of time.(i) sin ωt + cos ωt is a periodic function, it can Fig. 13.4 shows the positions of a particle also be written as 2 sin (ωt + π/4). executing SHM at discrete value of time, each Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π) interval of time being T/4, where T is the period of motion. Fig. 13.5 plots the graph of x versus t, = 2 sin [ω (t + 2π/ω) + π/4] which gives the values of displacement as a The periodic time of the function is 2π/ω. continuous function of time. The quantities A, Reprint 2025-26 OSCILLATIONS 263 any loss of generality]. As the cosine function of time varies from +1 to –1, the displacement varies between the extremes A and – A. Two simple harmonic motions may have same ω and φ but different amplitudes A and B, as shown in Fig. 13.7 (a). While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of the particle at any time t is determined by the Fig. 13.4 The location of the particle in SHM at the discrete values t = 0, T/4, T/2, 3T/4, T, 5T/4. The time after which motion repeats itself is T. T will remain fixed, no matter what location you choose as the initial (t = Fig. 13.7 (a) A plot of displacement as a function of 0) location. The speed is maximum for zero time as obtained from Eq. (14.4) with displacement (at x = 0) and zero at the φ = 0. The curves 1 and 2 are for two extremes of motion. different amplitudes A and B. ω and φ which characterize a given SHM have standard names, as summarised in Fig. 13.6. argument (ωt + φ) in the cosine function. This Let us understand these quantities. time-dependent quantity, (ωt + φ) is called the The amplitutde A of SHM is the magnitude phase of the motion. The value of plase at t = 0 of maximum displacement of the particle. is φ and is called the phase constant (or phase [Note, A can be taken to be positive without angle). If the amplitude is known, φ can be determined from the displacement at t = 0. Two simple harmonic motions may have the same A and ω but different phase angle φ, as shown in Fig. 13.7 (b). Finally, the quantity ω can be seen to be related to the period of motion T. Taking, for simplicity, φ = 0 in Eq. (13.4), we have Fig. 13.5 Displacement as a continuous function of time for simple harmonic motion. x (t) : displacement x as a function of time t A : amplitude ω : angular frequency ωt + φ : phase (time-dependent) φ : phase constant Fig. 13.7 (b) A plot obtained from Eq. (13.4). The curves 3 and 4 are for φ = 0 and -π/4 respectively. The amplitude A is same for Fig. 13.6 The meaning of standard symbols both the plots. in Eq. (13.4) Reprint 2025-26 264 PHYSICS x(t) = A cos ωt (13.5) This function represents a simple harmonic motion having a period T = 2π/ω and a Since the motion has a period T, x (t) is equal to phase angle (–π/4) or (7π/4) x (t + T). That is, (b) sin2 ωt = ½ – ½ cos 2 ωt A cos ωt = A cos ω (t + T ) (13.6) The function is periodic having a period Now the cosine function is periodic with period T = π/ω. It also represents a harmonic 2π, i.e., it first repeats itself when the argument motion with the point of equilibrium ½ instead of zero. ⊳changes by 2π. Therefore, occurring at ω(t + T ) = ωt + 2π 13.4 SIMPLE HARMONIC MOTION AND UNIFORM CIRCULAR MOTION that is ω = 2π/ T (13.7) In this section, we show that the projection of uniform circular motion on a diameter of the ω is called the angular frequency of SHM. Its circle follows simple harmonic motion. A S.I. unit is radians per second. Since the simple experiment (Fig. 13.9) helps us visualise frequency of oscillations is simply 1/T, ω is 2π this connection. Tie a ball to the end of a string times the frequency of oscillation. Two simple and make it move in a horizontal plane about harmonic motions may have the same A and φ, a fixed point with a constant angular speed. but different ω, as seen in Fig. 13.8. In this plot The ball would then perform a uniform circular the curve (b) has half the period and twice the motion in the horizontal plane. Observe the frequency of the curve (a). ball sideways or from the front, fixing your attention in the plane of motion. The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the midpoint. You could alternatively observe the shadow of the ball on a wall which is perpendicular to the plane of the circle. In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing. Fig. 13.8 Plots of Eq. (13.4) for φ = 0 for two different periods. u Example 13.3 Which of the following functions of time represent (a) simple Fig. 13.9 Circular motion of a ball in a plane viewed harmonic motion and (b) periodic but not edge-on is SHM. simple harmonic? Give the period for each case. Fig. 13.10 describes the same situation (1) sin ωt – cos ωt mathematically. Suppose a particle P is moving (2) sin2 ωt uniformly on a circle of radius A with angular Answer speed ω. The sense of rotation is anticlockwise. (a) sin ωt – cos ωt The initial position vector of the particle, i.e., = sin ωt – sin (π/2 – ωt) the vector OP at t = 0 makes an angle of φ with = 2 cos (π/4) sin (ωt – π/4) the positive direction of x-axis. In time t, it will = √2 sin (ωt – π/4) cover a further angle ωt and its position vector Reprint 2025-26 OSCILLATIONS 265 u Example 13.4 The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. Fig. 13.10 will make an angle of ωt + φ with the +ve x-axis. Next, consider the projection of the position vector OP on the x-axis. This will be Answer OP′. The position of P′ on the x-axis, as the (a) At t = 0, OP makes an angle of 45o = π/4 rad particle P moves on the circle, is given by with the (positive direction of) x-axis. After x(t) = A cos (ωt + φ ) 2 π time t, it covers an angle t in thewhich is the defining equation of SHM. This T shows that if P moves uniformly on a circle, anticlockwise sense, and makes an angle its projection P′ on a diameter of the circle executes SHM. The particle P and the circle of 2 πt + π with the x-axis. on which it moves are sometimes referred to T 4 as the reference particle and the reference circle, The projection of OP on the x-axis at time t respectively. is given by, We can take projection of the motion of P on any diameter, say the y-axis. In that case, the  2π π  x (t) = A cos t +displacement y(t) of P′ on the y-axis is given by  T 4  y = A sin (ωt + φ) For T = 4 s, which is also an SHM of the same amplitude as that of the projection on x-axis, but differing  2π π  x(t) = A cos t +by a phase of π/2.  4 4  In spite of this connection between circular motion and SHM, the force acting on a particle which is a SHM of amplitude A, period 4 s, in linear simple harmonic motion is very πdifferent from the centripetal force needed to and an initial phase* = . keep a particle in uniform circular motion. 4 * The natural unit of angle is radian, defined through the ratio of arc to radius. Angle is a dimensionless quantity. Therefore it is not always necessary to mention the unit ‘radian’ when we use π, its multiples or submultiples. The conversion between radian and degree is not similar to that between metre and centimetre or mile. If the argument of a trigonometric function is stated without units, it is understood that the unit is radian. On the other hand, if degree is to be used as the unit of angle, then it must be shown explicitly. For example, sin(150) means sine of 15 degree, but sin(15) means sine of 15 radians. Hereafter, we will often drop ‘rad’ as the unit, and it should be understood that whenever angle is mentioned as a numerical value, without units, it is to be taken as radians. Reprint 2025-26 266 PHYSICS (b) In this case at t = 0, OP makes an angle of where the negative sign shows that v (t) has a π direction opposite to the positive direction of 90o = with the x-axis. After a time t, it x-axis. Eq. (13.9) gives the instantaneous 2 2π velocity of a particle executing SHM, where covers an angle of t in the clockwise T displacement is given by Eq. (13.4). We can, of  π 2π  course, obtain this equation without using sense and makes an angle of  2 − T t  geometrical argument, directly by differentiating (Eq. 13.4) with respect of t: with the x-axis. The projection of OP on the x-axis at time t is given by d v(t) = x (t ) (13.10)  π 2π  d t x(t) = B cos  2 − T t  The method of reference circle can be similarly used for obtaining instantaneous acceleration  2π  of a particle undergoing SHM. We know that the = B sin  T t  centripetal acceleration of a particle P in uniform For T = 30 s, circular motion has a magnitude v2/A or ω2A, and it is directed towards the centre i.e., the  π  direction is along PO. The instantaneous x(t) = B sin  15 t  acceleration of the projection particle P′ is then (See Fig. 13.12)  π π  a (t) = –ω2A cos (ωt + φ) Writing this as x (t) = B cos  15 t − 2 , and comparing with Eq. (13.4). We find that this = –ω2x (t) (13.11) represents a SHM of amplitude B, period 30 s, π and an initial phase of − . ⊳ 2

27.3 days which is also roughly equal to the Which is approximately 85 minutes. rotational period of the moon about its own axis. ⊳ Example 7.5 The planet Mars has twoSince, 1957, advances in technology have enabled moons, phobos and delmos. (i) phobos hasmany countries including India to launch artificial a period 7 hours, 39 minutes and an orbitalearth satellites for practical use in fields like radius of 9.4 ×103 km. Calculate the masstelecommunication, geophysics and meteorology. of mars. (ii) Assume that earth and mars We will consider a satellite in a circular orbit move in circular orbits around the sun,of a distance (RE + h) from the centre of the earth, with the martian orbit being 1.52 timeswhere RE = radius of the earth. If m is the mass the orbital radius of the earth. What isof the satellite and V its speed, the centripetal the length of the martian year in days ?force required for this orbit is mV 2 Answer (i) We employ Eq. (7.38) with the sun’s F(centripetal) = (7.33) ( R E + h ) mass replaced by the martian mass Mm directed towards the centre. This centripetal force 2 4 π 2 3 T = Ris provided by the gravitational force, which is GM m G m M E 4 π 2 R 3 F(gravitation) = 2 (7.34) Mm = 2 ( R E + h ) G T where ME is the mass of the earth. 2 3 18 Equating R.H.S of Eqs. (7.33) and (7.34) and 4 × ( 3.14 ) × ( 9.4 ) × 10 = -11 2cancelling out m, we get 6.67 × 10 × ( 459 × 60 ) 2 G M E 2 3 18 V = (7.35) 4 × ( 3.14 ) × ( 9.4 ) × 10 ( R E + h ) M m = 2 -5 6.67 × ( 4.59 × 6 ) × 10 Thus V decreases as h increases. From = 6.48 × 1023 kg. equation (7.35),the speed V for h = 0 is (ii) Once again Kepler’s third law comes to our V 2 (h = 0) = GM / R E = gR E (7.36) aid, where we have used the relation T M2 R MS3 2 2 = 3 g = GM / R E . In every orbit, the satellite T E R ES Reprint 2025-26 138 PHYSICS where RMS is the mars -sun distance and RES is − 13  1 2   1    d the earth-sun distance. = 10    ( 24 × 60 × 60 ) 2  ( 1 / 1000 ) 3 km 3  ∴ TM = (1.52)3/2 × 365 = 1.33 ×10–14 d2 km–3 = 684 days Using Eq. (7.38) and the given value of k, We note that the orbits of all planets except the time period of the moon is Mercury and Mars are very close to being 2 T = (1.33 × 10-14)(3.84 × 105)3 circular. For example, the ratio of the semi- T = 27.3 d ⊳ minor to semi-major axis for our Earth is, Note that Eq. (7.38) also holds for elliptical b/a = 0.99986. ⊳ orbits if we replace (RE+h) by the semi-major axis ⊳ of the ellipse. The earth will then be at one of Example 7.6 Weighing the Earth : You the foci of this ellipse. are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the moon R 7.10 ENERGY OF AN ORBITING SATELLITE = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the Using Eq. (7.35), the kinetic energy of the satellite mass of the Earth ME in two different ways. in a circular orbit with speed v is 1 m v 2Answer From Eq. (7.12) we have K i E = 2 g R E2 M E = Gm M E G = , (7.40) 2( R E + h ) 6 2 Considering gravitational potential energy at 9.81 × ( 6.37 × 10 ) = -11 infinity to be zero, the potential energy at distance 6.67 × 10 (Re+h) from the centre of the earth is = 5.97× 1024 kg. The moon is a satellite of the Earth. From G m M E P .E = − (7.41)the derivation of Kepler’s third law [see Eq. ( R E + h ) (7.38)] The K.E is positive whereas the P.E is 2 4 π2R 3 negative. However, in magnitude the K.E. is half T = G M E the P.E, so that the total E is E 4 π2R 3 E = K .E + P .E = − G m M ME = G T 2 2( R E + h ) (7.42) 4 × 3.14 × 3.14 × ( 3.84 ) 3 × 10 24 The total energy of an circularly orbiting = -11 2 satellite is thus negative, with the potential 6.67 × 10 × ( 27.3 × 24 × 60 × 60 ) energy being negative but twice is magnitude of = 6.02 × 1024 kg the positive kinetic energy. Both methods yield almost the same answer, When the orbit of a satellite becomes the difference between them being less than 1%. elliptic, both the K.E. and P.E. vary from point ⊳ to point. The total energy which remains constant is negative as in the circular orbit case. ⊳ Example 7.7 Express the constant k of Eq. This is what we expect, since as we have (7.38) in days and kilometres. Given discussed before if the total energy is positive or k = 10–13 s2 m–3. The moon is at a distance zero, the object escapes to infinity. Satellites of 3.84 × 105 km from the earth. Obtain its are always at finite distance from the earth and time-period of revolution in days. hence their energies cannot be positive or zero. Answer Given k = 10–13 s2 m–3 Reprint 2025-26 GRAVITATION 139 The change in the total energy is⊳ Example 7.8 A 400 kg satellite is in a circular ∆E = Ef – Ei orbit of radius 2RE about the Earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in  E E = G M E m =  the kinetic and potential energies ?  G M2  m R 8 R E  R E  8 Answer Initially, g m R E = 9.81 × 400 × 6. 37 × 106 = 3.13 × 10 9 J ∆ E = G M E m 8 8 E i = − 4 R E The kinetic energy is reduced and it mimics While finally ∆E, namely, ∆K = Kf – Ki = – 3.13 × 109 J. The change in potential energy is twice the G M E m E f = − change in the total energy, namely 8 R E ∆V = Vf – Vi = – 6.25 × 109 J ⊳ SUMMARY 1. Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude m 1m 2 F = G 2 r where G is the universal gravitational constant, which has the value 6.672 ×10–11 N m2 kg–2. 2. If we have to find the resultant gravitational force acting on the particle m due to a number of masses M1, M2, ….Mn etc. we use the principle of superposition. Let F1, F2, ….Fn be the individual forces due to M1, M2, ….Mn, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force FR is then found by vector addition n FR = F1 + F2 + ……+ Fn = ∑ Fi i = 1 where the symbol ‘Σ’ stands for summation. 3. Kepler’s laws of planetary motion state that (a) All planets move in elliptical orbits with the Sun at one of the focal points (b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved. (c) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related by 2  4 π 2  3 T =   R  G M s  where Ms is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major axis, a. 4. The acceleration due to gravity. (a) at a height h above the earth’s surface G M E g ( h ) = 2 ( R E + h ) G M E  2 h  ≈ 2 1 − for h << RE R E  R E  Reprint 2025-26 140 PHYSICS  2 h  G M E g (h ) = g ( 0 ) 1 − where g ( 0 ) = 2  R E  R E (b) at depth d below the earth’s surface is d   1 − g (d ) = G M2 E 1 − d  = g ( 0 ) R E  R E   R E  5. The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is given by G m1 m 2 V = − r where V is taken to be zero at r → ∞. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition. 6. If an isolated system consists of a particle of mass m moving with a speed v in the vicinity of a massive body of mass M, the total mechanical energy of the particle is given by 1 G M m E = m v 2− 2 r That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion. 7. If m moves in a circular orbit of radius a about M, where M >> m, the total energy of the system is G M m E = − 2a with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are G M m K = 2a G M m V = − a 8. The escape speed from the surface of the earth is 2 G M E ve = = 2 gR E R E and has a value of 11.2 km s–1. 9. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere. 10. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle. Reprint 2025-26 GRAVITATION 141 POINTS TO PONDER 1. In considering motion of an object under the gravitational influence of another object the following quantities are conserved: (a) Angular momentum (b) Total mechanical energy Linear momentum is not conserved 2. Angular momentum conservation leads to Kepler’s second law. However, it is not special to the inverse square law of gravitation. It holds for any central force. 3. In Kepler’s third law (see Eq. (7.1) and T2 = KS R3. The constant KS is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)]. 4. An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in “free fall” towards the Earth. 5. The gravitational potential energy associated with two particles separated by a distance r is given by G m 1 m 2 V = – + constant r The constant can be given any value. The simplest choice is to take it to be zero. With this choice G m 1 m 2 V = – r This choice implies that V → 0 as r → ∞. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant. 6. The total mechanical energy of an object is the sum of its kinetic energy (which is always positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative. 7. The commonly encountered expression m g h for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above. 8. Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central. 9. The gravitational force on a particle inside a spherical shell is zero. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible. EXERCISES 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ? Reprint 2025-26 142 PHYSICS 7.2 Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth. 7.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ? 7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun. 7.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly. 7.6 Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence. 7.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? 7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. 7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem. 7.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0. Fig. 7.11 7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. 7.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m). 7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km. 7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ? 7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? Reprint 2025-26 GRAVITATION 143 7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ? 7.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G). 7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ? Reprint 2025-26