5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. ⊳ Example 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

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