5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳

5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x Fig. 5.3(a) Reprint 2025-26 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x →0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is Reprint 2025-26 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg . g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts

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