Oscillations PDF

CHAPTER THIRTEEN OSCILLATIONS 13.1 INTRODUCTION In our daily life we come across various kinds of motions. You have already learnt about some of them, e...

CHAPTER THIRTEEN OSCILLATIONS 13.1 INTRODUCTION In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear 13.1 Introduction motion and motion of a projectile. Both these motions are 13.2 Periodic and oscillatory non-repetitive. We have also learnt about uniform circular motions motion and orbital motion of planets in the solar system. In 13.3 Simple harmonic motion these cases, the motion is repeated after a certain interval of 13.4 Simple harmonic motion time, that is, it is periodic. In your childhood, you must have and uniform circular enjoyed rocking in a cradle or swinging on a swing. Both motion these motions are repetitive in nature but different from the 13.5 Velocity and acceleration periodic motion of a planet. Here, the object moves to and fro in simple harmonic motion about a mean position. The pendulum of a wall clock executes 13.6 Force law for simple a similar motion. Examples of such periodic to and fro harmonic motion motion abound: a boat tossing up and down in a river, the 13.7 Energy in simple harmonic piston in a steam engine going back and forth, etc. Such a motion motion is termed as oscillatory motion. In this chapter we 13.8 The simple pendulum study this motion. Summary The study of oscillatory motion is basic to physics; its Points to ponder Exercises concepts are required for the understanding of many physical phenomena. In musical instruments, like the sitar, the guitar or the violin, we come across vibrating strings that produce pleasing sounds. The membranes in drums and diaphragms in telephone and speaker systems vibrate to and fro about their mean positions. The vibrations of air molecules make the propagation of sound possible. In a solid, the atoms vibrate about their equilibrium positions, the average energy of vibrations being proportional to temperature. AC power supply give voltage that oscillates alternately going positive and negative about the mean value (zero). The description of a periodic motion, in general, and oscillatory motion, in particular, requires some fundamental concepts, like period, frequency, displacement, amplitude and phase. These concepts are developed in the next section. 2024-25 260 PHYSICS 13.2 PERIODIC AND OSCILLATORY MOTIONS Very often, the body undergoing periodic Fig. 13.1 shows some periodic motions. Suppose motion has an equilibrium position somewhere an insect climbs up a ramp and falls down, it inside its path. When the body is at this position comes back to the initial point and repeats the no net external force acts on it. Therefore, if it is 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 the frequency is small, we call it oscillation (like, 2 the oscillation of a branch of a tree), while when 1 2 the frequency is high, we call it vibration (like, h = ut – gt for upward motion, 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, 2024-25 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 13.2.2 Displacement in an A C circuit, is also a displacement variable. In section 3.2, we defined displacement of a In the same way, pressure variations in time in particle as the change in its position vector. In the propagation of sound wave, the changing this chapter, we use the term displacement electric and magnetic fields in a light wave are in a more general sense. It refers to change examples of displacement in different contexts. with time of any physical property under The displacement variable may take both consideration. For example, in case of rectilinear positive and negative values. In experiments on motion of a steel ball on a surface, the distance oscillations, the displacement is measured for from the starting point as a function of time is different times. 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 2024-25 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 2π 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 The same result is obviously correct if we period 2π/4ω = T0/4. The period of the first consider a sine function, f (t ) = A sin ωt. Further, term is a multiple of the periods of the last a linear combination of sine and cosine functions two terms. Therefore, the smallest interval like, of time after which the sum of the three 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 T. Taking, (iii) The function e – ω t is not periodic, it 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. f (t) = D sin (ωt + φ ) , (13.3d) (iv) The function log ( ω t) increases 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, D = A 2 + B 2 and φ = tan –1  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 (i) sin ωt + cos ωt is a periodic function, it can displacement is a sinusoidal function of time. 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, 2024-25 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 Fig. 13.6 The meaning of standard symbols respectively. The amplitude A is same for in Eq. (13.4) both the plots. 2024-25 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 changes by 2π. Therefore, occurring at ½ instead of zero. ⊳ 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 (2) sin2 ωt mathematically. Suppose a particle P is moving 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 2024-25 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 ω t + 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 ) which is the defining equation of SHM. This time t, it covers an angle 2πt in the 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 2πt + 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 x (t) = A cos  2π π displacement y(t) of P′ on the y-axis is given by t+   T 4 y = A sin (ωt + φ ) For T = 4 s, which is also an SHM of the same amplitude x(t) = A cos  as that of the projection on x-axis, but differing 2π π by a phase of π/2. t+   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 different from the centripetal force needed to and an initial phase* = π. 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. 2024-25 266 PHYSICS (b) In this case at t = 0, OP makes an angle of where the negative sign shows that v (t) has a 90o = π with the x-axis. After a time t, it direction opposite to the positive direction of 2 x-axis. Eq. (13.9) gives the instantaneous covers an angle of 2π t in the clockwise velocity of a particle executing SHM, where T displacement is given by Eq. (13.4). We can, of course, obtain this equation without using sense and makes an angle of  − π 2π   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) dt x(t) = B cos  − π 2π   2 T  t The method of reference circle can be similarly used for obtaining instantaneous acceleration = B sin  t  2π of a particle undergoing SHM. We know that the  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 x(t) = B sin  t  direction is along PO. The instantaneous π  15  acceleration of the projection particle P′ is then (See Fig. 13.12) Writing this as x (t) = B cos  t −  , and π π  15 a (t) = –ω2A cos (ωt + φ) 2 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 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 = ωA (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 geometry of Fig. 13.11, it is clear that the velocity Fig. 13.12 The acceleration, a(t), of the particle P′ is the projection of the acceleration a of the of 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 (13.12) a (t ) = v (t ) dt 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 2024-25 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 [(2 π 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 = 10 π × 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 [(2 π 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. 2024-25 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 situation the spring on the left side gets K= mv 2 2 1 = m ω 2 A 2 sin 2 (ωt + φ ) 2 1 = k A 2 sin2 (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 position) U= k x2 (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) The net force, F, acting on the mass is then 1 U(x) = k x2 given by, 2 F = –2kx 1 Hence the force acting on the mass is = k A 2 cos2 (ωt + φ ) (13.17) proportional to the displacement and is directed 2 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 = 2π displacements. 2k ⊳ 2024-25 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 E =U+K positive in Fig. 13.16. Kinetic energy can, of course, be never negative, since it is proportional to the square of speed. Potential 1 1 energy is positive by choice of the undermined = k A 2 cos2 (ωt + φ ) + k A 2 sin2 (ωt + φ ) 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 = k A 2 cos2 (ωt + φ ) + sin 2 (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 E= k A2 (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 oscillator is thus independent of time as pulled to a distance x = 10 cm from its expected for motion under any conservative equilibrium position at x = 0 on a frictionless 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) 2024-25 270 PHYSICS Or cos (7.07t) = 0.5 and hence let it go. The stone executes a to and fro motion, 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 = m v2 2 = ½[1kg × (0.6123 m s–1 )2 ] = 0.19 J (a) The P.E. of the block, 1 = k x2 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 about the support. The tangential force = ½(50 N m–1 × 0.1 m × 0.1 m ) 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 13.8 The Simple Pendulum the string is fixed to a rigid support. The bob It is said that Galileo measured the periods of a oscillates in a plane about the vertical line swinging chandelier in a church by his pulse through the support. Fig. 13.17(a) shows this beats. He observed that the motion of the system. Fig. 13.17(b) is a kind of ‘free-body’ 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 2024-25 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 work with torque about the support since the Equation (13.24) is mathematically, identical to radial force gives zero torque. Torque τ about Eq. (13.11) except that the variable is angular the support is entirely provided by the tangental displacement. Hence we have proved that for component of force small q, the motion of the bob is simple harmonic. From Eqs. (13.24) and (13.11), τ = –L (mg sinθ ) (13.19) This is the restoring torque that tends to reduce mgL angular displacement — hence the negative ω = sign. By Newton’s law of rotational motion, I τ = Iα (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 gL α = − sin θ (13.22) L I T = 2π (13.26) We can simplify Eq. (13.22) if we assume that g 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 α = − mgL θ L = (13.24) 4π 2 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(s2 ) = sin θ is nearly the same as θ expressed 4π 2 in radians. =1m ⊳ 2024-25 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 2024-25 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. 2024-25 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 2024-25 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. 2024-25 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 ? 2024-25 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 ρ1g 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. 2024-25