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5. Oscillations 12th Science 12th - Physics Syllabus 1. Introduction 2. Explanation of periodic motion 3. Linear simple harmonic motion (S.H.M.) 4. Differential equation of S.H.M. 5. Acceleration (a), velocity (v) and displacement (x) of S.H.M. 6. Amplitude (A), period (T) and frequency (n) of S.H.M. 6.1 Amplitude of S.H.M. 6.2 Period of S.H.M. 6.3 Frequency of S.H.M. 7. Reference Circle Method 8. Phase in S.H.M. 9. Graphical Representation of S.H.M. 10. Composition of two S.H.M.s having same period and along the same path 11. Energy of a Particle Performing S.H.M. 12. Simple Pendulum 13. Angular S.H.M. and its Differential Equation 13.1 Magnet Vibrating in Uniform Magnetic Field 14. Damped Oscillations 15. Free Oscillations, Forced Oscillations and Resonance Theory Notes Q. Define periodic motion. Periodic Motion i " : ‘ val of 1) The motion (or process) which repeats itself in equal inter time is called as periodic motion. nt is called as Q. What is oscillatory motion? Q. Define oscillation. Q. What is linear simple harmonic motion? Q. Define angular simple harmonic motion. Q. Define frequency of the periadic motion. Q. Explain SHM. 2) The time taken for one such a set of moveme: periodic time or period. 3) Ex. i) Motion of tip of hand of clock. ii) Motion of planets around the sun. iii) Motion of blades of fan. pe Oscillatory Motion ic 1) The to and fro motion of a body about the mean position 1S called as oscillatory motion. 2) Ex. _ £i) Motion of pendulum of wall clock. s Motion of piston of engine of car. tal iii) Motion of prongs of tuning fork. j “ iv) Motion of needle of sewing machine. v) Vibration of string of musical instrument. Oscillation A set of movement which start from one path the body comes back to the original oscillation. ; (The repeating to and fro motion of the body along the same path in equal interval of time is called as oscillatory motion. | ——— same point and after completing point is called one Linear Simple Harmonic Motion (Linear S.H.M.) . 1) The motion in which object moves ‘to and fro’ along the straight line is called as linear simple harmonic motion. 2) Ex. i) Motion of pendulum of clock. ii) Motion of needle of saving machine. iii) Motion of spring loaded with mass. * Every oscillatory motion is periodic but every periodic motion need not to be oscillatory. e.g. circular motion is periodic but it is not oscillatory. | Angular Simple Harmonic Motion The motion in which object moves to and fro along an arc of a circle is called angular simple harmonic motion. Frequency (n) The number of oscillations completed per unit time is called the frequency of periodic motion. Liner Simple Harmonic Motion (Linear S.H.M.) OR Oscillatory Motion 1) The to and fro motion of a body along straight line is called as linear simple harmonic motion. 2) Consider a body of mass 'm' attached to one end of an ideal spring of force constant k. The other end of the spring is fixed at rigid support. The spring is free to move over a frictionless horizontal surface. 3) As long as the spring remains in equilibrium (fig b), it exerts no force on the body. 4) Now if the body is displaced towards the right from its equilib- rium position, then the force exerted by the spring on the body is directed towards the left (fig a). force = mass acceleration of = MG eo (2) From equation (1) and (2) We have “a= aR Put this value in (3) _ ax _ ake ‘at m [dix ke In £2 + ka = 0) ae tin = Ojor mae z EL, gi [EE + ate = O...n fe £= 0%} Where is angular frequency. These are the differential equations of linear SHM. Q. From differential equa- tion of linear S.H.M. obtain expression for acceleration of Particle performing S.H.M. Expression for Acceleration in SHM ; The differential equation of linear SHM is given by - ax ke _ Where a acceleration of particle x is displacement of particle from mean position m is mass of particle k is force constant. We have — =’, Wherew is angular velocity. Therefore equation (1) becomes 2 dx fig 2x dP +ox=0 ea | “Gp = Tax “. acceleration of S.H.M. (a) =—@" x] Negative sign shows that the acceleration opposite directions. In magnitude, and displacement are in lacceleration(a x| Maximum and Minimum Acceleration i) At extreme position: When particle is at extreme position, x =+ A 3G = (+A) + dn = FO" A This is the maximum acceleration ii) At mean position: When particle is at mean position, x = 0 2 “a=0X0=0 min = 0 This is the minimum acceleration. ee ©. From differential equa- tion of linear S.H.M. obtain expression for velocity of particle performing SAM. | Expression for Velocity in SHM Acceleration of particle performing S.H.M. is given as- But We have ax _ Put in equation (1) de 8 ax _dv_ dv, dx dt dt dx” dt & = (velocity) ax _ dv ode dx ye =-w’x wo vdv. =o xdx Integrating above equation on both sides- Where C is the constant of inte applying fvav= [-o'xde=—0 [ xax gration. Value of C is obtained by boundary value condition. At extreme position, x =+A,v = 0 wi? v0 9 +6 wa cata Put the value of C in equation (2) , 1 pty? 242 vi =x’ , wA 8 2 veo 4+ oA? ww = aA? 2") Taking square root on both sides Sign +s tof/A-x hows that when particle passes through a point in positive direction xthen v is +ve and when it passes though the same point in negative direction of x then v is +ve. This is the equation for velocity of particle performing SHM. Maximum and minimum velocity i) At extreme position: When the particle is at extreme position, x =+ A 2. Vmin = Wy A? A? Vain = YO os Vin = 0 This is the minimum velocity ii) At mean position: When the particle is at mean position, x = 0 Vw =ta/ A= 0 6 Vinay =to/A 2. Vinx = AW This is the maximum velocity. eee a) If the particle starts from extreme position- At extreme position x =+ A at t= 0 Therefore equation (1) becomes +A=A(0+¢)=Asing «sing =41 v @=sin (41) p= 5 or 3B Put value of ¢ in equation (1) jest sin(wt +3) =+4 Acoswt ..x =+ Acoswt This is the expression for displacement of particle when it starts from extreme position. b) If the particle starts from mean position- At mean position x = 0 at r= 0 | Therefore equation (1) becomes | 0 = Asin(0+¢)=Asing “sing =0 “G@=0 Put value of ¢ in equation (1) ox =+ Asin(wr+ 0) [x= Asinot This is the expression for displacement of particle when it starts from mean position. Q. Define amplitude of SHM. | Amplitude of S.H.M. (A) 1) The maximum displacement of the particle in S.H.M. from the mean position or equilibrium position is called as amplitude of S.H.M. It is always away from mean position. 2) Its SI unit is meter (m) and dimensions are [M°L'7"]. Q. What is period of SHMP Period of SHM. 1) The time taken by the particle to complete one oscillation is *Q. Define period of S.H.M. called as period of SHM. At extreme position x = A coswt at t=0,a@t=0 -. coswt = cos0 = | fot =A atr=22, of = 0x22 = 27 o 3) -. COSW = cos 27 = 1 Tt HA Thus the particle takes a time an to return to the extreme position from which motion is started. Hence the period of SHM is pa 2h 2n 1 @ Dag gpm = Dat} 2) But w= /£ Q. Define frequency of S.HM. Q. What is series combina- tion and parallel combination of springs? Q. Derive the equation for the effective spring constant for the series combination of Springs. »_ 2m _ dn __ 20 wo k f m mx But .. £ = acceleration 2n [acl no acceleration per unit displacement Vv x . 0 7 0epu 3) SI unit of period is sec and dimensions are [M°L'T"]. Frequency of SHM (n or f) . Feat 1) The number of oscillations performed by the particle per orming S.H.M. per unit time is called as frequency of S.H.M. Frequency (n) = ‘ad bad ae 2) Its SI unit is hertz(Hz) or c.p.s. or r.p.s. and dimensions are [M°L'T 4] Series Combination of Springs . If two or more springs are connected one after the other forming a single chain, then this combination is called series combination of springs. Parallel Combination of Springs If two or more springs are connected between same two points (support and the stretching force), then this combination is called parallel combination of springs. Effective Spring Constant for the Series Combination of Springs Consider two springs of spring constant k, and k, connected in series as shown below If the springs are massless, each spring has the same stretching force as f. Let e and @ are the extensions of the springs. Restoring force in each spring will be same. “f=ha= he weal _f Me and e, = lo Total extension is e=Hate cent yt ORT 2g HL | Ant y If k, is the effective spring constant, then Explaiy method, : oo n reference circle | The S.H.M. as projection of UCM along any diamete; 1) If an object performs U.C.M., Its projection along any Giamey, executes S.H.M. ing U.C.M. along the o " 7 ticle P performing U.C. lor circum, °) cons ee anticlockwise direction. Let 'A' be the ra, fag Tene circle and w be the constant angular speed of the Particie”. the y : F H 3) When the particle P moves along the circumference of the Circle, its projection particle at point M moves to and fro along the hori- zontal diameter GE.The particle P is called reference Particle and its circle is called reference circle. This method is reference circle method. Y H 4) The x—component of displacement, velocity and acceleration of M is always same as that of P. 5) Suppose that the particle starts from initial position with initial phase ¢ (angle between radius OP and x— axis at f=0).In time t, the angle between OP and x —axis is (wt +b). Here @ is the constant angular velocity. -. cos(wt +h) = " “= Acos(wt+ )........1) This is the expression for the di isplacement of particle M at time . 6) The velocity of particle is given as the rate of change 0 displacement. = dx .. ~~ «. velocity(v) = ho £ (Acos(w+ 6) “.V=— Awsin(wt + 6) 10 | L Q. Define phase of SHM. (OCT. 14, MARK 1) Q. Define epoch of S.H.M. This is the expression for the velocity of particle. 7) The acceleration of particle is given as the rate of change of velocity «. acceleration(a) = 2% = # [-Aasin(wr + )] dt dt «. a= Aw’ cos(wt + ¢) = WX... 4 Acos(wt + 6) = x} This is the expression for acceleration of particle Thus the acceleration of particle M is directly proportional to its displacement and is opposite to that of the displacement. 8) Therefore when the particle P performs U.C.M., its projection particle M performs S.H.M. along a diameter of the circle. Phase of S.H.M. (9) The physical quantity which describes the state of oscillations (magnitude and direction of displacement of particle) at given instant is called phase of S.H.M. The equation of displacement of S.H.M. is- x = Asin(wt + 6) The quantity (wi + @) is called as phase angle or phase of S.H.M. Phase angle 6 = (wi +) Spacial Cases i 0=0 The particle is at the mean position and moving to the positive during the first oscillation. Second oscillation will begin at @ = 360° or 2x". ii) 0 = 180° or z° The particle is at the mean position and moving to the negative during the first oscillation. Second oscillation will begin at 6 = (360 + 180)’ or (27+ xf. seer A — Ge on ZH) iii) 9 = 90 or (Z) The particle is at the positive extreme position during the first oscillation. - second oscillation will begin at 0 = (360 + 90)° or (2+ ) . iv) 0 = 270° or (32) The particle is at the negative extreme position during the first oscillation. TC Second oscillation will begin at 6 = (360+ 270) or (27+ al Epoch of S.H.M. _ 1) The physical quantity which describes the state of oscillaiton of particle performing S.H.M. at the start of the motion is called as epoch of S.H.M. 2) In the term phase (wf + d),@ is the phase of the S.H.M. at the start ie. /= 0. 3) It is also known as starting p constant or epoch of S.H.M. hase or initial phase or phase Q. Particle performing S.H.M. starts from extreme position. Plot a graph of displacement, velocity and acceleration against time. time (1) Fig. Variation of acceleration with time Conclusions ; “aed and ancel> 1) From graphs it is clear that the displacement, velocity and acce eration of S.H.M. are periodic functions of time. : - time curve are sine rve and acceleration s cosine wave. Displacement - time cui nd velocity is 2 waves and velocity - time curve i 3) The phase difference between displacement ai (% radian. 4) The phase difference between velocity and acceleration is (F)radian. 5) The phase difference between displacement and acceleration is 7 radian. ; 6) All curves repeat same path after 27 radian. (B) Starting from positive extreme position - If the particle is starting S.H.M. from the positive extreme position then- ¢ = a i) Its displacement is (x) = Acos wi ii) Its velocity is (v) = & = rae cos wt) =~ Aw sin wt iii) Its acceleration is x = 4(-Aesin wt) =— Aw’ cos wt Time 0 T T 3° T 5T (t) 4 2 4 4 Phase [0 a (a 3x 2x Sx (av) 2 2 2 Dis- A 0 -A 0 A 0 place- ment (x) Velocity | 0 -Aw = }0 Aw ) —Aw (v) Accel- |—Aw’? | 0 ~Aw? |0 —Aw |0 eration (a) Graphs: Fig. Variation of acceleration with time time (t) Fig- Variation of velocity with time +A -Ao’| Fig- Variation of acceleration with time Conculusions 1) From graphs it is clear that the displacement, velocity and accel- eration of S.H.M. are periodic functions of time, 2) Displacement - time curve and acceleration - time curve are cosine curves and velocity - time curve is sine curve, 3) Phase difference between displacement and velocity is (F radian. 4) Phase difference between velocity and acceleration is (z radian, 5) Phase difference between displacement and acceleration is 7 radian. 6) All curves repeat same path after 27 radian. 14 Resultant phase 5 of the resultant motion - Divide the equation (2) by (8) - Rsind _ Arsing + Aosin ReosS Arcos: + Arcos 2 A,sin d + Aosin db» «. tan8 = —— 7a, Axcos Qi + A. cos 2 _f Aising + Aosin d» «8 = tan ||——7 7 cos bs A,cos$, + A2cos 2 This is the expression for the resultant phase of S.H.M. Spacial Cases The equation of resultant amplitude is given as- R= VAl+ Alt 2AAscos(d, ~ $2) Case 1 (in phase) If the phase difference between two S.H.M.s is zero ie. b:— 2 = 0 Then - «R= VA} + Ait 2A,A2c08 0 = fAR+ At 2A Ad { cos0 = 1} = (A+ Ax) wR=At+A A=A=A “R=2A Case 2 (out phase) _ If the phase difference between two SHMs is “radian ies b-b=F “R= [AP + Al + AAcosy R= IBFD oxen d'e cos = of Then - “R= VATA ff A=A=A R=V2A “R=141A Case 3 (out phase) If the phase difference between two S.H.M.sis radian i.e. 6: — $2 then- R= Ai t+ Ai + AiArcos 7 “R= JAi+ Aim 2A Ad “R= MAH Ay {ts cosa =~ 1} R=AW~AD If A =A =A “R=0 Q. Obtain an expression for the potential energy of a particle performing simple harmonic motion. (Mar. 15, Mark 2) Q. State the expressions for the kinetic energy (KE) and potential energy (PE) at a displacement X for a particle performing linear SHM. Represent them graphicaily. (Mar. 14, Mark 2) Q. Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M. Q. Show that energy of S.H.M. is directly propor- tional to - i) square of amplitude ii) square of frequency Kinetic energy, Potential energy and Total e performing SHM The diagram shows the mea performing SHM. nergy of a particle n and extreme position of the particle Fig. - Particle performing S.H.M. Let 'm' is the mass of particle 'O' is the mean position of the particle N and M are the extreme positions of the particle Je at a distance x from the mean P is the position of the partic’ tance x } of the particle is given as- position then the K.E. and T.E. (A) Kinetic energy in S.H.M. The instantaneous velocity of the particle perfor distance x from the mean position is given aS~ yetoVa=x rming S.H.M. at a ve wlA-2) But cla KE.= amv 2 RE. = Aa (AP = x) 1) Also ko = m k=mo 2 RE = YMA? ¥en 2) The equations (1) and (2) represent K.E. of the particle performing S.H.M. (B) Potential energy in S.H.M. When the particle is displaced from the mean position by distance dxthen dw work is done and it is given by - dw =— fdx But faa kx “dw =—(~kx)dx «dw = kx.dx The total work done is given by integrating the above equation between limits 0 to x as- | x Ves Cory We'[" dw= [‘kedst= Ek] xdv 5 »wetne - 1 Ws yhe Wetmore oe Ke ot bk Sey? 1 WE IW score og = Ook = mo This work done is stored in the form of P.E. o PBL= She PE.=4mo'x or 7 PE =e PEL = yk = = Ly 42(?) = 1 pa? KE. = kA? 0") = 7 kA TE.=KE.+PE.=4 Thus at mean position re KE. (B) At extreme position- At extreme position, velocity of particle in S. displacement is maximum i.e. x =+A PB =1tKe 1 PE. = kA H.M. is minimum and K.E.= ma (A? — A’) =0 TE.=KE.+PE.=0+44A" = kA" Thus at extreme position |7-E. Conclusions 1) The PE. of particle at mean position is zero. 2) The P.E. of particle at extreme position is maximum 3) The K.E. of particle at mean position is maximum. 4) The K.E. of particle at extreme position zero. 5) At mean position T.E. is purely K.E. 6) At extreme position T.E. is purely P. E. 7) T.E. at any position of the particle is always tka’. Hence the total energy of the particle performing S.H.M. is always conserved. The graph is plotted between variation of K.E. , P.K. w.rt instanta- neous displacement as shown below- energy (E) T.E.=constant “A displacement (x) The nature of graph of KE., P.K. against displacement is parabolic. Q. Find the displacement at which K.E. is equal to P.B. (Mar. 14, Mark 1) Q. Find K.E. and P.E. at + = tA for a particle performing linear S.H.M. Q. What is ideal simple pendulum? Explain the motion of simple pendulum as SHM. Q. Define an ideal simple pendulum. Show that, under certain conditions, a simple pendulum performs linear simple harmonic motion. (Mar. 13, Mark 4) Q. Derive an expression for the period of motion of a simple pendulum. On which factors does it depend? (Oct. 15, Mark 4) Ans. If K.E. = PE. —_— Sine’ (A? 3°) = yx" wy= tA wa a : +A - -£ -= 74 KE = PE= Thus at x 7) 2 ee Ans. = tA Atx 2 _1,2_1,(A?)—1(Lyy?)=£ PB Soke =5K(4 )=a(z) 4 E=PE.+KE. «KE. E-PE.=E-£= Thus at z=3A 3. 4 E , total energy consists of 25% P.E. and 75% K.E. Ideal Simple Pendulum A point mass (heavy particle) suspe' y sible and twistless string from perfectly rigid sup ideal simple pendulum. Practical simple pendulum is a small heavy sphere (bob) suspended by light and inextensible string from a rigid support. Motion of simple pendulum as S.H.M. . sed to some height from its 1) If an ideal simple pendulum is rai : | ‘mean position and released from that height then it performs same sets of repetitive motion again and again called oscillations. nded by a weightless, inexten- port is known as 10) Fig- Simple pendulum 2) Consider a bob of mass mi fixed to a rigid support by an inexten- sible string of length L. If the mass is displaced by small angle 7 (angular amplitude) then it oscillates about the mean position. Here two forces are acting on the bob- i) Its weight mg acting vertically in downward direction. This weight is resolved into two components a) mgcos@ along the string (radial component) b) mgsin@ perpendicular to the string (tangential component) ii) The tension 7’ along the string in upward direction. 20 
