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Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION Q.1-What is Simple Harmonic Motion? A periodic motion confined between two fixed points about a mean position such that acceleration produced is directly proportional to the displacement , but taking place in opposite direction in known as simple harmonic motion. Eg – Motion of a simple pendulum. Q.2- Prove that projection of uniform circular motion along its diameter is SHM or Deduce the equation of motion of SHM. Let us consider a particle under going uniform circular motion with uniform angular velocity in anticlockwise direction along the circumference of a circle of radius r . When the particle goes from X to Y, Y to X/, X/ to Y/, Y/ to X/, the projection of the particle moves from O to Y, Y to O, O to Y/, Y/ to O, respectively along the diameter YO Y/ . So the projection of a uniform circular motion , along a diameter of the circular path is (i) taking place symmetrically about the mean position O (ii) is confined between two extreme position Y and Y/ (iii) is periodic , as = constant as = 2π/T Let at any instant t, it be at P such that OM = y and POX = From ∆OPM, Sin = OM/OP = y/R => y = r sin . => y = r sin t -------------- (i) = /t so = t 1 Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION Differentiating equation (i) with respect to time we have velocity as => v = dy d r sin t = - r cos t -------- (ii) dt dx Differentiating equation (ii) with respect to time we have acceration as => a = dv d r cos t. = - 2r sin t dt dt => a = - 2y …………………(iii) As w is constant => a – y ---- (iv) Hence, motion of the projection of uniform circular motion along any of its diameter is SHM in nature. (iii) => a + 2y = 0 -------- (iii) This is the equation of motion of SHM. d2y 2 In differential form the equation of motion of SHM => 2 + y = 0 …(v) dt Q.3- Give the condition of maximum and minimum velocity and acceleration of a body during SHM . We know , displacement of SHM , y = rsint ……………..(i) And velocity of SHM , v = r cost ……………..(ii) And acceleration of SHM , a = - r2sint ……………..(i) y 2 r2 y2 r2 y2 (ii) => v r 1 sin wt r 1 ( ) r 2 r r 2 (iii)=> a 2y At extreme position, y = r , so v = 0 and a = 2 r At mean position , y =0 , so v = r and a = 0 ----------------------------------------------------------- 2 Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION Q. 1. Prove that motion of a simple pendulum is simple Harmonic in nature. Calculate its frequency & time period. A simple pendulum is a metallic bob suspended by means of a mass less inextensible string from a rigid support. As shown in the diagram. Let us consider a simple pendulum of mass (m) be suspended by means of are string OA of length (l) from a rigid support, and in equilibrium position. Let it be displaced from A to B such that its displacement, AB= x. Let is the angle made by the string with the vertical. Resolving its weight mg at B, we have – 1. mg sin , towards the mean position 2.mg cos , in a direction opposite to the tension T in the string., As, mg cos T, they will cancelled each other . Hence driving force in the bob, F = mg sin -------------- (i) As is small so from AOB, Sin = AB x ---------- (ii) OA L 3 visit www.physics365.com Tridib’s Physics Tutorials NCERT-XI / Unit- 14 – OSCILLATION (i)=> F = mg(- x/L) (negative signal indicates that force and displacement are always opposite to each other) => ma = - mg x/L => a = - g x/L => a = - g/Lx ------------- (iii) as g/L is constant a – x. So the motion of the simple pendulum is simple harmonic in nature. Comparing equation (iii) with standard eqn of simple harmonic motion, a 2 x , we have 2 g L g v l 1 2π 2π T g l T 2 π L g g L Q. 3. State the laws of vibration of a pendulum Ans:- From the explain for time period of a simple pendulum, T 2π L g , we have 1. Law of length – Time period of a simple pendulum is directly proportional to the square root of effective length of the pendulum, provided g remains constant. 2. Law of Gravity of acceleration due to Gravity – Time period of a simple pendulum is inversely proportional to the sqr root of acceleration dire to gravity, providing its length remaining the same. 3. Law of mass – Time period of simple pendulum is independent of the mass of the bob. 4 Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION Q.4.What is second pendulum ? calculate its length. Ans:- The simple pendulum of which time period equal to 2 seconds, is known as second’s pendulum. We know that T 2π L 2 2 2 2 g => T = 4 L/g =>L = GT /4 . Here T = 2 seconds. So L = 9.8 x 4 /4 2. = 9.8/9.87 = 0.9927 m = 99.27cm. Q. 2. Prove that motion of a vertically loaded spring is SHM in nature and also calculate its frequency and time period. Ans:- . Let us consider a spring of spring constant, k. be suspended from a visit support in a vertically downward direction. Let it be elongated by a amount, l by suspending a weight at it free end so that restoring force acting on the spring, F = - k l -------------- (i) Let the spring be elongated further by a small amount, x by applying a force, F/ so that restoring force is, F/ = - k (l + x) ------- (ii) 5 Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION The acceleration of the body = Net force acting on the spring / mass of the body a = (F/ - F)/m = [- k (l + x) + kl]/m => a = - k x/m ---------- (iii) So, k/m is constant , a - x. So, the motion of a vertically loaded spring is simple harmonic in nature. Comparing eq(iii) with the standard equation oh SHM , a 2 x , we have k 2π k k T 2π k m m m m T 1 k v m 2π 2 Q.5.Calculate the energy of a body in SHM Or –Prove that a body in SHM obeys the principal of conservation of energy. Ans :- Let us consider a body of mass ‘m’ , connected to a horizontal spring of spring constant ,k on a smooth horizontal surface , undergoing S.H.M. in between X & X l about the mean position O, with r as its amptitude. To calculate its Potential energy (U) – Let P be a point at a distant of x from the mean position at which we have to calculate its potential energy. The magnitude of restoring force at P is given as – F = kx ---------------- (i) Let Q be a point very much closer to P so that PQ = dx. The amount of work done in moving it from P to Q. dW = F. dx ----------- (ii) 6 Tridib’s Physics Tutorials visit www.physics365.com NCERT-XI / Unit- 14 – OSCILLATION Total amount of work done in moving it from O to P. W W dW 0 x x Fdx 0 Kxdx 0 x2 x 1 k[ ] 0 kx 2 _(iii) 2 2 This amount of work done will be converted into potential energy at P, => UP = ½ k x2 --------------- (iv) But k = m 2. so => UP = ½ m2 x2 -------- (v) To calculate its Kinetic energy (K) :We know , velocity of the particle at P, v = r 2 x 2 So KP = ½ mv2 = ½ m 2 (r 2 – x2) ------ (vi) So total energy at P , EP = UP + KP = ½ m2x2 + ½ m 2 ( r 2 – x2) EP = ½ m2 r 2 ---------- (vii) This is the total energy of a particle executing SHM at any point. At O , x= 0 so, EO = UO + KO = 0 + ½ m 2 ( r 2 – 0) = ½ m2 r 2 At X , x= r so, EX = UX + KX = ½ m2r 2 + ½ m 2 ( r 2 – r2) = ½ m2 r 2 So , EP = EO = EX = ½ m2 r 2 , which is found to be constant at any point on its path . Thus principle of conservation of energy is obeyed by the body executing SHM Drawing the graph between KE & PE with distance , it is found that graph between total energy & distance is a straight line. 7