PHYSICS (CLASSS XII) - Chapter 5 : Oscillations

PHYSICS (CLASSS XII) - Chapter 5 : Oscillations

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  1 CLASS XII PHYSICS CHAPTER 5 Oscillations
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  2 Introduction Oscillatory motion isa periodic motion. Explanation of Periodic Motion Any motion which repeats itself after a definite interval of time is called periodic motion. The time taken for one such set of movements is called its period or periodic time. Circular motion is periodic but it is not oscillatory. The smallest interval of time after which the to and fro motion is repeated is called its period (T) and the number of oscillations completed per unit time is called the frequency (n) of the periodic motion. Linear Simple Harmonic Motion (S.H.M.) The block will begin its to and fro motion on either side of its equilibrium position. This motion is linear simple harmonic motion. If x is the displacement, the restoring force f is, f = - k x where, k is a constant that depends upon the elastic properties of the spring. It is called the force constant. The acceleration (a) is, a = 𝑓 𝑚 = − ( 𝑘 𝑚 )𝑥 Linear S.H.M. is defined as the linear periodic motion of a body, in which force is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position. Differential Equation of S.H.M. f = - k x According to Newton’s second law of motion, f = ma ∴ 𝑚𝑎 = −𝑘 𝑥 The velocity of the particle is, v = 𝑑𝑥 𝑑𝑡 and its acceleration, a = 𝑑𝑣 𝑑𝑡 = 𝑑2𝑥 𝑑𝑡2 𝑚 𝑑2𝑥 𝑑𝑡2 = - k x ∴ 𝑑2𝑥 𝑑𝑡2 + 𝑘 𝑚 𝑥 = 0
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  3 Substituting 𝑘 𝑚 = 𝑤2 , whereω is the angular frequency, 𝑑2𝑥 𝑑𝑡2 + 𝑤2 𝑥 = 0 Acceleration (a), Velocity (v) and Displacement (x) of S.H.M. 𝑑2𝑥 𝑑𝑡2 + 𝑤2 𝑥 = 0 𝑑2𝑥 𝑑𝑡2 = - 𝑤2 𝑥 a = - 𝑤2 𝑥 This is the expression for acceleration in terms of displacement x. 𝑑2𝑥 𝑑𝑡2 = - 𝑤2 𝑥 ∴ 𝑑 𝑑𝑡 ( 𝑑𝑥 𝑑𝑡 ) = - 𝑤2 𝑥 ∴ 𝑑𝑣 𝑑𝑡 = - 𝑤2 𝑥 ∴ 𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑡 = - 𝑤2 𝑥 ∴ 𝑣 𝑑𝑣 𝑑𝑥 = - 𝑤2 𝑥 ∴ v dv = - 𝑤2 𝑥 𝑑𝑥 Integrating both the sides, ∫ 𝑣 𝑑𝑣 = −𝑤2 ∫ 𝑥 𝑑𝑥 ∴ 𝑣2 2 = − 𝑤2 𝑥2 2 + 𝐶 where C is the constant of integration. Let A be the maximum displacement (amplitude) of the particle in S.H.M. When the particle is at the extreme position, velocity (v) is zero. Thus, at x = ±𝐴, 𝑣 = 0 0 = − 𝑤2 𝐴2 2 + 𝐶 ∴ 𝐶 = + 𝑤2 𝐴2 2 Using C in Equation, Substituting v = 𝑑𝑥 𝑑𝑡 Here φ is the constant of integration. This is the general expression for the displacement (x) of a particle performing linear S.H.M. at time t.
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  4 Case (i) Ifthe particle starts S.H.M. from the mean position, x = 0 at t = 0 φ = 𝑠𝑖𝑛−1 ( 𝑥 𝐴 ) = 0 𝑜𝑟 𝜋 x = ± 𝐴 si n (𝑤𝑡) Case (ii) If the particle starts S.H.M. from the extreme position, x = ±𝐴 at t = 0 Expressions of displacement (x), velocity (v) and acceleration (a) at time t Extreme values of displacement (x), velocity (v) and acceleration (a) 1) Displacement: The general expression for displacement x in S.H.M. is, x = A sin (𝜔𝑡 + 𝜙) At the mean position, (𝜔𝑡 + 𝜙) = 0 or 𝜋. Thus, at the mean position, the displacement of the particle performing S.H.M. is minimum (i.e., zero). (wt + Φ) = 𝜋 2 0r 3𝜋 2 Thus, at the extreme position the displacement of the particle performing S.H.M. is maximum. 2) Velocity: According to Equation of the magnitude of velocity of the particle performing S.H.M. is, v = ± 𝜔 √𝐴2 − 𝑥2. At the mean position, x = 0. ∴ 𝑉 𝑚𝑎𝑥 = ±𝐴 𝜔. Thus, the velocity of the particle in S.H.M. is maximum at the mean position. At the extreme position, x = ± 𝐴. ∴ 𝑉𝑚𝑖𝑛 = 0. Thus, the velocity of the particle in S.H.M. is minimum at the extreme positions. 3) Acceleration: The magnitude of the acceleration of the particle in S.H.M is 𝑤2 𝑥. At the mean position x = 0, so that the acceleration is minimum. ∴ 𝑎𝑚𝑖𝑛 = 0. At the extreme positions x = ± 𝐴, so that the acceleration is maximum, 𝑎𝑚𝑎𝑥 = ∓ 𝜔2 𝐴. Amplitude(A), Period(T) and Frequency (n) of S.H.M. Amplitude of S.H.M. Fig. S.H.M. of a particle. The displacement of the particle is, The particle will have its maximum displacement when i.e., when x = ± 𝐴. This distance A is called the amplitude of S.H.M. The maximum displacement of a particle performing S.H.M. from its mean position is called the amplitude of S.H.M.
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  5 Period of S.H.M. Thetime taken by the particle performing S.H.M. to complete one oscillation is called the period of S.H.M. Displacement of the particle at time t is, After a time, t = (𝑡 + 2𝜋 𝑤 ) the displacement will be 𝜔2 = 𝑘 𝑚 = 𝑓𝑜𝑟𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 ∴ 𝑇 = 2 𝜋 √𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 Also, 𝑇 = 2 𝜋 √ 𝑚 𝑘 Frequency of S.H.M. The number of oscillations performed by a particle performing S.H.M. per unit time is called the frequency of S.H.M. In time T, the particle performs one oscillation. Hence in unit time it performs 1 𝑇 oscillations. Hence, frequency n of S.H.M. is, 𝑛 = 1 𝑇 = 𝜔 2 𝜋 = 1 2 𝜋 √ 𝑘 𝑚 Reference Circle Method A rod rotating along a vertical circle in the x-y plane. Fig: Projection of a rotating rod. Its angular positions are decided with the reference OX. It means, at F it is 900 = 𝜋 2 𝑐 , at G it is 1800 = 𝜋𝑐 , and so on. Let 𝑟 ⃗ = OP be the position vector of this particle. Fig: S.H.M. as projection of a U.C.M. Projection of displacement: The position vector OM = OP sin 𝜃 = y = r sin (wt + 𝜙)
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  6 Fig: Projection ofvelocity. Projection of velocity: 𝑉 𝑦 = 𝑟 𝜔 cos 𝜃 = 𝑟 𝜔 cos(𝜔𝑡 + 𝜙) Projection of acceleration: Its projection on the reference diameter will be The projection of a U.C.M. on any diameter is an S.H.M. Phase in S.H.M. Phase in S.H.M. is basically the state of oscillation. In order to know the state of oscillation in S.H.M. - displacement (position) - direction of velocity - and the oscillation number at that instant of time. The angular displacement 𝜃 = (𝑤𝑡 + Φ) Special cases (i) Phase θ = 0 indicates that the particle is at the mean position. Phase angle 𝜃 = 3600 𝑜𝑟 𝜋𝑐 is the beginning of the second oscillation. (ii) Phase 𝜃 = 1800 or 𝜋𝑐 indicates that during its first oscillation, the particle is at the mean position and moving to the negative. Similar state in the second oscillation will have phase 𝜃 = (360 + 180)0 or (2𝜋 + 𝜋)𝑐 . (iii) Phase 𝜃 = 900 or ( 𝜋 2 ) 𝑐 indicates that the particle is at the positive extreme position. For the second oscillation it will be 𝜃 = (360 + 90)0 or (2𝜋 + 𝜋 2 ) 𝑐 . (iv) Phase 𝜃 = 2700 or ( 3𝜋 2 ) 𝑐 indicates that the particle is at the negative extreme position. For the second oscillation it will be 𝜃 = (360 + 270)0 or (2𝜋 + 3𝜋 2 ) 𝑐 . Graphical Representation of S.H.M. (a) Particle executing S.H.M., starting from mean position, towards positive: As the particle starts from the mean position, towards positive, φ = 0 ∴ displacement x = A sin wt, Velocity v = A ω cos wt, Acceleration a = - A 𝑤2 sin wt Conclusions from the graphs: • Displacement, velocity and acceleration of S.H.M. are periodic functions of time. • Displacement time curve and acceleration time curves are sine curves and velocity time curve is a cosine curve. • There is phase difference of 𝜋 2 radian between displacement and velocity. • There is phase difference of 𝜋 2 radian between velocity and acceleration.
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  7 • There isphase difference of π radian between displacement and acceleration. • Shapes of all the curves get repeated after 2π radian or after a time T. Fig.: (a) Variation of displacement with time, (b) Variation of velocity with time, (c) Variation of acceleration with time. (b) Particle performing S.H.M., starting from the positive extreme position. As the particle starts from the positive extreme position, 𝜙 = 𝜋 2 ∴ 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡, 𝑥 = 𝐴 sin (𝜔𝑡 + 𝜋 2 ) = 𝐴 cos 𝜔𝑡 Velocity, v = 𝑑𝑥 𝑑𝑡 = 𝑑(𝐴 cos𝜔𝑡) 𝑑𝑡 = −𝐴 𝜔 sin(𝜔𝑡) Acceleration, a = 𝑑𝑣 𝑑𝑡 = 𝑑 (−𝐴 ω sin (𝜔𝑡)) 𝑑𝑡 = −𝐴 𝜔2 cos (𝜔𝑡) Fig.: (a) Variation of displacement with time, (b) Variation of velocity with time, (c) Variation of acceleration with time. Composition of two S.H.M.s having same period and along the same path
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  8 Equations of displacementof the two S.H.M.s along same straight line (x-axis) are The resultant displacement (x) at any instant (t) is, 𝑥 = 𝑥1 + 𝑥2 𝑥 = 𝐴1si n(𝜔𝑡 + Φ1) + 𝐴2si n(𝜔𝑡 + Φ2) 𝑥 = 𝐴1si n 𝜔𝑡 𝑐𝑜𝑠 Φ1 + 𝐴1cos ωt 𝑠𝑖𝑛 Φ1 + 𝐴2si n 𝜔𝑡 𝑐𝑜𝑠Φ2 + 𝐴2cos ωt 𝑠𝑖𝑛 Φ2 𝐴1, 𝐴2, 𝜙1, 𝜙2 are constants and 𝜔𝑡 is variable. 𝑥 = (𝐴1 cos Φ1 + 𝐴2 cos Φ2) sin 𝜔𝑡 + (𝐴1si𝑛 Φ1 + 𝐴2 𝑠𝑖𝑛 Φ2) cos 𝜔𝑡 As 𝐴1, 𝐴2, 𝜙1 𝑎𝑛𝑑 𝜙2 are constants, we can combine them in terms of another convient constants R and 𝛿 as R cos 𝛿 = 𝐴1 𝑐𝑜𝑠 𝜙1 + 𝐴2 𝑐𝑜𝑠 ϕ2 and R sin 𝛿 = 𝐴1 𝑠𝑖𝑛 𝜙1 + 𝐴2 𝑠𝑖𝑛 ϕ2 ∴ 𝑥 = 𝑅 (si n 𝜔𝑡 co s 𝛿 + co s 𝜔𝑡 si n 𝛿) ∴ 𝑥 = 𝑅 𝑠𝑖𝑛 (ωt + δ) Resultant amplitude, Initial phase (δ) of the resultant motion: Special cases: (i) If the two S.H.M.s are in phase, (ii) If the two S.H.M.s are 90° out of phase, (iii) If the two S.H.M.s are 180° out of phase, Energy of a Particle Performing S.H.M. Total energy of the particle performing an S.H.M. is thus the sum of its kinetic and potential energies. Fig.: Energy in an S.H.M. Velocity of the particle in S.H.M. is,
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  9 where x isthe displacement of the particle performing S.H.M. and A is the amplitude of S.H.M. Thus, the kinetic energy, This is the kinetic energy at displacement x. At time t, it is Thus, with time, it varies as 𝑐𝑜𝑠2 𝜃. The external work done (dW) during this displacement is The total work done on the particle to displace it from O to P is, This should be the potential energy (P.E.) 𝐸𝑃 of the particle at displacement x. Thus, with time, it varies as 𝑠𝑖𝑛2 𝜃. The total energy of the particle is the sum of its kinetic energy and potential energy. As m, ω and A are constant, the total energy of the particle at any point P is constant. If n is the frequency in S.H.M., 𝜔 = 2𝜋𝑛. Thus, the total energy in S.H.M. is directly proportional to (a) the mass of the particle (b) the square of the amplitude (c) the square of the frequency (d) the force constant, and inversely proportional to square of the period. Special cases: (i) At the mean position, x = 0 and velocity is maximum. Hence, and potential energy (ii) At the extreme positions, the velocity of the particle is zero and 𝑥 = ±𝐴. Hence and kinetic energy (iii) If K E.= P. E.,
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  10 Thus at 𝑥= ±𝐴 √2 , the K.E. = P.E. = 𝐸 2 for a particle performing linear S.H.M. (iv) At x = ±𝐴 2 , P.E. = 1 2 𝑘𝑥2 = 1 4 ( 1 2 𝑘𝐴2) = 𝐸 4 ∴ 𝐾. 𝐸. = 3(𝑃. 𝐸) Thus, at 𝑥 = ±𝐴 2 , the energy is 25% potential and 75% kinetic. Fig.: Energy in S.H.M. Simple Pendulum An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support. The distance between the point of suspension and centre of gravity of the bob is called the length of the pendulum. Fig.: Simple pendulum In the displaced position (extreme position), two forces are acting on the bob. (i) Force T' due to tension in the string, directed along the string, towards the support (ii) Weight mg, in the vertically downward direction. Weight mg is resolved into two components; (i) The component mg cos θ along the string, which is balanced by the tension T ' and (ii) The component mg sin θ perpendicular to the string is the restoring force acting on mass m tending to return it to the equilibrium position. ∴ 𝑅𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒, 𝐹 = −𝑚 𝑔 sin 𝜃 As θ is very small (θ <10°), sin𝜃 ≅ 𝜃𝑐 ∴ 𝐹 ≅ −𝑚 𝑔 𝜃 The small angle 𝜃 = 𝑥 𝐿 ∴ 𝐹 = −𝑚 𝑔 𝑥 𝐿 As m, g and L are constant, F ∝- x The period T of oscillation of a pendulum = 2 𝜋 𝜔 𝑇 = 2𝜋 √𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝐹 = −𝑚 𝑔 𝑥 𝐿 ∴ 𝑚 𝑎 = −𝑚 𝑔 𝑥 𝐿 ∴ 𝑎 = − 𝑔 𝑥 𝐿
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  11 ∴ 𝑎 𝑥 = − 𝑔 𝐿 = 𝑔 𝐿 (𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒) Substituting inthe expression for T, 𝑇 = 2 𝜋 √ 𝐿 𝑔 The Equation gives the expression for the time period of a simple pendulum. (i) The amplitude of oscillations is very small. (ii) The length of the string is large and (iii) During the oscillations, the bob moves along a single vertical plane. Frequency of oscillation n of the simple pendulum is 𝑛 = 1 𝑇 = 1 2𝜋 √ 𝑔 𝐿 (a) The period of a simple pendulum is directly proportional to the square root of its length. (b) The period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity. (c) The period of a simple pendulum does not depend on its mass. (d) The period of a simple pendulum does not depend on its amplitude. These conclusions are also called the 'laws of simple pendulum'. Second’s Pendulum A simple pendulum whose period is two seconds is called second’s pendulum. 𝑃𝑒𝑟𝑖𝑜𝑑 𝑇 = 2 𝜋√ 𝐿 𝑔 ∴ 𝐹𝑜𝑟 𝑎 𝑠𝑒𝑐𝑜𝑛𝑑′ 𝑠𝑝𝑒𝑛𝑑𝑢𝑙𝑢𝑚, 2 = 2 𝜋 √ 𝐿𝑠 𝑔 where 𝐿𝑆 is the length of second’s pendulum, having period T = 2 s. ∴ 𝐿𝑠 = 𝑔 𝜋2 Experimentally, if 𝐿𝑆 is known, it can be used to determine acceleration due to gravity g at that place. Angular S.H.M. and its Differential Equation If the disc is slightly twisted about the axis along the wire, and released, it performs rotational motion partly in clockwise and anticlockwise sense. Such oscillations are called angular oscillations or torsional oscillations. Fig.: Torsional (angular) oscillations Thus, for the angular S.H.M. of a body, the restoring torque acting upon it, for angular displacement θ, is 𝜏 ∝ − 𝜃 𝑜𝑟 𝜏 = − 𝑐 𝜃 The constant of proportionality c is the restoring torque per unit angular displacement. If I is the moment of inertia of the body, the torque acting on the body is, 𝜏 = 𝐼𝛼 Where α is the angular acceleration. Using this in Equation, 𝐼𝛼 = −𝑐𝜃 ∴ 𝐼 𝑑2 𝜃 𝑑𝑡2 + 𝑐 𝜃 = 0 𝛼 = 𝑑2 𝜃 𝑑𝑡2 = − 𝑐 𝜃 𝐼 Since c and I are constants. This oscillatory motion is called angular S.H.M.
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  12 Angular S.H.M. isdefined as the oscillatory motion of a body in which the torque for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement. The time period T of angular S.H.M. is, 𝑇 = 2𝜋 𝜔 𝑇 = 2𝜋 √𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 Magnet Vibrating in Uniform Magnetic Field If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field. Fig.: Magnet vibrating in a uniform magnetic field The magnitude of this torque is 𝜏 = 𝜇 𝐵 sin 𝜃 If 𝜃 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙, sin𝜃 ≅ 𝜃𝑐 ∴ 𝜏 = 𝜇 𝐵 𝜃 For clockwise angular displacement θ, the restoring torque is in the anticlockwise direction. ∴ 𝜏 = 𝐼 𝛼 = − 𝜇 𝐵 𝜃 where I is the moment of inertia of the bar magnet and α is its angular acceleration. ∴ 𝛼 = − ( 𝜇 𝐵 𝐼 ) 𝜃 Since μ, B and I are constants. 𝑇 = 2𝜋 √𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑇 = 2𝜋 √ 𝛼 𝜃 ∴ 𝑇 = 2𝜋 √ 𝐼 𝜇𝐵 Damped Oscillations Fig.: A damped oscillator Periodic oscillations of gradually decreasing amplitude are called damped harmonic oscillations and the oscillator is called a damped harmonic oscillator.
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  13 Where b isthe damping constant and negative sign indicates that 𝐹𝑑 opposes the velocity. For spring constant k, the force on the block from the spring is 𝐹𝑠 = −𝑘𝑥. A𝑒 −𝑏𝑡 2𝑚 is the amplitude of the damped harmonic oscillations. Fig.: Displacement against time graph The term cos (𝜔′ 𝑡 + 𝜙) shows that the motion is still an S.H.M. The damping increases the period (slows down the motion) and decreases the amplitude. Free Oscillations, Forced Oscillations and Resonance Free Oscillations: If an object is allowed to oscillate or vibrate on its own, it does so with its natural frequency. which is called its natural frequency and the oscillations are free oscillations. Fig: Forced oscillations For unequal natural frequencies on either side, the energy absorbed is less. The peak occurs when the forced frequency matches with the natural frequency. Fig: Resonant frequency