CLASS XII - CHAPTER 5: OSCILLATION (PHYSICS - MAHARASHTRA STATE BOARD)

CLASS XII
CHAPTER 5: OSCILLATION
(PHYSICS - MAHARASHTRA STATE BOARD)

1. CHAPTER 5 Oscillations CLASS XII PHYSICS MAHARASHTRA STATE BOARD

2. Can you recall? 1. What do you mean by linear motion and angular motion? 2. Can you give some practical examples of oscillations in our daily life? 3. What do you know about restoring force? 4. All musical instruments make use of oscillations, can you identify, where? 5. Why does a ball floating on water bobs up and down, if pushed down and released?

3. MOTION “Motion is the phenomenon in which an object changes its position over time.” It is described by displacement, distance velocity, speed acceleration time

4. “A force acting opposite to displacement to bring the system back to equilibrium i.e. at rest position.” RESTORING FORCE Periodic motion “Any motion which repeats itself after a definite interval of time is called periodic motion.”

5. OSCILLATION “Oscillation is defined as the process of repeating vibrations of any quantity or measure about its equilibrium value in time.” Or “Oscillation refers to any periodic motion at a distance about the equilibrium position and repeat itself and over for a period of time.” Oscillation is periodic motion Displacement, acceleration and velocity for oscillatory motion can be defined by Harmonic function. Sine Cosine

6. Linear simple harmonic motion ( s.h.m .) When we pull block right side from mean position the spring will pull object toward itself i.e. force produced by spring is opposite. f f = - k x f = m a Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.  

7. A complete oscillation is when the object goes from one extreme to other and back to the initial position. The conditions required for simple harmonic motion are: Oscillation of the particle is about a fixed point. The net force or acceleration is always directed towards the fixed point. The particle comes back to the fixed point due to restoring force. Harmonic oscillation is that oscillation which can be expressed in terms of a single harmonic function, such as x = a sin wt or x = a cos wt Non-harmonic oscillation is that oscillation which cannot be expressed in terms of single harmonic function. It may be a combination of two or more harmonic oscillations such as x = a sin ωt + b sin 2ωt , etc.

8. Differential Equation of S.H.M. Consider, f Restoring force, x Displacement done by the block. f = - k x ……….( i ) According to newtons second law of motion, f = m a m a = - k x ……….(ii) Also, velocity Rate of change of displacement v = Acceleration Rate of change of velocity a = = a = m x = - k x m + k x = 0 i.e. + x = 0, Where, =  

9. Example A body of mass 0.2 kg performs linear S.H.M. It experiences a restoring force of 0.2 N when its displacement from the mean position is 4 cm. Determine ( i ) force constant (ii) period of S.H.M. and (iii) acceleration of the body when its displacement from the mean position is 1 cm. Solution: Force constant, k = f / x = (0.2)/ 0.04 = 5 N/m Period T = 2 = 2 = 0.4 s Acceleration a = - x = = 0.04 = - 1 m  

10. TERMS FOR S.H.M. + x = 0 = - x a = x For velocity, + x = 0 = - x = - x . = - x . v = - x Integrating both side, = = …………..( i )  

11. Now, if object is at extreme position x = A, v = 0 C = From equation ( i ) = v = For displacement, We know that, v = v = =  

12. Case ( i ) If the particle starts S.H.M. from the mean position, x = 0 at t = 0 i.e. x = Case (ii) If the particle starts S.H.M. from the extreme position, x = A at t = 0 A 1 = or  

13. Expressions of displacement (x), velocity (v) and acceleration (a) at time t v = .(w+0) v = Aw cos a = Extreme values of displacement (x), velocity (v) and acceleration (a): Displacement: At mean position, = 0 or At extreme position, =  

14. 2) Velocity : v = At mean position, = 0 At extreme position, = 3) Acceleration: a = At mean position, = 0 At extreme position, = Amplitude The maximum displacement of a particle performing S.H.M. from its mean position is called the amplitude of S.H.M. For maximum displacement = 1 i.e. x =  

15. Period of S.H.M. The time 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, After some time, Where k = m  

16. Now, T = T = 2 X 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. n = PHASE IN S.H.M. Phase in S.H.M. is basically the state of oscillation. Requirements to know the state of oscillation Position of particle (displacement) Direction of velocity Oscillation number  

17. PHASE IN S.H.M. Phase in S.H.M. is basically the state of oscillation. Requirements to know the state of oscillation Position of particle (displacement) Direction of velocity Oscillation number Commonly, Expressions of displacement (x), velocity (v) and acceleration (a) at time t v = = a =  

18.   Phase = 0 = Phase = = Phase = = Phase = =  

19.   Consider, two S.H.M having same period and along same path. are displacements of both S.H.M. Composition of two S.H.M. x = x = x = x = + ……………….( i ) R cos = ………….(ii) R sin = ………….(iii) x = R cos . sin x = R [cos . sin ] x = R sin (  

20. RESULTANT AMPLITUDE R = From equation (ii) and (iii) R = SPECIAL CASES If the two S.H.M are in phase, = , = 1 R = = If, (ii) If the two S.H.M.s are out of phase, = , = 0 R = If,  

21. SPECIAL CASES (iii) If the two S.H.M.s are out of phase, = , = -1 R = I I If, Initial Phase ( ) Dividing equation (ii) and (iii)  

22. Energy of a Particle Fig.: Energy in an S.H.M. When particle performing S.H.M. then it passes both kinetic and potential energy. Velocity of particle performing S.H.M = A Kinetic Energy: ………( i ) Displacement x ………….(ii) External work done ( dw ) dW = f (-dx) dW = - kx (-dx) dW = kx dx  

23. Total work done on the particle, W = W = = ………….(iii) Total energy = E E E E For frequency, E = E =  

24. Simple Pendulum An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support. A practical simple pendulum is a small heavy (dense) sphere (called bob) suspended by a light and inextensible string from a rigid support. In the displaced position (extreme position), two forces are acting on the bob. Force T' due to tension in the string, directed along the string, towards the support and Weight mg, in the vertically downward direction.

25. Simple Pendulum At the extreme positions, there should not be any net force along the string. The component of mg can only balance the force due to tension. Thus, weight mg is resolved into two components; The component mg cos θ along the string, which is balanced by the tension T ' and 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 = - mg sin θ As θ is very small (θ < 10°), sin θ ≅ Small angle, = - m g  

26. Simple Pendulum = - m g For time period, T = T = T = = 2 For frequency, n =  

27. Second’s Pendulum A simple pendulum whose period is two seconds is called second’s pendulum. T = 2 For a second s pendulum, 2 = 2 Where,  

28. Angular S.H.M. and its Differential Equation Thus, for the angular S.H.M. of a body, the restoring torque acting upon it, for angular displacement θ, is ……….( i ) The constant of proportionality c is the restoring torque per unit angular displacement. Where, = Since c and I are constants, the angular acceleration α is directly proportional to θ and its direction is opposite to that of the angular displacement. Hence, this oscillatory motion is called angular S.H.M.  

29. Angular S.H.M. is defined 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 given by, T = T = Magnet Vibrating in Uniform Magnetic Field If a bar magnet is freely suspended in the plane of a uniform magnetic field. Consider, μ be the magnetic dipole moment and B the magnetic field.  

30. The magnitude of this torque is, If θ is small, Here, restoring torque is in anticlockwise Where, I – Moment of inertia T = = T =