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Transcript
A motion that repeats itself in equal interval of time is called periodic motion. The motion of the earth about its own axis, the motion of the earth round the sun, the motion of electrons in an atom are few common examples of periodic motion. When a particle in periodic motion moves to and fro along the same path, the motion is called as oscillatory or vibratory motion. The oscillation of a simple pendulum, the oscillation of a mass attached to a spring etc. are the few examples of oscillatory motion. It may be said from the above discussion that any oscillatory motion should always be periodic, but a periodic motion may not always be oscillatory in nature. The time required to complete one oscillation is called time period and the number of oscillations made per second is called the frequency of the motion. So, 1 the relation between time period and frequency is given by, The position at which on an oscillatory particle is called its equilibrium or mean position. The distance of a particle at any instant from the equilibrium position is known as displacement of the particle. The maximum displacement on either side of the equilibrium position is called amplitude of the motion. If a particle moves to and fro in such a way that the acceleration of the particle is always directed towards a fixed point in its path of motion and is proportional to the displacement of the particle from the fixed point at any instant, the motion of the particle is called simple harmonic motion (S.H.M.). The graphical representation of simple harmonic motion is given in Fig. 1. 1 2 A Textbook of Basic Engineering Physics If a particle of mass ‘ ’ executes S.H.M. and be the displacement of it from the equilibrium position, the acceleration of the particle is given by, 1.1 where is a proportionality constant. Hence, the force ‘ ’ acting on the particle is given by, 1.2 Here, is called of the particle. representing The characteristics of S.H.M are, The motion is linear. The motion is periodic and oscillatory. The restoring force acting on the body ( the acceleration of the body) is proportional and oppositely directed to its displacement measured from a fixed point ( mean position) in the path of the motion. If be the mass of the particle executing S.H.M. and be its displacement at any instant from the mean position ‘ ’ (as shown in Fig. 2), the force acting on the particle at any point is given by, 2 1.3 2 where is the constant of proportionality (force constant) and negative sign is taken, as the restoring force is opposite to the direction of displacement. The equation 1.3 may be written as, 2 2 or, or, 2 2 2 2 2 2 , 2 , where = natural angular frequency 0 This is 1.4 . The time period of a S.H.M. is given by, 2 1.5 3 Simple Harmonic Motion Multiplying both sides of equation 1.4 by 2 2 2 2 2 2 0 , we get, 2 or, 2 2 0 Integrating , we get, 2 2 2 Now, the velocity displacement , where is the constant of integration. 1.6 of the simple harmonic particle is zero at its maximum at (amplitude of the particle). 2 2 Hence, we get from the equation 1.6 , 0 So, the equation 1.6 takes the form, 2 2 2 2 or, 2 or, Proceeding with the positive sign, we have, 2 2 2 2 2 On integration, 2 or, sin or, 2 1 [where = integration constant] sin sin or, 1.7 Here is the phase of the vibrating particle and is the epoch, or initial phase of the particle ( it is the phase 0 ). Here the natural cyclic frequency of vibration 0 of the particle at . Here if 2 sin we proceed with negative sign, the solution will be of the same form . To express the solution of equation 1.7 1.7 as, = = = sin cos sin cos cos sin sin sin cos cos This is the general solution of equation sin in a more general form, let us write relation are two constants such that 1 1.5 of a S.H.M., where 2 2 and cos and tan 1 Phase is the state of motion of the particle executing S.H.M. and it indicates about the position and direction of motion at any time. 4 A Textbook of Basic Engineering Physics Now, we get from equation 1.5 by ignoring negative sign, 2 2 2 or, 1.8 acceleration , displacement [ time period of oscillation] displacement acceleration 2 We know, the displacement given by, 1.9 of a particle executing S.H.M. at any time is sin Velocity = or, cos 1.10 When the particle is at its mean position, cos 1 at time 2 2, 2 2 0, where 0. 0 so, sin , cos 1 =2 2 2 1.11 Now, when the particle is at the mean position of the path for 0, max At the extreme positions of the path for We have, from equation 1.10 , cos , min 0. 2 sin Acceleration 1.12 When the particle is at the mean position, 0. 2 sin 2 or, [ at 0, 0 so, When the particle is at extreme positions ( max 2 2 for ] ), the acceleration (neglecting the negative sign) When the particle is at the mean position ( min sin 0 0 for 0 ), the acceleration, 1.13 Simple Harmonic Motion 5 We know that the displacement of a particle executing S.H.M. at any instant of time sin is 1 velocity cos Comparing equations 1 sin and 2 2 2 we get that, the volocity of the particle executing S.H.M. at any instant leads the displacement by a phase difference in quadrature) Now differentiating eqn. 2 2 radian ( two are we get acceleration cos 2 sin 2 sin 3 Comparing equations 2 and 3 we find that the acceleration at any instant leads the velocity by radians in phase. 2 Similarly comparing equations 1 and displacement by radian. [ antiphase] 3 we find that acceleration leads the A particle executing a simple harmonic motion possesses both potential and kinetic energy at any instant. Kinetic energy ( with velocity ) The kinetic energy of a particle of mass executing S.H.M. is given by, 1 2 But, and moving 2 sin cos 1 2 1 2 1 2 2 2 2 2 2 2 cos 1 2 2 2 1.14 Potential energy ( ) The potential energy of a vibrating particle of mass at any instant is calculated from the total amount of work done in overcoming the 2 . The negative sign is taken restoring force through a distance . Here, force = as the force is opposite to the direction of motion of the vibrating particle. So, the potential energy, 2 0 1 2 2 2 1.15 6 A Textbook of Basic Engineering Physics So, we may conclude from equation 1.14 and 1.15 that, for min 0 , but 1 2 max 2 2 1 2 2 , but min 2 So, K.E. is maximum when P.E. is minimum and vice versa. Similarly, for 0, max Total energy Now, the total energy, or, = or, 1 2 2 2 2 1 2 2 2 2 1 2 2 0 = 2 2, 2 2 , where 2 constant 1.16 So, the total energy is constant in S.H.M. and is proportional to the square of the amplitude. In case of free vibration, the restoring force is proportional to its displacement and it is the only internal force acting on the body. Hence, we can say from 1.16 that the equation 1.7 and maximum amplitude of free vibration and the total energy of the (free) simple harmonic motion remain constant with time [Fig. 4]. 1 The displacement of a particle of mass 0.2 kg executing S.H.M. is indicated by 10 sin m. Calculate 3 12 i the amplitude iv the maximum velocity and maximum acceleration v the epoch ii the angular velocity iii the time period vi the phase difference after 1 second vii the displacement, velocity and acceleration of the particle after one second 7 Simple Harmonic Motion viii the kinetic energy and potential energy of the particle at displacement 0.04 m . ix the total energy of the particle. The general equation of a simple harmonic motion is given by, sin 1 10 sin Here, 3 2 12 Comparing equations 1 and 2 , we get i 10 m ii iii 3 2 , 2 6s 3 iv 2 executing S.H.M. is, 0 for max 10 3.14 3 10 3 The 10.46 m s 2 for 3.14 3 2 1 2 , max 2 10 10 3 2 10.95 m s v 3 Now, is the initial phase for We get from equation 2 , vi The phase after when 0. 12 second 1 , phase 1 3 4 1s = So, vii 3 12 12 12 4 12 4 3 = 12 = 2 180 = = = 30 12 6 6 1 second, 10 sin The 10 sin 12 10 2 4 5 2m 1 second 2 The 1 3 2 3 10 2 5 2 1 second 2 3 2 50 2 3 7.4 m s 5 2 1 7.75 m s 2 8 A Textbook of Basic Engineering Physics 1 2 viii K.E. 2 2 2 So, at displacement 1 2 0.2 3.14 3 P.E. = 1 2 0.2 3.14 3 1 2 2 2 0.04 m K.E. ix 1 2 and P.E. 2 2 2 2 1 2 10 2 0.04 0.04 2 10.95 joule 0.0001 joule 3.14 3 0.2 2 2 10 2 10.955 joule 2 The differential equation of motion of a freely oscillating body is given by 2 2 2 2 18 0 . Calculate the natural frequency of the body. According to the problem, 2 2 9 3 2 or, 2 3 or, 2 18 2 2 or, 2 0 2 or, 2 9 2 3 1.5Hz . 3 A weight glass tube is floating in liquid with 20 cm of its length immersed. It is pushed down a certain distance and then released. Compute the time period of its vibration. Let the density of the liquid be the glass tube be . g cm 3 and the area of cross-section of The volume of the liquid displaced by the glass tube = mass of the displaced liquid = 20 Mass of the tube = cm 3 20 g Now, if we consider the tube is pushed by a further length of upthrust acting on the glass tube, 2 or, or, cm, the additional 2 2 2 2 , This equation implies that the motion is simple harmonic. The time period, 2 2 =2 20 =2 20 980 0.89s .