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751 60 Simple Harmonic Motion E3 16 Periodic and Oscillatory (Vibratory) Motion ID Simple Harmonic Motion (ii) Harmonic oscillation is that oscillation which can be expressed in terms of single harmonic function (i.e. sine or cosine function). Example : y  a sin t or y  a cos  t D YG U (iii) Non...

751 60 Simple Harmonic Motion E3 16 Periodic and Oscillatory (Vibratory) Motion ID Simple Harmonic Motion (ii) Harmonic oscillation is that oscillation which can be expressed in terms of single harmonic function (i.e. sine or cosine function). Example : y  a sin t or y  a cos  t D YG U (iii) Non-harmonic oscillation is that oscillation which can not be expressed in terms of single harmonic function. It is a combination of two Example : or more than two harmonic oscillations. y  a sin t  b sin2 t. (1) A motion, which repeat itself over and over again after a regular interval of time is called a periodic motion Revolution of earth around the sun (period one year), Rotation of earth about its polar axis (period one day), Motion of hour’s hand of a clock (period 12-hour) etc are common example of periodic motion. U (2) Oscillatory or vibratory motion is that motion in which a body moves to and fro or back and forth repeatedly about a fixed point in a definite interval of time. In such a motion, the body is confined with in welldefined limits on either side of mean position. Oscillatory motion is also called as harmonic motion. (i) Common examples are ST (a) The motion of the pendulum of a wall clock (b) The motion of a load attached to a spring, when it is pulled and then released. (c) The motion of liquid contained in U-tube when it is compressed once in one limb and left to itself. (d) A loaded piece of wood floating over the surface of a liquid when pressed down and then released executes oscillatory motion. Simple Harmonic Motion (1) Simple harmonic motion is a special type of periodic motion, in which a particle moves to and fro repeatedly about a mean position. (2) In linear S.H.M. a restoring force which is always directed towards the mean position and whose magnitude at any instant is directly proportional to the displacement of the particle from the mean position at that instant i.e. Restoring force  Displacement of the particle from mean position. F  – x  F = – kx Where k is known as force constant. Its S.I. unit is Newton/meter and dimension is [MT ]. –2 (3) In stead of straight line motion, if particle or centre of mass of body is oscillating on a small arc of circular path, then for angular S.H.M. Restoring torque ()  – Angular displacement () Some Important Definitions (1) Time period (T) : It is the least interval of time after which the periodic motion of a body repeats itself. S.I. unit of time period is second. (2) Frequency (n) : It is defined as the number of oscillations executed by body per second. S.I unit of frequency is hertz (Hz). (3) Angular Frequency () : Angular frequency of a body executing periodic motion is equal to product of frequency of the body with factor 2 . Angular frequency  = 2  n 752 Simple Harmonic Motion Its unit is rad/sec. (4) Phase () : Phase of a vibrating particle at any instant is a physical quantity, which completely express the position and direction of motion, of the particle at that instant with respect to its mean position. In oscillatory motion the phase of a vibrating particle is the argument of sine or cosine function involved to represent the generalised equation of motion of the vibrating particle. y  a sin  a sin( t   0 ) (i) y  a sin t when the time is noted from the instant when the vibrating particle is at mean position. when the time is noted from the instant when (ii) y  a cos  t the vibrating particle is at extreme position. 0 t (4) If the projection of P is taken on X-axis then equations of S.H.M. can be given as 60  = Initial phase or epoch. It is the phase of a vibrating particle at t =  2  x  a cos ( t   )  a cos  t     a cos (2n t   )  T  t=0 t 0 2 t  a sin 2 n t  a sin( t   ) T (iii) y  a sin( t   ) when the vibrating particle is  phase leading or lagging from the mean position. here,    t   0 = phase of vibrating particle. –a +a –a E3 0. y  a sin t  a sin x = a sin  t x = – a sin  t x = – a cos  t (A) Fig. 16.2 x = a cos  t (B) Fig. 16.4 (5) Direction of displacement is always away from the equilibrium position, particle either is moving away from or is coming towards the equilibrium position. ID (1) Same phase : Two vibrating particle are said to be in same phase, if the phase difference between them is an even multiple of  or path difference is an even multiple of ( / 2) or time interval is an even multiple of (T / 2) because 1 time period is equivalent to 2 rad or 1 wave length (). +a Velocity in S.H.M. (1) Velocity of the particle executing S.H.M. at any instant, is defined as the time rate of change of its displacement at that instant. U (2) Opposite phase : When the two vibrating particles cross their respective mean positions at the same time moving in opposite directions, then the phase difference between the two vibrating particles is 180. o multiple of  (say (T / 2). D YG Opposite phase means the phase difference between the particle is an odd multiple of  (say , 3, 5, 7…..) or the path difference is an odd  3 2 , 2 ,.......) or the time interval is an odd multiple of (3) Phase difference : equation are So v  dy 2  a cos  t  a 1  sin  t   a 2  y 2 dt [As sin t = y/a] If two particles performs S.H.M and their y 1  a sin( t  1 ) and y 2  a sin( t   2 ) then phase difference   ( t   2 )  ( t  1 )   2  1 U Displacement in S.H.M. (1) The displacement of a particle executing S.H.M. at an instant is defined as the distance of particle from the mean position at that instant. ST (2) Simple harmonic motion is also defined as the projection of uniform circular motion on any diameter of circle of reference. (3) If the projection is taken on y-axis. then from the figure X (2) In case of S.H.M. when motion is considered from the equilibrium position, displacement y  a sin t Y P N y O Y Fig. 16.3 M max (4) At extreme position (y =  a and  = t =/2), velocity of oscillating particle is zero i.e. v = 0. v2 (5) From v   a 2  y 2  v 2   2 (a2  y 2 )   v2 a 2 2  y2 1 a2 2 X  a2  y 2 v a This is the equation of ellipse. Hence the graph between v and y is an ellipse. For  = 1, graph between v and y is a circle. a  = t (3) At mean position or equilibrium position (y = 0 and  = t = 0), velocity of particle is maximum and it is v = a. y a Fig. 16.5 (6) Direction of velocity is either towards or away from mean position depending on the position of particle. Acceleration in S.H.M. (1) The acceleration of the particle executing S.H.M. at any instant, is defined as the rate of change of its velocity at that instant. So acceleration Simple Harmonic Motion A 753 or dv d 2  (a cos  t)   a sin t   2 y dt dt v   a2  y 2 [As y  a sin t ] Acceleration Acceleration   2 y (3) In S.H.M. acceleration is maximum at extreme position (at y =  a). Hence Amax   2a when sin t  maximum  1 i.e. at t   2. From equation (ii) | Amax |   2a T 4 T/2 O or Fmax F  m y T/2 O A –a y Fig. 16.6  U  dU   dW   U Comparative Study of Displacement Velocity and Acceleration (i) The restoring force F = – ky against which work has to be done. Hence potential energy U is given by  – 2a 2 Energy in S.H.M. ID +a Slope of the line = –  D YG (1) All the three quantities displacement, velocity and acceleration show harmonic variation with time having same period. (ii) Also U  times the displacement (4) In S.H.M. the velocity is ahead of displacement by a phase angle  (5) In S.H.M. the acceleration is ahead of velocity by a phase angle  / U 2 ST (6) The acceleration is ahead of displacement by a phase angle of  Table 16.1 : Various physical quantities in S.H.M. at different position : Graph Formula At mean position O y=0 At extreme position U max  O T 1 m 2y 2 2 [As  2  k / m ] when y  0 ;  t  0 ; t  0 U min  0 (2) Kinetic energy : This is because of the velocity of the particle 1 1 mv 2  m  2 (a 2  y 2 ) 2 2 [As v   a 2  y 2 ] v  a cos  t T/2 0 (iv) Potential energy is minimum at mean position (i)  a sin( t  1 2 ky + U 2 T 1 2 1 ka  m  2 a 2 when y  a ;  t   / 2 ; t  2 2 4 y=a –a vmax ky dy  (iii) Potential energy maximum and equal to total energy at extreme positions T Velocity y 0 Hence potential energy varies periodically with double the frequency of S.H.M. Kinetic Energy K  y  a sin  t T/2  [As y  a sin t ] Displacement +a Fdx  1 1 m  2 a 2 sin2  t  m  2 a 2 (1  cos 2 t) 2 4 (2) The velocity amplitude is  times the displacement amplitude /2 x 0 0 0 (3) The acceleration amplitude is  amplitude  where U = Potential energy at equilibrium position. If U = 0 then U  2 m  2a (1) Potential energy : This is an account of the displacement of the particle from its mean position. 2a Graph between acceleration (A) and displacement (y) is a straight line as shown T E3 (ii) Acceleration is always directed towards the mean position and so is always opposite to displacement Fmax  Fmin = 0 2 T or  t  . From equation (ii) A min  0 when y  0 2 A  y |Amax| = 2a F   m  2 a sin  t Force when y  a. From equation (i) Amin  0 when sin t  0 i.e. at t  0 or i.e., T or (i) In S.H.M. acceleration is minimum at mean position t  a 2 sin( t   ) Amin = 0 or A   2y Amax equations of translatory motion can not be applied. t  A  a 2 sin  t is not constant. So 60 (2) In S.H.M. as  2 vmax =a ) vmin = 0 Also K  1 1 m  2 a 2 cos 2  t  m  2 a 2 (1  cos 2t) 2 4 [As v  a cos  t ] Hence kinetic energy varies periodically with double the frequency of S.H.M. 754 Simple Harmonic Motion (ii) Kinetic energy is maximum at mean position and equal to total energy at mean position. Step 1 : When particle is in its equilibrium position, balance all forces acting on it and locate the equilibrium position mathematically. 1 1 1 m  2 (a 2  y 2 )  m  2 y 2  m  2 a 2 2 2 2 Step 2 : From the equilibrium position, displace the particle slightly by a displacement y and find the expression of net restoring force on it. Step 3 : Try to express the net restoring force acting on particle as a proportional function of its displacement from mean position. The final expression should be obtained in the form. 60 when y  a ; t  T / 4 ,  t   / 2 (3) Total mechanical energy : Total mechanical energy always remains constant and it is equal to sum of potential energy and kinetic energy i.e. E UK E [As c = Restoring torque constant and I = Moment How to Find Frequency and Time Period of S.H.M. (iii) Kinetic energy is minimum at extreme position. K min  0 c I of inertia] 1 m  2 a 2 when y  0 ; t  0 ;  t  0 2 K max  where  2  F   ky Total energy is not a position function. Here we put – ve sign as direction of F is opposite to the displacement y. If a be the acceleration of particle at this displacement, we have (4) Energy position graph k  a   y m  E3 Energy Total energy (E) Step 4 : Comparing this equation with the basic differential equation Potential energy (U) of S.H.M. we get  2  y =+ a y=0 y =– a of oscillation can be given as T  Fig. 16.7 (i) At y = 0; U = 0 and K = E U a E 3E ; U  and K  2 4 4 a 2 D YG (iv) At y   E 2 ; UK Average Value of P.E. and K.E. The average value of potential energy for complete cycle is given by Uaverage  1 T  T U dt  0 1 T  T1 2 0 m  2 a 2 sin2 ( t   )  1 m  2a2 4 The average value of kinetic energy for complete cycle 1 T  K dt  1 T  T1 m  2a 2 cos 2  t dt  U Kaverage  T 0 0 2 1 m  2a2 4 ST Thus average values of kinetic energy and potential energy of harmonic oscillator are equal and each equal to half of the total energy 1 1 Kaverage  Uaverage  E  m  2 a 2. 2 4 Differential Equation of S.H.M. For S.H.M. (linear) Acceleration  – (Displacement) A  y or A   y or 2 or m d 2y dt 2  ky  0 For angular S.H.M. 2  m k  2 (i) In different types of S.H.M. the quantities m and k will go on taking different forms and names. In general m is called inertia factor and k is called spring factor. (ii) At y =  a; U = E and K =0 (iii) At y   k m As  is the angular frequency of the particle in S.H.M., its time period ID Kinetic energy (K) k 1 or n  m 2 k  m d 2y dt 2 [As      c   y 2 Thus T  2 1 Inertia factor or n  2 Spring factor (ii) In linear S.H.M. the spring factor stands for force per unit displacement and inertia factor for mass of the body executing S.H.M. and in Angular S.H.M. k stands for restoring torque per unit angular displacement and inertial factor for moment of inertia of the body executing S.H.M. For linear S.H.M. T  2 m  k m Force/Displacement d 2   2  0 dt 2  2 Displacement Accelerati on Simple Pendulum (1) An ideal simple pendulum consists of a heavy point mass body (bob) suspended by a weightless, inextensible and perfectly flexible string from a rigid support about which it is free to oscillate. (2) But in reality neither point mass nor weightless string exist, so we can never construct a simple pendulum strictly according to the definition. (3) Suppose simple pendulum of length l is displaced through a small angle  from it’s mean (vertical) position. Consider mass of the bob is m and linear displacement from mean position is x S k ] m and Spring factor Inertia factor  T l y O mg sin  Fig. 16.8 P  mg mg cos  Simple Harmonic Motion 1 (4) Effect of g : T  755 i.e. as g increase T decreases. g (i) As we go high above the earth surface or we go deep inside the mines the value of g decrease, hence time period of pendulum (T) increases. Restoring force acting on the bob or (When  is small sin  ~   g So T  2 1 T F mg   k (Spring factor) x l 1 g T g Inertia factor m  2  2 Spring factor mg / l l g 1 g T Factor Affecting Time Period of Simple Pendulum 0 Fig. 16.10 (5) Effect of temperature on time period : If the bob of simple pendulum is suspended by a wire then effective length of pendulum will increase with the rise of temperature due to which the time period will increase. l  l0 (1    ) (If  is the rise in temperature, l0  initial ID (1) Amplitude : The period of simple pendulum is independent of amplitude as long as its motion is simple harmonic. But if  is not small, sin    then motion will not remain simple harmonic but will become oscillatory. In this situation if  is the amplitude of motion. Time period T E3  (iii) Different graphs x Arc OP = = ) Length l l 60 F  mg sin (ii) If a clock, based on simple pendulum is taken to hill (or on any other planet), g will decrease so T will increases and clock will become slower. x F  mg  mg l length of wire, l = final length of wire)    2 l  1 2 0   .......   T0 1  0 1  2 sin  g 16 2  2       T  T0 U T  2 (2) Mass of the bob : Time period of simple pendulum is also independent of mass of the bob. This is why D YG (i) If the solid bob is replaced by a hollow sphere of same radius but different mass, time period remains unchanged. (ii) If a girl is swinging in a swing and another sits with her, the time period remains unchanged. (3) Length of the pendulum : Time period T  l where l is the distance between point of suspension and center of mass of bob and is called effective length. (i) When a sitting girl on a swinging swing stands up, her center of mass will go up and so l and hence T will decrease. ST U (ii) If a hole is made at the bottom of a hollow sphere full of water and water comes out slowly through the hole and time period is recorded till the sphere is empty, initially and finally the center of mass will be at the center of the sphere. However, as water drains off the sphere, the center of mass of the system will first move down and then will come up. Due to this l and hence T first increase, reaches a maximum and then decreases till it becomes equal to its initial value. T2 T 1  1    T0 2 i.e. T 1    T 2 Oscillation of Pendulum in Different Situations (1) Oscillation in liquid : If bob a simple pendulum of density  is made to oscillate in some fluid of density  (where 

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