Tiêu đề 03A.SHM-PHY LEVEL - V ( 97 -114 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 97 JEE ADVANCED - VOL - III JEE MAINS - CW - VOL - I SHM NISHITH Multimedia India (Pvt.) Ltd., SYNOPSIS NOTE: Read this synopsis after mains synopsis. * Phase diagram :- 1) It is visual representation of a motion of simple harmonic oscillator 2) It is represented by a circle in XY-plane centered at origin and having radius A which is amplitude of S.H.M. 3) The radius vector is making an angle,  , with positive X-axis at t = 0 and is called initial phase. 4) This radius is rotating with angular velocity,  , in anti-clockwise direction. Here  represents angular frequency of S.H.M. 5) The projection of the tip of the radius on y- axis will be equal to the displacement of the body executing S.H.M. From its position.    x A t sin    6) This phase diagram is used a) to compare the motion of two bodies executing S.H.M. b) to find out the phase when velocity of body executing SHM change abruptly. General motion near equilibrium :- 1) For SHM, the restoring force acting on oscillator when it is in disturbed condition must be towards mean position 2) Let us analyses following case The corresponding potential energy curve is : From above, it is clear that A is stable equilibrium point and B is at unstable equilibrium point. And at A, the curve is at near parabolic in nature. Hence at A, body performs S.H.M.  At A ie : 1 x x  , the potential energy can be written as :sssss   2 1 U P Q x x P Q    ; , are constants and force, dU F dx   conservative    2Q x x  1     k x x  1  particle executes S.H.M. about 1 x x  SIMPLE HARMONIC MOTION
SHM 98 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., * Problem solving strategy in S.H.M I. Force and Torque method : a) identification of mean position b) determine small displacement from mean position c) draw free body diagram in that disturbed position d) apply newton’s law to find acceleration of oscillator. i.e:   F a for linear m    (or)  for angular I      e) compare these equations with S.H.M. equations : i.e. 2 a x   (or) 2      f) determine  and hence time period from these comparisons II. Energy method : a) Determine total energy of oscillator when it is in disturbed position i.e., Kinetic, potential in both translational & rotational as applicable. b) As total energy of oscillator constant, use   0 d total energy dt  c) From that condition, obtain S.H.M. differential equation as 2 2 2 0 d x x dt    (or) 2 2 2 0 d dt      d) find  & hence time period, T. III. Superposition of two SHMs :- Case-I :- Two SHMs are of same frequency and moving in same direction:- Let : 1 1 x A t  cos x A t 2 2   cos  ; where  is the phase difference between x x 1 2 & The resultant wave equation is 1 2 x x x     A t R sin   . where 2 2 1 2 1 2 2 cos A A A A A R     and 1 2 1 2 sin tan cos A A A             Case-II :- Two SHMs are of different frequencies and moving in same direction :- Let 1 1 1 2 2 2 x A t x A t   cos ; cos    The resultant motion is : 1 2 x x x   1 1 2 2   A t A t cos cos   1 2 1 2 2 cos cos 1 2 2 A t t                          1 2 .cos 2 R x A t           The motion is said to harmonic having frequency of 1 2 2 t         Here amplitude is getting fluctuations in between A A 1 2  as A A 1 2  with frequency of   1 2   Case-III :- Two SHMs moving in perpendicular directions :- Let : x A t  cos ,  along X-axis y B t   cos ,    along Y-axis  Resultant path equation is represented as : 2 2 2 2 2 2 cos sin x y xy A B AB      This represents ellipse whose axes are rotated with respect to coordinate axes special cases have been mentioned below
NISHITH Multimedia India (Pvt.) Ltd., 99 JEE ADVANCED - VOL - III JEE MAINS - CW - VOL - I SHM NISHITH Multimedia India (Pvt.) Ltd., * Damped oscillations :- Here some non-conservative forces are acting on oscillator,. The differential equation of damped oscillation is : 2 2 0 d x dx m b kx dt dt    ; b = damping constant, k = spring constant, m = mass of oscillator The possibilities are : 1) Underdamped motion :- Here b is small enough, such that 2 b k m m  displacement is     2 cos b t m x t Ae t            Where 2 2 0 2 b m          ; 0 k m   ;   phase constant Energy of oscillator is: 0 t E E e    ; 2 2 0 0 1 ; 2 m E m A b     Quality factor is : 0 0 m Q b      The fractional energy loss per one period is: E E  in the period T 2 Q     2 cycle Q E E           Power dissipation is 2 ; dE P bv dt    v instantaneous velocity. 2) Overdamped case :- Here 0 2 b m   displacement     2 b t t t m x t Ae Be e              where 2 2 0 2 b m            ; A, B are constants 3) Critically damped motion :- Here 0 2 b m   displacement is :     2 b t m x t C Dt e         ; C, D are constants Plots :
SHM 100 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., LEVEL-V SINGLE ANSWER QUESTIONS 1. A plank of negligible mass oscillates with a body of mass m as governed by y t t   sin 3 cos   in a vertical plane. The time at which the mass just looses its contact with plank is : (A) g  3 2 (B) g 2 3  (C) g  3 2 (D) 6 2  g 2. Simple Pendulum attached to the ceiling of a stationary lift has a time period T. If the lift moves upward with distance covered as y = (1 m/s2)t2, the time period of the pendulum will be (take g = 10.0 m/s2) A) 0.6T B) 5 6 T C) T D) 6 5 T 3. A block of mass m moves with a speed v towards the right block which is in equilibrium with a spring attached to rigid wall. If the surface is frictionless and collisions are elastic, the frequency of collisions between the masses will be : K m v m L (A) m K L v  1 2  (B)        m K L v  1 2 2 (C)        K m v L  2 2 (D) none of these 4. A mass of 0.98 kg suspended using a spring of constant k= 300 N 1 m is hit by a bullet of 20gm moving with a velocity 3m/s vertically. The bullet gets embedded and oscillates with the mass ina vertical plane. The amplitude of oscillation will be : (A) 0.15cm (B) 0.12 cm (C) 1.2cm (D) 12 m 5. Two simple pendulam of length 1 m and 16 m respectively are given small displacement at the same time in the same direction. The number of oscillations ‘N’ of the smaller pendulum for them to be in phase again is : (A) 3 4 (B) 4 3 (C) 4 (D)16 1 6. Driver of a car records a period of 3  seconds a pendulum of 1m hung from the roof. The acceleration of the car is (A) 2 10  ms (B) 2 15  ms (C) 2 17.2  ms (D) 34.5 ms–2 . 7. A uniform rod of length l is mounted so as to rotate about a horizontal axis perpendicular to the rod and at a distance x from the centre of mass. The time period will be the least when x is (A) 4 l (B) 2 l (C) 3 l (D) 12 l 8. A bob of mass M is hung using a string of length l. A mass m moving with a velocity u pierces through the bob and emerges out with velocity 3 u horizontally. The frequency of small oscillations of the bob, considering A as amplitude is, (A) 1 3 2 2 mu  MA (B) MA m 3 2 2 1  (C)        1 2mu 2 3MA (D)        1 3mu 2 2MA 9. A particle oscillating as given by U(y) = 3 K y with force constant K has an amplitude A. The maximum velocity during the oscillation is proportional : (A) to A (B) proportional to 3 A (C) 2 3 A m K (D) 2m A K