Chapter 14 PDF

E3 60 170 Simple Harmonic Motion 15.1 Periodic Motion. ID A motion, which repeat itself over and over again after a regular interval of time is called a periodic motion and the fixed interval of time after which the motion is repeated is called period of the motion. U Examples : (i) Revolution of ea...

E3 60 170 Simple Harmonic Motion 15.1 Periodic Motion. ID A motion, which repeat itself over and over again after a regular interval of time is called a periodic motion and the fixed interval of time after which the motion is repeated is called period of the motion. U Examples : (i) Revolution of earth around the sun (period one year) D YG (ii) Rotation of earth about its polar axis (period one day) (iii) Motion of hour’s hand of a clock (period 12-hour) (iv) Motion of minute’s hand of a clock (period 1-hour) (v) Motion of second’s hand of a clock (period 1-minute) (vi) Motion of moon around the earth (period 27.3 days) U 15.2 Oscillatory or Vibratory Motion. ST 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. Example : (i) The motion of the pendulum of a wall clock. (ii) The motion of a load attached to a spring, when it is pulled and then released. Simple Harmonic Motion 171 (iii) The motion of liquid contained in U- tube when it is compressed once in one limb and left to itself. (iv) A loaded piece of wood floating over the surface of a liquid when pressed down and then released 60 executes oscillatory motion. 15.3 Harmonic and Non-harmonic Oscillation. sine or cosine function). Example : y  a sin  t or y  a cos  t E3 Harmonic oscillation is that oscillation which can be expressed in terms of single harmonic function ( i.e. Non-harmonic oscillation is that oscillation which can not be expressed in terms of single harmonic 15.4 Some Important Definitions. ID function. It is a combination of two or more than two harmonic oscillations. Example : y  a sin  t  b sin 2 t S.I. units of time period is second. U (1) Time period : It is the least interval of time after which the periodic motion of a body repeats itself. D YG (2) Frequency : It is defined as the number of periodic motions 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 S.I. units of  is Hz [S.I.]  also represents angular velocity. In that case unit will be rad/sec. U (4) Displacement : In general, the name displacement is given to a physical quantity which undergoes a change with time in a periodic motion. ST Examples : (i) In an oscillation of a loaded spring, displacement variable is its deviation from the mean position. (ii) During the propagation of sound wave in air, the displacement variable is the local change in pressure (iii) During the propagation of electromagnetic waves, the displacement variables are electric and magnetic fields, which vary periodically. (5) 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. 172 Simple Harmonic Motion 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. here,    t   0 = phase of vibrating particle. (i) Initial phase or epoch : It is the phase of a vibrating particle at t = 0. In    t   0 , when t = 0;    0 here,  0 is the angle of epoch. 60 y  a sin   a sin( t   0 ) E3 (ii) 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 / ID 2) because 1 time period is equivalent to 2 rad or 1 wave length () (iii) 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 U Opposite phase means the phase difference between the particle is an odd multiple of  (say , 3, 5, / 2). D YG 7…..) or the path difference is an odd multiple of  (say  3 2 , 2 ,.......) or the time interval is an odd multiple of ( T (iv) Phase difference : If two particles performs S.H.M and their equation are y 1  a sin( t  1 ) and y 2  a sin( t   2 ) then phase difference   ( t   2 )  ( t  1 )   2   1 U 15.5 Simple Harmonic Motion. ST Simple harmonic motion is a special type of periodic motion, in which a particle moves to and fro repeatedly about a mean position under 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. 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]. Simple Harmonic Motion 173 15.6 Displacement in S.H.M.. The displacement of a particle executing S.H.M. at an instant is defined as the distance of particle from the 60 mean position at that instant. As we know that simple harmonic motion is defined as the projection of uniform circular motion on any diameter of circle of reference. If the projection is taken on y-axis. Y P N E3 then from the figure y  a sin  t X 2 y  a sin t T ID y  a sin 2 n t y  a sin( t   ) y O a  =t M X Y U where a = Amplitude,  = Angular frequency, t = Instantaneous time, T = Time period, n = Frequency and  = Initial phase of particle D YG If the projection of P is taken on X-axis then equations of S.H.M. can be given as x  a cos ( t   )  2  x  a cos  t   T  x  a cos (2n t   ) U Important points ST (i) y  a sin  t (ii) y  a cos  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 the vibrating particle is at extreme position. (iii) y  a sin( t   ) when the vibrating particle is  phase leading or lagging from the mean position. (iv) Direction of displacement is always away from the equilibrium position, particle either is moving away from or is coming towards the equilibrium position. (v) If t is given or phase ( ) is given, we can calculate the displacement of the particle. 174 Simple Harmonic Motion T  (or   ) 2 4 Similarly if t  then from equation y  a sin 2 2 T    a sin    a t , we get y  a sin T T 4 2 T (or    ) then we get y  0 2 Sample problems based on Displacement A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from E3 Problem 1. x  A to x  A / 2 is (a) T / 6 (b) T / 4 (c) T / 3 [CBSE 1992; SCRA 1996] (d) T / 2 Because the S.H.M. starts from extreme position so y  a cos t form of S.H.M. should be used. ID Solution : (a) 60 If t  A  2 2  A cos t  cos  cos t  t T /6 2 3 T T A mass m = 100 gms is attached at the end of a light spring which oscillates on a friction less horizontal U Problem 2. table with an amplitude equal to 0.16 meter and the time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement x = – 0.16 meter. The expression for the displacement of the D YG mass at any time (t) is (a) x  0.16 cos ( t) Solution : (b) (c) x  0.16 cos( t   ) (d) x  0.16 cos( t   ) Standard equation for given condition x  a cos 2 t  x  0.16 cos( t) T [As a = – 0.16 meter, T = 2 sec] The motion of a particle executing S.H.M. is given by x  0.01 sin 100 (t .05). Where x is in meter and U Problem 3. (b) x  0.16 cos( t) [MP PMT 1995] time t is in seconds. The time period is ST (a) 0.01 sec Solution : (b) (c) 0.1 sec (d) 0.2 sec By comparing the given equation with standard equation y  a sin( t   )   100  so T  Problem 4. (b) 0.02 sec 2   2  0. 02 sec 100  Two equations of two S.H.M. are x  a sin( t   ) and y  b cos( t   ). The phase difference between the two is (a) 0o Solution : (c) [MP PMT 1985] (b) o (c) 90o x  a sin( t   ) and y  b cos( t   ) = b sin( t     / 2) (d) 180o Simple Harmonic Motion 175  Now the phase difference = ( t    )  ( t   )   / 2  90 o 2 15.7 Velocity in S.H.M.. 60 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. In case of S.H.M. when motion is considered from the equilibrium position E3 y  a sin  t dy  a cos  t dt  v  a cos t or v  a 1  sin 2  t or v   a2  y 2 Important points ……(i) ID v [As sin  t = y/a] …..(ii) U so D YG (i) In S.H.M. velocity is maximum at equilibrium position. From equation (i) vmax  a when from equation (ii) vmax  a when cos  t =1 i.e. = t=0 i.e  t  y 0 (ii) In S.H.M. velocity is minimum at extreme position. U From equation (i) From equation (ii) vmin  0 when cos  t = 0 v min  0 when y = a  2 ST (iii) Direction of velocity is either towards or away from mean position depending on the position of particle. Problem 5. Sample problems based on Velocity A body is executing simple harmonic motion with an angular frequency 2 rad/sec. The velocity of the body at 20 mm displacement. When the amplitude of motion is 60 mm is (a) 40 mm/sec Solution : (c) (b) 60 mm/sec v   a2  y 2  2 (60)2  (20)2 = 113 mm/sec (c) 113 mm/sec [AFMC 1998] (d) 120 mm/sec 176 Simple Harmonic Motion Problem 6. A body executing S.H.M. has equation y  0.30 sin(220 t  0.64 ) in meter. Then the frequency and maximum velocity of the body is [AFMC 1998] Solution : (a) (b) 45 Hz , 66 m / s (c) 58 Hz , 113 m / s By comparing with standard equation y  a sin( t   ) we get a  0.30 ;   220  2n  220  n  35 Hz so v max  a  0.3  220  66 m / s A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its E3 Problem 7. (d) 35 Hz , 132 m / s 60 (a) 35 Hz , 66 m / s speed is half of the maximum speed. Its displacement y is (a) A/2 (b) A / 2 (c) (d) 2 A / 3 A 3 /2 a a2   a2  y 2   a2  y 2  y  2 4 3A 2 v max a ]  2 2 v   a2  y 2  Problem 8. A particle perform simple harmonic motion. The equation of its motion is x  5 sin(4 t  [As v  U ID Solution : (c) displacement. If the displacement of the particle is 3 units then its velocity is (b) 5 / 6 D YG (a) 2 / 3 (c) 20  6 ). Where x is its [MP PMT 1994] (d) 16 Solution : (d) v   a2  y 2  4 5 2  3 2 = 16 [As  = 4, a = 5, y = 3] Problem 9. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude (A) and time period (T). The speed of the pendulum at x  A 3 (b) T U (a) v   a2  y 2  v  ST Solution : (a) Problem 10. 2 T A will be 2 A T A2  A 3 A2  4 T [MP PMT 1987] (c) A 3 2T (d) 3 2 A T [As y = A/2] A particle is executing S.H.M. if its amplitude is 2 m and periodic time 2 seconds. Then the maximum velocity of the particle will be (a) 6 (b) 4 (c) 2 (d)  2 2 2  v max  2 T 2 Solution : (c) v max  a  a Problem 11. A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be (a) Solution : (d) 4a T v max  a  (b) a 2 T 2a T (c) 2 a T [MP PMT 1985] (d) 2a T Simple Harmonic Motion 177 Problem 12. A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm its maximum speed in cm/sec is [AIIMS 1982] (b)  (c) 2 (d) 3 60 (a)  / 2 2 2  v max   3 T 6 v max  a  a Problem 13. A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec, at a distance (a) 5 Solution : (c) (b) 5 2 E3 Solution : (b) (c) 5 3 [CPMT 1976] (d) 10 2 v max  a  100 cm / sec and a  10 cm so   10 rad / sec. ID  v   a2  y 2  50 = 10 10 2  y 2  y  5 3 15.8 Acceleration in S.H.M.. U 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 A  dv  d (a cos  t) dt dt ……(i) A   2 y ……(ii) D YG A   2 a sin  t [As y  a sin  t ] Important points (i) In S.H.M. as Accelerati on   2 y is not constant. So equations of translatory motion can not be applied. (ii) In S.H.M. acceleration is maximum at extreme position. sin  t  maximum  1 i.e. at t  when From equation (ii) | A max |   2 a when y  a ST U From equation (i) A max   2 a T or t   4 2 (iii) In S.H.M. acceleration is minimum at mean position From equation (i) A min  0 when sin  t  0 i.e. at t  0 or t  From equation (ii) A min  0 when y  0 T or  t   2 (iv) Acceleration is always directed towards the mean position and so is always opposite to displacement i.e., A  y 15.9 Comparative Study of Displacement, Velocity and Acceleration. 178 Simple Harmonic Motion y  a sin  t  Velocity Acceleration +a v  a cos  t  a sin( t  ) 2 y A  a 2 sin  t  a 2 sin( t   ) +a v From the above equations and graphs we can conclude –a – a 0 T 0 T +a2 that. (i) All the three quantities displacement, velocity and acceleration show harmonic variation with time having same y = a sin t 1 T 2 T Velocity v = a cos t Acceleration 2T A = – a2 cos t T 0 – a2 Displacement 3 T 2 2T Time ID period. 2T E3 a 2T 60 Displacement (ii) The velocity amplitude is  times the displacement amplitude U (iii) The acceleration amplitude is  2 times the displacement amplitude (iv) In S.H.M. the velocity is ahead of displacement by a phase angle  / 2 D YG (v) In S.H.M. the acceleration is ahead of velocity by a phase angle  / 2 (vi) The acceleration is ahead of displacement by a phase angle of  (vii) Various physical quantities in S.H.M. at different position : Physical quantities Equilibrium position (y = 0) Extreme Position (y =  a) Minimum (Zero) Maximum (a) Velocity v   a2  y 2 Maximum (a) Minimum (Zero) Minimum (Zero) Maximum (  2 a ) ST U Displacement y  a sin  t Acceleration A   2 y 15.10 Energy in S.H.M.. A particle executing S.H.M. possesses two types of energy : Potential energy and Kinetic energy (1) Potential energy : This is an account of the displacement of the particle from its mean position. The restoring force F  ky against which work has to be done Simple Harmonic Motion 179   x  y 1 2 ky 2 So U   dw    potential Energy U 1 m 2 y 2 2 [As  2  k / m ] U 1 m  2 a 2 sin 2  t 2 [As y  a sin  t ] 0 k y dy  60 Fdx  Important points E3 0 (i) Potential energy maximum and equal to total energy at extreme positions 1 2 1 ka  m  2 a 2 2 2 U max  when y  a ;  t   / 2 ; t  T / 4 ID (ii) Potential energy is minimum at mean position when y  0 ;  t  0 ; t  0 U min  0 K 1 mv 2 2 D YG Kinetic Energy U (2) Kinetic energy : This is because of the velocity of the particle K 1 ma 2 2 cos 2  t 2 [As v  a cos  t ] K 1 m  2 (a 2  y 2 ) 2 [As v   a 2  y 2 ] (i) Kinetic energy is maximum at mean position and equal to total energy at mean position. 1 m 2a2 2 U K max  when y  0 ; t  0 ;  t  0 (ii) Kinetic energy is minimum at extreme position. ST K min  0 when y  a ; t  T / 4 ,  t   / 2 (3) Total energy : Total mechanical energy = Kinetic energy + Potential energy E 1 1 1 m  2 (a 2  y 2 )  m  2 y 2  m  2 a 2 2 2 2 Total energy is not a position function i.e. it always remains constant. (4) Energy position graph : Kinetic energy (K)  1 m  2 (a 2  y 2 ) 2 Potential Energy (U) = Energy 1 m 2y 2 2 U K y =– a y=0 y =+ a 180 Simple Harmonic Motion Total Energy (E) = 1 m 2a2 2 It is clear from the graph that 60 (i) Kinetic energy is maximum at mean position and minimum at extreme position (ii) Potential energy is maximum at extreme position and minimum at mean position (5) Kinetic Energy Potential Energy E3 (iii) Total energy always remains constant. K 1 1 1 m  2 a 2 cos 2  t  m  2 a 2 (1  cos 2 t)  E(1  cos  ' t) 2 4 2 U 1 1 1 m  2 a 2 sin 2  t  m  2 a 2 (1  cos 2 t)  E(1  cos  ' t) 2 2 4 ID where  '  2 and E  1 m  2 a 2 i.e. in S.H.M., kinetic energy and potential S.H.M. (i.e. with time period T '  T / 2 ) T otalenergy  U energy vary periodically with double the frequency of Energy 2 1 2 2 2 m a E KE  D YG From the graph we note that potential energy or kinetic energy completes two vibrations in a time PE  0 1 2 during which S.H.M. completes one vibration. Thus the T T 1 2 1 2 2 2 2 m  a cos t 2 2 2 m  a sin t Time frequency of potential energy or kinetic energy double than that of S.H.M. U Sample problems based on Energy Problem 14. A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy ST changes into potential energy is (a) f / 2 (b) f (c) 2 f (d) 4 f Solution : (c) Problem 15. When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ‘ a’ is [CBSE 1993; MP PMT 1994; MP PET 1995, 96; MP PMT 2000] (a) a / 4 (b) a / 3 (c) a / 2 (d) 2a / 3 Simple Harmonic Motion 181 Solution : (c) According  problem potential energy = 1 maximum 4 Energy 1 1 1 a2  m  2 y 2   m  2a2   y 2  y  a/ 2 2 4 2 4  A particle of mass 10 grams is executing S.H.M. with an amplitude of 0.5 meter and circular frequency of 60 Problem 16. to 10 radian/sec. The maximum value of the force acting on the particle during the course of oscillation is (a) 25 N E3 [MP PMT 2000] (b) 5 N (c) 2.5 N (d) 0.5 N Maximum force = mass  maximum acceleration = m  2 a  10  10 3 (10)2 (0.5) = 0.5 N Problem 17. A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 ID Solution : (d) meters. There is no friction and the collision with the walls are elastic. The motion of the body is (b) Periodic but not simple harmonic U (a) Not periodic (c) Periodic and simple harmonic (d) Periodic with variable time period Since there is no friction and collision is elastic therefore no loss of energy take place and the body strike D YG Solution : (b) [MP PMT 1999] again and again with two perpendicular walls. So the motion of the ball is periodic. But here, there is no restoring force. So the characteristics of S.H.M. will not satisfied. Problem 18. Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is U (a) 30o (c) 90o (d) 120o Let two simple harmonic motions are y  a sin  t and y  a sin( t   ) ST Solution : (d) (b) 60o In the first case a  a sin  t  sin  t  1 / 2 2 In the second case   cos  t  3 2 a  a sin ( t   ) 2 1    [sin  t. cos   cos  t sin  ]  1   1 cos   3 sin   2 2  2 2   1  cos   3 sin   (1  cos  )2  3 sin2   (1  cos )2  3(1  cos 2  ) By solving we get i.e. cos   1 or cos   1 / 2  0 or   120 o 182 Simple Harmonic Motion Problem 19. The acceleration of a particle performing S.H.M. is 12 cm/sec2 at a distance of 3 cm from the mean position. Its time period is (b) 1.0 sec (c) 2.0 sec 2 2 12     3. 14  2 ; but T   2 3 (d) 3.14 sec 60 (a) 0.5 sec [MP PET 1996; MP PMT 1997] Solution : (d) A   2y    Problem 20. A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude A  y (a) 37.5 2 erg Problem 21. (c) 375  2 erg (d) 0.375  2erg 2 Kinetic energy  1 m  2 (a 2  y 2 )  1 10 4 (10 2  5 2 )  375  2 ergs. 2 2 4 The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude is (b) E / 4 (c) 3 E / 4 U (a) E / 2 ID Solution : (c) (b) 3.75  2 erg E3 of 10 cm. Its kinetic energy when it is at 5 cm. From its equilibrium position is 2  1 3E  m  2 (a 2  y 2 )  1 m  2  a 2  a   3  1 m  2 a 2  =   2 4 2 4  4 2   [RPET 1996] (d) 3E / 4 a ] 2 Kinetic energy  Problem 22. A body executing simple harmonic motion has a maximum acceleration equal to 24 m/sec2 and maximum D YG Solution : (c) [As y  velocity equal to 16 meter/sec. The amplitude of simple harmonic motion is (a) Solution : (a) 32 meters 3 (b) 3 meters 32 (c) 1024 meters 9 Maximum acceleration  2 a  24 …..(i) and maximum velocity a  16 ….(ii) U Dividing (i) by (ii)  [MP PMT 1995] (d) 64 meters 9 3 2 ST Substituting this value in equation (ii) we get a  32 / 3meter Problem 23. The displacement of an oscillating particle varies with time (in seconds) according to the equation. y(cm )  sin t 1   . The maximum acceleration of the particle approximately 2 2 3 (a) 5.21 cm/sec2 Solution : (d) (b) 3.62 cm/sec2 (c) 1.81 cm/sec2 By comparing the given equation with standard equation, y  a sin( t   ) We find that a  1 and    / 4  2 Now maximum acceleration   2 a    4    3.14  2   0.62 cm / sec 2   4   (d) 0.62 cm/sec2 Simple Harmonic Motion 183 Problem 24. The potential energy of a particle executing S.H.M. at a distance x from the mean position is proportional to (a) (b) x x 60 [Roorkee 1992] (c) x 2 (d) x 3 Solution : (c) The kinetic energy and potential energy of a particle executing S.H.M. will be equal, when displacement is E3 Problem 25. (amplitude = a) Solution : (c) (b) a 2 According to problem Kinetic energy = Potential energy   a2  y 2  y 2  y  a / 2 The phase of a particle executing S.H.M. is (a) Maximum velocity (d) a 2 3 1 1 m  2 (a 2  y 2 )  m  2 y 2 2 2  when it has 2 U Problem 26. (c) a / 2 ID (a) a / 2 [MP PMT 1987; CPMT 1990] (b) Maximum acceleration(c) Maximum energy [MP PET 1985] (d) Maximum displacement Problem 27. D YG Solution : (b, d) Phase  / 2 means extreme position. At extreme position acceleration and displacement will be maximum. The displacement of a particle moving in S.H.M. at any instant is given by y  a sin  t. The acceleration T after time t  is (where T is the time period) [MP PET 1984] 4 (a) a Solution : (d) (c) a 2 (d)  a 2 A particle of mass m is hanging vertically by an ideal spring of force constant k, if the mass is made to U Problem 28. (b) a oscillate vertically, its total energy is ST (a) Maximum at extreme position (b) Maximum at mean position (c) Minimum at mean position (d) Same at all position Solution : (d) 15.11 Time Period and Frequency of S.H.M.. For S.H.M. restoring force is proportional to the displacement Fy or F  ky For S.H.M. acceleration of the body …(i) A   2 y …(ii) where k is a force constant. 184 Simple Harmonic Motion F  mA  m  2 y …(iii)  Time period (T )  2  2 m  k or Frequency (n)  1 1  2 T 60 k m From (i) and (iii) ky  m  2 y    k m E3  Restoring force on the body In different types of S.H.M. the quantities m and k will go on taking different forms and names. T  2 or n Inertia factor Spring factor U Thus ID In general m is called inertia factor and k is called spring factor. Spring factor Inertia factor 1 2 D YG 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 n 1 2 Acceleration Dispalcement U or m  k m  Displacement m  2  2 Force/Displacement m  Acceleration  1 2 A y ST 15.12 Differential Equation of S.H.M.. For S.H.M. (linear) Acceleration  – (Displacement) A  y or A   2 y or d 2y   2 y dt 2 or m d 2y  ky  0 dt 2 [As   k ] m Displaceme nt  2 Acceleration y A Simple Harmonic Motion 185 For angular S.H.M.   c 2 and d    2  0 dt 2 c [As c = Restoring torque constant and I = Moment of inertia] I 60 where  2  Sample problems based on Differential equation of S.H.M. A particle moves such that its acceleration a is given by a  bx. Where x is the displacement from E3 Problem 29. equilibrium position and b is a constant. The period of oscillation is [NCERT 1984; CPMT 1991; MP PMT 1994; MNR 1995] (a) 2 b (b) 2 (c) Solution : (b) ID b 2 b We know that Acceleration =   2 (displacement) and d 2y dt 2 2 k (b) 2k Standard equation m So, T  2   2 b  ky  0 where k is a positive constant. The time period of the D YG motion is given by Solution : (c) 2 U The equation of motion of a particle is (a) b a  bx (given in the problem) Comparing above two equation  2  b    b  Time period T  Problem 30. (d) 2  d 2y dt 2 (c) 2 (d) 2 k k  ky  0 and in a given equation m =1 and k = k 2 m  k k U 15.13 Simple Pendulum. ST An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible and perfectly flexible string from a rigid support about which it is free to oscillate. But in reality neither point mass nor weightless string exist, so we can S never construct a simple pendulum strictly according to the definition.  Let mass of the bob = m Length of simple pendulum = l T l y Displacement of mass from mean position (OP) = x O P  mg sin mg mg cos 186 Simple Harmonic Motion When the bob is displaced to position P, through a small angle  from the vertical. Restoring force acting on the bob F  mg  (When  is small sin  ~   or F  mg  F  mg   k (Spring factor) x l So time period T  2 x l OP x Arc = = ) l l Length E3 or 60 F  mg sin  Important points ID l Inertia factor m  2  2 g mg / l Spring factor U (i) 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 D YG situation if 0 is the amplitude of motion. Time period T  2   0 2  l  1 2 0   T  1  sin ....... 1     0   g 16  22  2    (ii) Time period of simple pendulum is also independent of mass of the bob. This is why (a) If the solid bob is replaced by a hollow sphere of same radius but different mass, time period remains U unchanged. ST (b) If a girl is swinging in a swing and another sits with her, the time period remains unchanged. (iii) Time period T  l where l is the distance between point of suspension and center of mass of bob and is called effective length. (a) 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. (b) 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 Simple Harmonic Motion 187 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. 1 1  l R so T  2 (b) If l  R( )1 / l  1 / R so T  2 1 1  g   R l l g E3 (a) If l  R , then 1 60 (iv) If the length of the pendulum is comparable to the radius of earth then T  2 R 6.4  10 6  2  84.6 minutes g 10 so T  2 R  1hour 2g U (c) If l  R ID and it is the maximum time period which an oscillating simple pendulum can have (v) If the bob of simple pendulum is suspended by a wire then effective length of pendulum will increase D YG with the rise of temperature due to which the time period will increase. (If  is the rise in temperature, l0  initial length of wire, l = final length of l  l0 (1    ) wire) 1 l  (1    )1 / 2  1    2 l0 T  T0 1 T  1    2 T0 U So i.e. T 1    T 2 ST (vi) If bob a simple pendulum of density  is made to oscillate in some fluid of density  (where 