# Simple Harmonic Motion-01-Theory Page( 3-25)

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26 de May de 2023
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### Simple Harmonic Motion-01-Theory Page( 3-25)

• 1. S SI IM MP PL LE E H HA AR RM MO ON NI IC C M MO OT TI IO ON N P PE ER RI IO OD DI IC C M MO OT TI IO ON N: : Periodic motion of a body is that motion which is repeated identically after a fixed interval of time. Examples (i) The revolution of earth around the sun is a periodic motion. Its period of revolution is one year. (ii) The motion of hands of a clock is a periodic motion. The period of motion of hour’s hand of a clock is 12 hours, of minute’s hand of a clock is 1 hour and of second’s hand of a clock is one minute. (iii) Uniform circular motion is a periodic motion. O OS SC CI IL LL LA AT TO OR RY Y M MO OT TI IO ON N: : Oscillatory or Vibratory motion is that motion in which a body moves to and fro or back and forth repeatedly about a fixed point (called mean position), in a definite interval of time. In such a motion, the body is confined within well defined limits (called extreme positions) on either side of mean position. Thus a periodic and bounded motion of a body about a fixed point is called an oscillatory or vibratory motion. (i) The motion of the bob of a simple pendulum when it is displaced once from its mean position and left to itself, is oscillatory motion. (ii) The motion of liquid contained in U-tube when it is compressed once in one limb and left to itself, is oscillatory motion. S SI IM MP PL LE E H HA AR RM MO ON NI IC C M MO OT TI IO ON N: : (a) LINEAR SHM (b) ANGULAR SHM Important one among all oscillatory motion is the simple harmonic motion. A particle executing linear simple harmonic motion oscillates in straight line periodically in such a way that the acceleration is proportional to its displacement from a fixed point (called equilibrium), and is always directed towards that point. If a body is describing rotational motion in such a way that direction of its angular velocity changes periodically and torque acting on it is must always be directed opposite to the angular displacement and magnitude of the torque is directly proportional to the angular displacement, then its motion is called angular SHM. P PR RO OJ JE EC CT TI IO ON N O OF F U UN NI IF FO OR RM M C CI IR RC CU UL LA AR R M MO OT TI IO ON N O ON N A A D DI IA AM ME ET TE ER R: : Let at 0  t the particle is at the point A and after time t it is at B . The foot of the perpendicular from the point on the diameter oscillates about O which is S.H.M. in nature. The displacement of the projection from centre is given by 0 sin( )     y R t …(i) The maximum displacement of the projection from the centre is called amplitude. If amplitude is denoted by A then  R A  0 sin( ) y A t     …(ii) and 0 cos( )     x A t …(iii) O Y y A N x Y X X’ 0 t B A At t=0 At t=t
• 2. B BA AS SI IC C D DE EF FI IN NI IT TI IO ON NS S: : D Di is sp pl la ac ce em me en nt t i in n S SH HM M: : In figure, the magnitude of the displacement of N , from the mean position, at any instant is given by cos   x A …(iv) In right triangle cos    ONP x A where A is the radius of the reference circle and  is the angle covered by the reference particle in time t . If  be the uniform angular velocity of the reference particle, then cos   x A t …(v)           t O Y X    y N’ A N x Y X P Expression for displacement If the projection  N of the reference particle is taken on the diameter  YOY , then sin   y A or sin   y A t . Note: In general, x or y could be a linear displacement, an electrical voltage etc. So, equation (ii) or (iii) can be used in a more general sense. When to use different equations? (i) X = A sin t  (ii) X = A cos t  (iii) X = - A sin t  (iv) X = - A cos t  A Am mp pl li it tu ud de e i in n S SH HM M: : The quantity ‘ A ’ in equation (iii) represents the maximum magnitude of displacement. x or y varies between ‘ A ’ and ‘ A’, the quantity ‘ A ’ is called the amplitude of the harmonic oscillation. It is also known as ‘displacement amplitude’. The amplitude of a vibrating particle is its maximum displacement from the mean position to one extreme position. V Ve el lo oc ci it ty y i in n S SH HM M: : Differentiating sin   y A t w.r.t. time t , we get ( ) ( sin )   d d y A t dt dt or velocity, cos    v A t or cos    v A   t     or 2 2    A y v A A or 2 2    v A y O Y y A N x Y X X’   2 2 A y 
• 3. At the mean position, 0  y .    v A (maximum value) At the extreme position, y A   2 2    v A A or 0  v Conclusion. A particle in SHM has maximum velocity at mean position and zero velocity at the extreme position. A Ac cc ce el le er ra at ti io on n i in n S SH HM M: : Differentiating cos    v A t w.r.t. time t , we get ( ) ( cos )    d d v A t dt dt or acceleration 2 2 (cos ) sin ( sin )           d A t A t A t dt or acceleration 2   y At the mean position, 0  y  acceleration = zero At the extreme position,  y A  acceleration 2   A (maximum value) Conclusion. A particle in simple harmonic motion has zero acceleration at the mean position and maximum acceleration at the extreme position. A An n A Al lt te er rn na at ti iv ve e M Me et th ho od d f fo or r F Fi in nd di in ng g V Ve el lo oc ci it ty y a an nd d A Ac cc ce el le er ra at ti io on n i in n S SH HM M: : Let V be the velocity of the reference particle at P . Resolve velocity V into two rectangular components cos V parallel to  YOY and sin  V perpendicular to  YOY . The velocity v of the projection N is clearly cos V .  cos cos      v V A t or 2 1 sin     v A t or 2 2 1    y v A A or 2 2    v A y . The centripetal acceleration 2 V A of the particle at P can be resolved into two rectangular components 2 cos   V A perpendicular to  YOY and 2 sin  V A anti-parallel to  YOY . Acceleration of 2 sin    V N A or Acceleration 2 2 2 ( sin ) ( sin )       V A A t A . or Acceleration 2   y . O Y X  X P    V Vcos Vsin Y N y A 2 cos V A O Y X  X P   V Y 2 sin V A y N A   T Ti im me e P Pe er ri io od d o or r P Pe er ri io od di ic c T Ti im me e o of f S SH HM M: : It is the smallest interval of time at which the details of motion repeat. It is generally represented by T . Let us replace t by  t T in equation (iii). 0 0 2 ' ' at ( ) cos 2 cos 2                        t T t x t T A A T T It is clear from here that the details of motion repeat after time T. Time period may also be defined as the time taken by the oscillating particle to complete one oscillation. It is equal to the time taken by the reference particle to complete one revolution. In one revolution, the angle traversed
• 4. by the reference particle is 2 radian and T is the time taken. If  be the uniform angular velocity of the reference particle, then 2   T or 2   T . F Fr re eq qu ue en nc cy y: : It is the number of oscillations (or vibrations) completed per unit time. It is denoted by f. In time T second, one vibration is completed. In 1 second , 1 T vibrations are completed. or 1 f T  or 1 fT  Also, 2 1 2 2 f T T        So, equation (iii) may also be written as under: 0 cos(2 ) x A ft     …(vi) The unit of f is 1  s or hertz or ‘cycles per second’ (cps).  1       s Hz cps . A An ng gu ul la ar r F Fr re eq qu ue en nc cy y: : It is frequency f multiplied by a numerical quantity 2. It is denoted by  so that 2 2 f T      Equation (vi) may be written as 0 cos( )     x A t P Ph ha as se e: : Phase of a vibrating particle at any instant is the state of the vibrating particle regarding its displacement and direction of vibration at that particular instant. The argument of the cosine in equation 0 cos( )     x A t gives the phase of oscillation at time t . It is denoted by  .  0 2      t T or 0      t It is clear that phase  is a function of time t . The phase of a vibrating particle can be expressed in terms of fraction of the time period that has elapsed since the vibrating particle left its mean position in the positive direction. Again, 0 2      t t T So, the phase change in time t is 2t T . The phase change in T second will be 2 which actually means ‘no change in phase’. Thus, time period may also be defined as the time interval during which the phase of the vibrating particle changes by 2. Illustration 1. If the displacement of a moving point at any time be given by an equation of the form ( ) cos sin     y t a t b t show that the motion is simple harmonic. If 3 , 4 and 2;     a m b m determine the period, amplitude, maximum velocity and maximum acceleration. Solution : The particle is moving along y-axis. ( ) cos sin     y t a t b t 2 2 0 sin ( )      y a b t
• 5. where 0 tan /   a b 2 2 1 sin( tan / )      y a b t a b comparing with sin( )     y a t , 2 2 2        T sec 2 2 2 2 3 4 5      A a b m max 2 5 10 /      v A m s 2 2 max 4 5 20 /      a A m s . Differential equation of acceleration of a particle executing S.H.M. is 2   a x i.e., 2 2 2   d x x dt S So ol lu ut ti io on n o of f t th he e E Eq qu ua at ti io on n o of f S S. .H H. .M M. . : : The necessary and sufficient condition for a motion to be simple harmonic is that the net restoring force (or torque) must be linear i.e.,    F ma kx (Where k is a constant)  2 2    d x k a x m dt where x is the instantaneous displacement. Multiplying both sides by dx dt and integrating with respect to t ,                     d dx dx k x dx dt dt dt m dt  . . .     dv k x dx v dt m dt         dx v dt 2 2 . 2 2    v k x c m We get 2 2 1 2 2          dx k x C dt m …(i) Where C is a constant of integration. Now when x is maximum dx dt will be zero. The maximum displacement max X of the particle from the mean position is called amplitude and is represented by A , then the value of C comes out to be 2 2  k A C m Here 2 2 2 1 ( ) 2 2         dx k A x dt m Putting 2   k m we get 2 2    dx A x dt …(ii) This equation gives the velocity of the particle in S.H.M. Again 2 2 dt    dx A x …(iii) Integrating this equation with respect to t . we get
• 6. 1 sin     x t A Where  is another constant of integration which depends on initial conditions. Thus sin( )     x A t …(iv) Here is called ANGULAR FREQUENCY and  is called INITIAL PHASE OR EPOCH CONSTANT, whose value depends upon initial conditions. If the time is recorded from the instant when x is zero and is increasing then according to equation (iv),  must be equal to zero. Now, the periodic time or time required to complete one vibration or the time to increase the phase angle of the particle by 2 is given by 2 2      m T k …(v) T is called the time period of oscillation. The frequency f , the number of complete vibration per second is given by 1 1 2 2       k f T m …(vi) If we put / 2    in place of  then cos( )     x A t …(vii) Thus S.H.M. may also be represented by a cosine function but initial phase is different in two cases. S SI IM MP PL LE E P PE EN ND DU UL LU UM M: : A small mass (bob) suspended by a light, long and inextensible string forms simple pendulum. Length of the simple pendulum is the distance between the point of suspension and the centre of mass of the suspended mass. Consider the bob when string deflects through a small angle  . Force acting on the bob are tension T in the string and weight mg of the bob. Torque on the bob about point O is sin 0        mg T mg l     mgl as  is very small. …(i)  M.I. of the bob about the point O is 2  I ml Hence 2 2 2    d ml dt …(ii) O l O T  l mg sin mg mg cos As torque  and  are oppositely directed hence from (i) and (ii), we get 2 2 2  d ml dt    mg l  2 2 ( / )     d g l dt Comparing with the equation 2 2 2   d x x dt , we get   g l Since 2 /    T  2   l T g Note: If a simple pendulum is made to oscillate in a non inertial frame of reference, the torque due to pseudo force must be taken into account. Illustration 2. A simple pendulum is executing simple harmonic motion with a time period T. If the length of the
• 7. pendulum is increased by 21%. Find the percentage increase in time period. Solution : 2   l T g 2 2 1 1 1.21 1.1     T l l T l l 2 1 1.1  T T T TH HE E S SP PR RI IN NG G- -M MA AS SS S S SY YS ST TE EM M: : Let us find out the time period of a spring-mass system oscillating on a smooth horizontal surface as shown in the figure. At the equilibrium position the spring is relaxed. When the block is displaced through a distance x towards right, it experiences a net restoring force   F kx towards left. The negative sign shows that the restoring force is always opposite to the displacement. That is, when x is positive, F is negative, the force is directed to the left. When x is negative, F is positive, the force always tends to restore the block to its equilibrium position 0  x .   F kx Applying Newton’s Second Law, 2 2    d x F m kx dt or 2 2 0         d x k x m dt Smooth m x k m k A block of mass m attached with a spring of stiffness k oscillates on a smooth horizontal plane. Comparing the above equation with, 2 2 2    d x a x dt we get 2   k m or 2   m T k Note: Time period is independent of the amplitude. For a given spring constant, the period increases with the mass of the block that means more massive block oscillates more slowly. For a given block, the period decreases as k increases. A stiffer spring produces quicker oscillations. Illustration 3: Solution: Find the period of oscillation of a vertical spring-mass system. Let 0 x be the deformation in the spring in equilibrium. Then 0  kx mg When the block is further displaced by x , the net restoring force is given by 0 [ ( ) ]     F k x x mg Using second law of motion, 2 2   d x m kx dt k x0 x m k k A block of mass m attached to a spring of stiffness k oscillates in a vertical plane.
• 8. or 2 2 0         d x k x m dt Thus, 2   k m or 2   m T k . kx0 mg k(x+x ) 0 mg Note: Gravity does not influence the time period of the spring-mass system, it merely changes the equilibrium position. S Se er ri ie es s a an nd d P Pa ar ra al ll le el l C Co om mb bi in na at ti io on n o of f S Sp pr ri in ng gs s: : (a) Series combination of spring When two spring are joined in series, the equivalent stiffness of the combination may be obtained as 1 2 1 1 1   k k k k1 k2 m m k (b) Parallel combination of spring When two springs are joined in parallel, the equivalent stiffness of the combination is given by 1 2   k k k m k1 k2 m k Illustration 4: A spring of stiffness constant k and natural length  is cut into two parts of length 3 4  and 4  respectively, and an arrangement is made as shown in the figure. If the mass is slightly displaced, find the time period of oscillation. Solution: The stiffness of a spring is inversely proportional to its length. Therefore the stiffness of each part is 1 4 3  k k and 2 4  k k Time period, 1 2 2    m T k k or 3 3 2 16 2     m m T k k . k k 1=(4/3) k k 2=4 P PH HY YS SI IC CA AL L P PE EN ND DU UL LU UM M: : Any rigid body suspended from a fixed support constitutes a physical pendulum. Consider the situation when the body is displaced through a small angle  . Torque on the body about O is given by sin    mgl …(i) where  l distance between point of suspension and centre of mass of the body. If I be the M.I. of the body about O. Then    I …(ii)
• 9. From (i) and (ii), we get 2 2 sin     d I mgl dt as  and 2 2  d dt are oppositely directed.  2 2     d mgl I dt Since  is very small. mg  G O Comparing with the equation 2 2 2     d dt , we get   mgl I  2   I T mgl . Illustration 5. What is the period of a pendulum formed by pivoting a metre stick so that it is free to rotate about a horizontal axis passing through the 75 cm mark ? Solution : Let m be the mass and  be the length of the stick. 100   cm The distance of the point of suspension from centre of gravity is d = 25 cm Moment of inertia about a horizontal axis through O is : 2   C I I md 2 2 12    m I md 2   I T mgd 2 2 12 2     m md T mgd O C d 2 2 12 2 12     d T gd 2 2 12(0.25) 2 153 12 9.8 0.25        s . E EN NE ER RG GY Y O OF F A A B BO OD DY Y I IN N S S. .H H. .M M. .: : The total energy of a harmonic oscillator consists of two parts, potential energy (P.E.) and Kinetic Energy (K.E.), the former being due to its displacement from the mean position and later due to its velocity. Since the position and velocity of the harmonic oscillator are continuously changing, P.E. and K.E. also change but their sum i.e., the total energy (T.E.) must have the same value at all times. Simple harmonic motion is defined by the equation   F kx . The work done by the force F during a displacement from x to  x dx is  dW F dx  kx dx . The work done in a displacement from 0  x to x is 2 0 1 ( ) 2      x W kx dx kx .
• 10. Let ( ) U x be the potential energy of the system when the displacement is x . As the change in potential energy corresponding to a force is negative of the work done by this force, 2 1 ( ) (0) 2     U x U W kx . Let us choose the potential energy to be zero when the particle is at the centre of oscillation 0  x . Then (0) 0  U and 2 1 ( ) 2  U x kx . This expression for potential energy is same as that for a spring and has been used so far in this chapter. As 2 ,     k k m m we can write 2 2 1 ( ) 2   U x m x …(i) The displacement and the velocity of a particle executing a simple harmonic motion are given by executing a simple harmonic motion are given by sin( )     x A t and cos( )      v A t The potential energy at time t is, therefore, 2 2 1 2   U m x 2 2 2 1 sin ( ) 2      m A t and the kinetic energy at time t is 2 1 2  K mv 2 2 2 1 cos ( ) 2      mA t The total mechanical energy at time t is K, U or E -A +A 2 2 1 E m A 2   2 2 1 PE m x 2   2 2 2 1 PE m (A x ) 2    x 2 A  2 A    E U K 2 2 2 2 1 [sin ( ) cos ( )] 2 m A t t          2 2 1 2   m A . ….(ii) We see that the total mechanical energy at time t is independent of t . Thus, the mechanical energy remains constant as expected. K, U or E 2 2 1 E m A 2   E T Time T v 2 2 2 1 PE m A sin t 2    2 2 2 1 KE m A cos t 2    1 2 E 2 A AV VE ER RA AG GE E V VA AL LU UE E O OF F P P. .E E. . A AN ND D K K. .E E. .: : By equation (i) P.E. at distance x is given by 2 2 2 2 2 1 1 sin ( ) 2 2        U m x m A t {since at time t, x = sin( )    A t } The average value of P.E. of complete vibration is given by 2 2 2 0 0 1 1 1 sin ( ) 2         T T average U U dt m A t T T 2 2 2 0 2sin ( ) 4       T m A t dt T
• 11. 2 2 1 4   m A because the average value of sine or of cosine function for the complete cycle is equal to zero. Now K.E. at x is given by 2 2 1 1 . . { sin( )} 2 2                  dx d K E m m A t dt dt 2 2 2 1 cos ( ) 2      m A t The average value of K.E. for complete cycle 2 2 2 0 1 1 . . cos ( ) 2       T average K E m A t dt T 2 2 0 {1 cos2( )} 4        T m A t dt T 2 2 2 2 1 . 4 4     m A T m A T Thus average values of K.E. and P.E. of harmonic oscillator are equal to half of the total energy. Illustration 6. A point particle of mass 0.1 kg is executing SHM with amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 3 8 10  joule. Obtain the equation of motion of this particle if the initial phase of oscillation is 45º. Solution : Given that 0.1  A m; m = 0.1 kg, 45   º = ( / 4)  rad, so the equation of SHM will be 0.1sin[ ( / 4)]     y t …(i) Now as in SHM, KE is given by 2 2 2 1 ( ) 2    K m A y which according to given problem is 3 8 10  J for 0  y so 3 2 2 2 1 8 10 0.1 (0.1 0 ), 2       i.e., 4   rad/s …(ii) substituting the value of  from Eqn. (ii) in (i), we get 0.1sin[4 ( / 4)]    y t .
• 12. O OB BJ JE EC CT TI IV VE E S SO OL LV VE ED D P PR RO OB BL LE EM MS S 1. A uniform cylinder of length L and mass M having cross-sectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density d at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is K, the frequency of oscillation of the cylinder is : (a) 1/ 2 1 2        K Adg M (b) 1/ 2 1 2         K dgL M (c) 1/ 2 1 2         K Adg M (d) 1/ 2 1 2         K Adg Adg . Ans. (c) Solution : Let the spring is further extended by y when the cylinder is given small downward push. Then the restoring forces on the spring are, (i) Ky due to elastic properties of spring (ii) upthrust = yAdg = weight of liquid displaced  Total restoring force ( )   K Adg y  ( ). M a K Adg y     y Comparing with a 2 ,   y we get 2          K Adg M or    K Adg M 1 2 2       K Adg f M . 2. A simple pendulum with length L and mass M of the bob is vibrating with amplitude a. Then the maximum tension in the string is : (a) Mg (b) 2 1                a Mg L (c) 2 1        a Mg L (d) 2 1 2        a Mg L . Ans. (b) Solution : Maximum tension in the string is at lowest position. Therefore 2   Mv T Mg L To find the velocity v at the lowest point of the path, we apply law of conservation of energy i.e. 2 1 (1 cos ) 2     Mv Mgh MgL [ , cos ]       h L x h L L or 2 2 (1 cos )    v gL ] or 2 (1 cos )    v gL L a  x h Mv 2 L Mg 2 (1 cos )      T Mg Mg 2 1 2 2sin 2                 T Mg
• 13. 2 1 4 2                  T Mg [ sin( / 2) / 2     for small amplitudes ] 2 [1 ]    T Mg From figure   a L  2 1                 a T Mg L . 3. A simple pendulum has a length l cm. Mass of the bob is m gm. The bob is given a charge of +q statcoulomb. The pendulum is suspended between the plates of a charged parallel plate capacitor. If E is the electric intensity between the plates as shown in the adjoining figure, then the time period T is given by : (a) 2         l T g (b) 2 ( / )          l T g Eq m +q (c) 2 ( / )          l T g Eq m (d) 2 2 1/ 2 2 { ( / ) }          l T g Eq m . Ans. (d) Solution : Forces acting on the bob are (i) Eq. = horizontal, due to electric field. (ii) mg  vertically down. Therefore, the net force on the bob 2 2 ) ( )   mg Eq . Effective value of acceleration is 2 2 2 2 2 ( ) ( ) mg Eq Eq g m m           mg +q Eq  1/ 2 2 2 2                    l T Eq g m . 4. A loaded vertical spring executes simple harmonic oscillations with period of 4 s. The difference between the kinetic energy and potential energy of this system oscillates with a period of : (a) 8 s (b) 1 s (c) 2 s (d) 4 s. Ans. (c) Solution : K.E. 2 2 2 1 cos 2    m a t P.E. 2 2 2 1 sin 2    m a t K.E. – P.E. 2 2 2 2 1 [cos sin ] 2      m a t t 2 2 1 .cos2 2    m a t  Angular frequency 2   and time period 2 2       2 2     T s.
• 14. 5. A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period : (a) 42 minutes (b) 1 day (c) 1 hour (d) 84.6 minutes. Ans. (d) Solution : Let the mass m be at a distance y from the centre of earth at any instant of time. Therefore force on it is 3 2 4 . . . 4 3 3 y G d m Gmd F y y       d = density of earth Negative sign indicates that the force is directed towards the centre of earth. The particle executes SHM because,   F y m y Now 4 3 Gdm F m a y      or 4 3 Gd a y   Comparing with a 2 ,   y we get 4 3    Gd or 2 3 2 4       T Gd = 3 3 2 2 4 . 4 5 e e e R M g G R      2 84.6    e R T g min. 6. The displacement of a particle is given at time t , by: 2 sin( 2 ) sin      x A t B t Then, (a) the motion of the particle is SHM with an amplitude of 2 2 4  B A (b) the motion of the particle is not SHM, but oscillatory with a time period of    T (c) the motion of the particle is oscillatory with a time period of 2    T (d) the motion of the particle is aperiodic. Ans. (a) Solution : The displacement of the particle is given by: 2 sin( 2 ) sin      x A t B t sin 2 (1 cos2 ) 2       B A t t ( sin 2 cos2 ) 2 2       B B A t t .
• 15. This motion represents SHM with an amplitude: 2 2 4  B A , and mean position 2 B . 7. A block of mass m , attached to a fixed position O on a smooth inclined wedge of mass M , oscillates with amplitude A and linear frequency f . The wedge is located on a rough horizontal surface. If the angle of the wedge is 60º, then the force of friction acting on the wedge is given by (coefficient of static friction   ). 60º M m A smooth rough O (a) ( )   M m g (b) 2 1 sin 2   m A t (c) 2 3 ( ) sin 2            M m g m A t (d) 2 ( ) sin     M m A t . Ans. (b) Solution : The small block oscillates along the inclined plane with an amplitude A . As a result the centre of mass of the system undergoes SHM along the horizontal direction: sin cos60º    cm mA t x m M 1 sin 2    m A t m M . The acceleration of the C.M. is 2   cm cm a x , along the horizontal while the net horizontal force is ( )   cm M m a , which is equal to the force of friction acing on it. 8. A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is (a) 2 m k (b) / 2    m m k k (c) 3 / 2  m k (d) 2    m m k k . k m k Ans. (d) Solution : When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another half oscillation in the upward direction. The total time period is: 2     m m T k k . 9. Two particles P and Q describe SHM with the same amplitude A and the same frequency f . The maximum distance separating the particles is observed to be A . The phase difference between the particles is (a) zero (b) 2  (c) 3  (d) 2 3  .
• 16. Ans. (c) Solution : It is easy to solve this problem using the equivalent circular motion. Suppose that both the particles are represented on the circle of reference. The maximum separation occurs when both the particles are symmetrically located with respect to the mean position O:     P Q PQ A .    OP Q is equilateral.  3      P OQ . P O Q P' Q' 10. A loop consists of two cords of lengths  and 2 , and their masses per unit length are their masses per unit length are  and 2 . It is placed in stable equilibrium over a smooth peg as shown in the figure. When slightly displaced, it executes SHM. The period of oscillation is (a) 5 2 3   g (b) 5 2  g (c) 5 2 2   g (d) 3 2  g . A B Peg Ans. (c) Solution : If the string is displaced slightly downward by x , we can write, the net (restoring)force ( 2 )2     x x g 2   xg  (5 ) 2        x xg or, 2 5      g x x  2 5    g . peg A' B' 2 x  or 2 5 2 T g       .
• 17. S SU UB BJ JE EC CT TI IV VE E S SO OL LV VE ED D P PR RO OB BL LE EM MS S 1. A uniform horizontal plank is resting symmetrically in a horizontal position on two cylindrical drums, which are spinning in opposite direction about their horizontal axes with equal angular velocity. The distance between the axes is 2L and the coefficient of friction between the plank and cylinder is  . If the plank is displaced slightly from the equilibrium position along its length and released, show that it performs simple harmonic motion. Calculate also the time period of motion. Solution : When the plank is situated symmetrically on the drums, the reactions on the plank from the drums will be equal and so the force of friction will be equal in magnitude but opposite in direction and hence, the plank will be in equilibrium along vertical as well as in horizontal direction. Now if the plank is displaced by x to the right, the reaction will not be equal. For vertical equilibrium of the plank   A B R R mg …(i) R fA fB A B mg RA RB x R 2L A B fA fB mg And for rotational of plank, taking moment about center of mass we have ( ) ( )    A B R L x R L x …(ii) Solving Eqns. (i) and (ii), we get 2         A L x R mg L and 2         B L x R mg L Now as   f R , so friction at B will be more than at A and will bring the plank back, i.e., restoring force here ( ) ( )         B A B A mg F f f R R x L As the restoring force is linear, the motion will be simple harmonic motion with force constant   mg k L So that 2 2      m L T k g . 2. A ball is suspended by a thread of length at the point O on an incline wall as shown. The inclination of the wall with the vertical is  . The thread is displaced through a small angle  away from the vertical and the ball is released. Find the period of oscillation of pendulum. Consider both cases. (a)    (b)    Assume that any impact between the wall and the ball is elastic. Q O P   l Solution : (a) If    , the ball does not collide with the wall and it performs full oscillations like a simple pendulum.  period 2    g (b) If    , the ball collides with the wall and rebounds with same speed. The motion of ball from A to Q is one part of a simple pendulum. time period of ball 2( )  AQ t . Q   l l l A
• 18. Consider A as the starting point ( 0)  t . Equation of motion is ( ) cos   x t A t ( ) cos ,     x t t because amplitude     A time from A to Q is the time t when x becomes    .        cost 1 1/ cos AQ t t              The return path from Q to A will involve the same time interval. Hence time period of ball 2  AQ t 1 1 2 cos 2 cos g                        1 2 2 cos              g g . 3. Consider a liquid which fills a uniform U-tube, as shown in the figure upto a height h . Find the frequency of small oscillations of the liquid in the U-tube. h Solution : Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, x . If the density of the liquid is  , then, the total mechanical energy of the liquid column is :   2 1 ( ) ( ) . 2            dx E A h x A h x dt ( ) ( ) 2 2                 h x h x A h x g A h x g h+x h-x   2 2 2 1 1 (2 ) 2 ( ) 2 2            dx Ah A g h x dt … (i) After differentiating the total energy and equating it to zero, one finds acceleration 2   x . The angular frequency of small oscillations, , is: 2 2      A g g Ah h … (ii) 4 A uniform plank of mass m , free to move in the horizontal direction only, is placed at the top of a solid cylinder of mass 2m and radius R . The plank is attached to a fixed wall by means of a light spring of spring constant k . There is no slipping between the cylinder and the plank, and between the cylinder and the ground. Find the time period of small k m R 2m
• 19. oscillations of the system. Solution : Suppose that the plank is displaced from its equilibrium position by x at time t , the centre of the cylinder is, therefore, displaced by 2 x .  the mechanical energy of the system is given by, . .  E K E (Plank) + P.E.(spring) + K.E. (cylinder) 2 2 2 1 1 1 2 2 2 2 2                      dx d x E m kx m dt dt 2 2 1 1 1 2 . 2 2 2                    d x m R R dt 2 2 1 7 1 ( ) 2 4 2 dx m kx dt         After differentiating the total energy and equating it to zero, one finds acceleration = 2 x   The angular frequency, 4 7 k m   . 5. A particle of mass m is allowed to oscillate on a smooth parabola: 2 4  x ay , 0  a , as shown in the figure. Find the angular frequency ( ) of small oscillations. y x g O Solution : Suppose that the particle is displaced from its equilibrium position at O , and that its x-coordinate at time t is given by x . The total energy of the particle at time t is given by, 2 2 1 2                          dx dy E m mgy dt dt … (i) Differentiating the equation of the curve, we get, 2 4  dx dy x a dt dt  2 2 2 2 1 1 2 4 4                dx x mg E m x dt a a .  The oscillations are very small, both x and dx dt are small. We ignore terms which are of higher order than quadratic terms in x or, dx dt or, mixed terms.  2 2 1 1 2 2 2              dx mg E m x dt a … (ii) After differentiating the total energy and equating it to zero, one finds acceleration = 2 x   The angular frequency of small oscillations is, consequently, 2 . 2    mg g a m a . … (iii)
• 20. 6. Consider a solid cylinder of density s , cross-sectional area A and height h floating in a liquid of density  , as shown in the adjacent figure ( )     s . It is depressed slightly and allowed to oscillate vertically. Find the frequency of small oscillations. h s l Solution : At equilibrium the net force on the cylinder is zero in the vertical direction: 0    net F B W ;  B the buoyancy and  W the weight of the cylinder. When the cylinder is depressed slightly by x , the buoyancy increases from B to   B B where: | |    B x Ag while the weight W remains the same.  the net force,      net F B B W  B | |   x Ag The equation of motion is, therefore, 2 2 s d x Ah x Ag dt     , the minus sign takes into account the fact that x and restoring force are in opposite directions.  2 2      s g d x x h dt and the angular frequency, , is      s g h . 7. A uniform rod of length  and mass m is hinged at its lowest point O and is connected at its highest point A by means of a spring of spring constant k as shown in the figure. Find the frequency of small oscillations. What is the condition for the system to perform small oscillations? k A O Solution : Suppose that the rod is displaced by a small angle  as shown in the figure. The total mechanical energy of the system is given by, 2 2 1 (1 cos ) 3 2         E m mg 2 1 ( ) 2    k 2 2 1 3     m 2 2 1 2 2            mg k … (i)  the angular frequency of small oscillations is, 2 2 3 3 2 1 2 3          mg k k g m m … (ii) The condition for the system to be oscillation is, 3 3 2   k g m or, 2   mg k … (iii) k A O mg  kx
• 21. 8. A block of mass m hangs by means of a string which goes over a pulley of mass m and moment of inertia I , as shown in the diagram. The string does not slip relative to the pulley. Find the frequency of small oscillations. k m Solution : Suppose that the block is depressed by x . The pulley (owing to the constraint) is depressed by 2 x . Suppose that the tension in the string are &  T T on both sides. We can write: For block:    mg T mx ...(i) For pulley: 0 ( ) 2        x T T mg k x x m … (ii) The angular acceleration of the pulley, / 2    x R … (iii) ( ) 2       x T T R I R … (iv) k m T x T’ mg From (i), (ii), (iii) and (iv) we get, 0 2 5 3 ( ) 2 2            m I mg k x x x R … (v) The frequency of small oscillation, 2 1 5 2 2 2    k f m I R . 9. A cylindrical pulley of mass M and radius R is fixed so that it can rotate about its axis O . A block of mass 1 m is hung from the pulley by means of a tightly wound string, while a particle of mass m attached to the rim of the pulley balances the weight of 1 m , as shown in the figure. (a) Find the angular position  of m w.r.t. the vertical line passing through O , when the system is in equilibrium. (b) If the system is displaced slightly from its equilibrium position, show that it will execute SHM and find the frequency of small oscillations. m  O m1 Solution : (a) At equilibrium, the net torque on the pulley is zero.  1 sin     m g R mg R … (i) or, 1 sin   m m or, 1 1 sin   m m … (ii) (b) If the system is displaced slightly from the equilibrium position, it oscillates. Suppose that the position of the particle is given by the angular variable  , at some instant. The total mechanical m  O m1 mg T m g 1
• 22. energy is given by: . . . .   E K E P E where, 2 2 2 2 2 2 1 1 1 1 1 . . ( ) 2 2 2 2                K E MR mR m R and, P.E.= loss in P.E. of 1 m + gain in P.E. of m 1 ( ) (cos cos )          m gR mgR m  O m1 1 ( ) 2 sin sin 2 2          m gR mgR . 1 2 sin sin 2 2              m gR mgR 2 1 2 sin 2 cos 2 2                m gR mgR mgR . where  is defined by the expression : ,       being a small quantity. Since the frequency depends only on terms which are quadratic in  , we can write, 2 2 2 1 1 1 1 cos ( ) 2 2 2               E M m m R mgR + terms linear in  or, constants. After differentiating the total energy and equating it to zero, one finds acceleration = 2 x    the angular frequency, 2 1 cos 1 2            mgR M m m R and the frequency, 1 1 cos 1 2 2            mg f M m m R . 10. A particle of mass 1 m rotates in a circle of radius r on a smooth horizontal surface, with an angular velocity 0  . It is connected by means of an inextensible string to a block 2 m the string passing through a hole B at the centre of the circle of rotation and the block of mass 2 m hanging by means of the string as shown in the figure. (a) Find the relationship between 0 1 2 , , ,  r m m so that the system remains in equilibrium. (b) If the block 2 m is slightly displaced downward and set free, it is found to execute small oscillations. Find the frequency of small oscillations. B A m1 m2 0 (a) Since the system is in equilibrium, we can write the tension in the string, T as: 2 1 0   T m r and, 2  T m g  2 1 0 2   m r m g … (i) (b) Suppose that the block 2 m is depressed by x . The radius of the circle of rotation is now given by,
• 23.    r r x . and the angular speed   is given by, 2 2 1 0 1( )      m r m r x or, 2 0 2 ( )      r r x … (ii) The free body diagram as well as the geometry of the problem are as shown in the adjacent figure. 2 2 1 1 2 ( ) ( )       d m r x m r x T dt …(iii) m g 2 B’ A’ ( +x) l (r–x) m g 1 T C’ T 1 2 1 2 ( ) m w r x  2 2 2 2 ( )     d m x m g T dt … (iv) The first term on the RHS of the equation (iii) can be rewritten as, 2 4 2 1 0 1 3 ( ) ( )       m r m r x r x 3 2 1 0 1           x m r r 2 1 0 3 1         x m r r (after binomial expansion and assuming  x r ) Equation (iii) and (iv) become 2 1 1 0 3 1             x m x m r T r 2 2     m x T m g . Adding, 2 1 2 1 0 2 3 ( ) 1              x m m x m r m g r  2 1 0 1 2 3      m x x m m … (v) Thus the angular frequency of small oscillations, , is given by, 1 0 1 2 3     m m m … (vi)