Physics 11th Simple Harmonic Motion

Simple Harmonic Motion

1. Periodic motion

  • If a particle moves such that it repeats its path regularly after equal intervals of time , it's motion is said to be periodic.
  • The interval of time required to complete one cycle of motion is called time period of motion.
  • If a body in periodic motion moves back and forth over the same path then the motion is said to be viberatory or oscillatory.
  • Examples of such motion are to and fro motion of pendulum , viberations of a tuning fork , mass attached to a spring and many more.
  • Every oscillatory motion is periodic but every periodic motion is not oscillatory for example motion of earth around the sun is periodic but not oscillatory.
  • Simple Harmonic Motion (or SHM) is the simplest form of oscillatory motion.
  • SHM arises when force on oscillating body is directly proportional to the displacement from it's equilibrium position and at any point of motion , this force is directed towards the equilibrium position.


2. Simple Harmonic Motion (or SHM)

  • SHM is a particular type of motion very common in nature.
  • In SHM force acting on the particle is always directed towards a fixed point known as equilibrium position and the magnitude of force is directly proportional to the displacement of particle from the equilibrium position and is given by
         F= -kx
    where k is the force constant and negative sign shows that fforce opposes increase in x. 
  • This force os known as restoring force which takes the particle back towards the equilibrium position , and opposes increase in displacement.
  • S.I. unit of force constant k is N/m and magnitude of k depends on elastic properties of system under consideration.
  • For understanding the nature of SHM consider a block of mass m whose one end is attached to a spring and another end is held stationary and this block is placed on a smooth horizontal surface shown below in the fig.

  • Motion of the body can be described with coordinate x taking x=0 i.e. origin as the equilibrium positionwhere the spring is neither stretched or compressed.
  • We now take the block from it's equilibrium position to a point P by stretching the spring by a distance OP=A and will then release it.
  • After we release the block at point P, the restoring force acts on the block towards equilibrium position O and the block is then accelerated from point P towards point O as shown below in the fig.

  • Now at equilibrium position this restoring force would become zero but the velocity of block increases as it reaches from point P to O.
  • When the block reaches point O it's velocity would be maximum and it then starts to move towards left of equilibrium position O.
  • Now this time while going to the left of equilibrium position spring is compressed and the block moves to the point Q where it's velocity becomes zero.

  • The compressed spring now pushes the block towards the right of equilibrium position where it's velocity increases upto point O and decreases to zero when it reaches point P.
  • This way the block oscillates to and fro on the frictionless surface between points P and Q.
  • If the distance travelled on both sides of equilibrium position are equal i.e. , OP=OQ then the maximum displacement on either sides of equilibrium are called the Amplitude of oscillations.

3. Equation of SHM

  • Consider any particle executing SHM with origin as it's equilibrium position under the influence of restoring force F=
  • kx , where k is the force constant and x is the displacement of particle from the equilibrium position.
  • Now since F= -kx is the restoring force and from Newton's law of motion force is give as F=ma , where m is the mass of the particle moving with acceleration a. Thus acceleration of the particle is
              a=F/m
               =-kx/m
    but we know that acceleration a=dv/dt=d2x/dt2
    ⇒           d2x/dt2=-kx/m          (1)
    This equation 1 is the equation of motion of SHM.
  • If we choose a constant φ=√(k/m) then equation 1 would become
              d2x/dt2=-φ2x          (2)
  • This equation is a differential equation which says that displacement x must be a funcyion of time such that when it's second derivative is calculated the result must be negative constant multiplied by the original function.
  • Sine and cosine functions are the functions satisfying above requirement and are listed as follows
              x=A sinωt                         (3a)
              x=A cosωt                         (3b)    
              x=A cos(ωt+φ)                    (3c)
    each one of equation 3a, 3b and 3c can be submitted on the left hand side of equation 2 and can then be solved for varification.
  • Convinently we choose equation 3c i.e., cosine form for representing displacement of particle at any time t from equilibrium position. Thus,
              x=A cos(ωt+φ)                    (4)
    and A , φ and φ are all constants.
  • Fig below shows the displacement vs. time graph for phase φ=0.


4. Characterstics of SHM
Here in this section we will learn about physical meaning of quantities like A, T, ω and φ.

(a) Amplitude
  • Quantity A is known as amplitude of motion. it is a positive quantity and it's value depends on how oscillations were started.
  • Amplitude is the magnitude of maximum value of displacement on either side from the equilibrium position.
  • Since maximum and minimum values of any sine and cosine function are +1 and -1 , the maximum and minimum values of x in equation 4 are +A and -A respectively.
  • Finally A is called the amplitude of SHM.


(b) Time period
  • Time interval during which the oscillation repeats itself is known as time period of oscillations and is denoted by T.
  • Since a particle in SHM repeats it's motion in a regular interval T known as time period of oscillation so displacement x of particle should have same value at time t and t+T. Thus,
              cos(ωt+φ)=cos(ω(t+T)+φ)
    cosine function cos(ωt+φ) will repeat it's value if angle (ωt+φ) is increased by 2π or any of it's multiple. As T is the pime period
              (ω(t+T)+φ)=(ωt+φ)+2π
    or,          T=2π/` = 2π√(m/k)               (5)
  • Equation 5 gives the time period of oscillations.
  • Now the frequency of SHM is defined as the numberof complete oscillations per unit time i.e., frequency is reciprocal of time period.
               f=1/T = 1/2π(√(k/m))                (6)
    Thus,            ω=2`/T = 2`f                     (7)
  • This quantity ω is called the angular frequency of SHM.
  • S.I. unit of T is s (seconds)
          f is Hz (hertz)
          ω is rad s-1 (radian per second)

(c) Phase
  • Quantity (ωt+φ) in equation (4) is known as phase of the motion and the constant φ is known as initial phase i.e., phase at time t=0, or phase constant.
  • Value of phase constant depends on displacement and velocity of particle at time t=0.
  • The knowledge of phase constant enables us to know how far the particle is from equilibrium at time t=0. For example,
    If φ=0 then from equation 4
              x=A cosωt
    that is displacement of oscillating particleis maximum , equal to A at t=0 when the motion was started. Again if φ=`/2 then from equation 4
              x=A cos(ωt+`/2)
               =Asinωt
    which means that displacement is zero at t=0.
  • Variation of displacement of particle executing SHM is shown below in the fig.



5.Velocity of SHM

  • We know that velocity of a particle is given by
         v=dx/dt
  • In SHM displacement of particle is given by
         x=A cos(ωt+φ)
    now differentiating it with respect to t
         v=dx/dt= Aω(-sin(ωt+φ))               (8)
  • Here in equation 8 quantity Aω is known as velocity amplitude and velocity of oscillating particle varies between the limits ±ω.
  • From trignometry we know that
         cos2θ + sin2θ=1

         A2 sin2(ωt+φ)= A2- A2cos2(ωt+φ)
    Or
         sin(ωt+φ)=[1-x2/A2]            (9)
    putting this in equation 8 we get,

  • From this equation 10 we notice that when the displacement is maximum i.e. ±A the velocity v=0, because now the oscillator has to return to change it’s direction.
  • Figure below shows the variation of velocity with time in SHM with initial phase φ=0.

6. Acceleration of SHM

  • Again we know that acceleration of a particle is given by
         a=dv/dt
    where v is the velocity of particle executing motion.
  • In SHM velocity of particle is give by,
         v= -ωsin(ωt+φ)
    differentiating this we get,

    or,
         a=-ω2Acos(ωt+φ)          (11)
  • Equation 11 gives acceleration of particle executing simple harmonic motion and quantity ω2 is called acceleration amplitude and the acceleration of oscillating particle varies betwen the limits ±ω2A.
  • Putting equation 4 in 11 we get
         a=-ω2x                    (12)
    which shows that acceleration is proportional to the displacement but in opposite direction.
  • Thus from above equation we can see that when x is maximum (+A or -A), the acceleration is also maximum(-ω2A or +ω2A)but is directed in direction opposite to that of displacement.
  • Figure below shows the variation of acceleration of particle in SHM with time having initial phase φ=0.

7. Total energy in SHM 

  • When a system at rest is displaced from its equilibrium position by doing work on it, it gains potential energy and when it is released, it begins to move with a velovity and acquires kinetic energy.
  • If m is the mass of system executing SHM then kinetic energy of system at any instant of time is
         K=(1/2)mv2                    (13)
    putting equation 8 in 13 we get,
  • From equation (14) we see that Kinetic Energy of system varies periodically i.e., it is maximum (= (1/2)mω2A2) at the maximum value of velocity ( ±ωA) and at this time displacement is zero.
  • When displacement is maximum (±A), velocity of SHM is zero and hence kinetic energy is also zero and at these extreme points where kinetic energy K=0, all the energy is potential.
  • At intermediate positions of lying between 0 and ±A, the energy is partly kinetic and partly potential.
  • To calculate potential energy at instant of time consider that x is the displacement of the system from its equilibrium at any time t. 
  • We know that potential energy of a system is given by the amount of work required to move system from position 0 to x under the action of applied force.
  • Here force applied on the system must be just enough to oppose the restoring force -kx i.e., it should be equal to kx.
  • Now work required to give infinitesimal displacement is dx=kx dx.
    Thus, total work required to displace the system from 0 to x is

    thus,

    where, from equation 5 ω=√(k/m) and displacement x=A cos(ωt+φ).
    -From equation 14 and 15 we can calculate total energy of SHM which is given by,
  • Thus total energy of the oscillator remains constant as displacement is regained after every half cycle.
  • If no energy is dissipated then all the potential energy becomes kinetic and vice versa.
  • Figure below shows the variation of kinetic energy and potential energy of harmonic oscillator with time where phase φ is set to zero for simplicity.

8. Some simple systems executing SHM


(A) Motion of a body suspended from a spring 

  • Figure (6a) below shows a spring of negligible mass, spring constant k and length l suspended from a rigid support.
  • When a body of mass m is attached to this spring as shown in figure 6(b), the spring elongates and it would then rest in equilibrium position such that upward force Fup exerted by spring is equal to the weight mg of the boby.
  • If the spring is extended by an amount Δl ater attachment of block of mass m then in its equilibrium position upward force equals
          Fup=kΔl
    also in this equilibrium position
          Fup=mg
    or,      kΔl=mg
  • Again the body is displaced in upwards direction such that it is at a distance x above equilibrium position as shown in figure 6(c).
  • Now extansion of spring would be (Δl-x), thus upward force now exerted on the body is
          Fup=k(Δl-x)
  • Weight of the body now tends to pull the spring downwards with a force equal to its weight. Thus resultant force on the body is
         F=k(Δl-x)-mg
          =mg-kx-mg
    or,
         F=-kx                    (17)
  • From equation 17 we see that resultant force on the body is proportional to the displacement of the body from its equilibrium position. 
  • If such a body is set into vertical oscillations it oscillates with an angular frequency
         ω=√(k/m)          (18)
(B) Simple pendulum

  • Simple pendulum consists of a point mass suspended by inextensible weightless string in a uniform gravitational field.
  • Simple pendulum can be set into oscillatory motion by pulling it to one side of equilibrium position and then releasing it.
  • In case of simple pendulum path ot the bob is an arc of a circle of radius l, where l is the length of the string.
  • We know that for SHM F=-kx and here x is the distance measured along the arc as shown in the figure below.
  • When bob of the simple pendulum is displaced from its equilibrium position O and is then released it begins to oscillate.
  • Suppose it is at P at any instant of time during oscillations and θ be the angle subtended by the string with the vertical.
  • mg is the force acting on the bob at point P in vertically downward direction.
  • Its component mgcosθ is balanced by the tension in the string and its tangential component mgsinθ directs in the direction opposite to increasing θ .
  • Thus restoring force is given by
         F=-mgsinθ               (19)
  • The restoring force is proportional to sinθ not to The restoring force is proportional to sinθ, so equation 19 does not represent SHM.
  • If the angle θ is small such that sinθ very narly equals θ then above equation 19 becomes
         F=-mgθ
    since x=lθ then,
         F=-(mgx)/l
    where x is the displacement OP along the arc. Thus,
         F=-(mg/l)x               (20)
  • From above equation 20 we see that restoring forcr is proportional to coordinate for small displacement x , and the constant (mg/l) is the force constant k.
  • Time period of a simple pendulum for small amplitudes is
  • Corresponding frequency relations are

    and angular frequency
         ω=√(g/l)          (23)
  • Notice that the period of oscillations is independent of the mass m of the pendulumand for small oscillations pperiod of pendulum for given value of g is entirely determined by its length.


(c) The compound pendulum

  • Compound pendulum is a rigid body of any shape, capable of oscillating about a horizontal axis passing through it.
  • Figure below shows vertical section of rigid body capable of oscillating about the point A.
  • Distance l between point A and the centre of gravity G is called length of the pendulum.
  • When this compound pendulum is given a small angulr displacement θ and is then released it begins to oscillate about point A.
  • At angular displacement θ its center of gravity now takes new position at G'.
  • Weight of the body and its reaction at the support constitute a reactive couple or torque given by
         τ=-mg G'B
          =-mglsinθ          (24)
  • Equation 24 gives restoring couple which tends to bring displaced body to its original position.
  • If α is the angular acceleration produced in this body by the couple and I is the moment of inertia of body about horizontal axis through A then the couple is
         Iα=-mglsinθ
    if θ is very small then we can replace sinθ≅θ, so that
         α=-(mgl/I)θ          (25)
  • From above equation (25) we se that pendulum is executing Simple Harmonic Motion with time period
9. Damped Oscillations

  • Fractional force, acting on a body opposite to the direction of its motion , is called damping force.
  • Damping force reduces the velocity and the Kinetic Energy of the moving body.
  • Damping or dissipative forces generally arises due to the viscosity or friction in the medium and are non conservative in nature.
  • When velocities of body are not high, damping force is found to be proportional to velocity v of the particle i.e.,
         Fd=-γv          (27)
    where, γ is the damping constant.
  • If we take damping into consideration for an oscillator then oscillator experiences
    (i) Restoring Force :- F=-kx
    (ii) Damping Force :- Fd=-γv
    where, x is th edisplacement of oscillating system and v is the velocity of this dispalcement.
  • Thus equation of motion of damped harmonic oscillator is

    where,r=(γ/2m) and ω2=k/m
  • Solution of above equation is of the form
         x=Ae-rtcos(ω't+φ)          (29)     
    where,
         ω'=√(ω2-r2)     (30)
    is the angular frequency of the damped oscillator. 
  • In equation 29 , x is a function of time but it is not a periodic function and because of the damping factor e-rt this function decreases continously with time. 


10. Driven or Forced Harmonic oscillator

  • If an extra periodic force is applied on a damped harmonic oscillator, then the oscillating system is called driven or forced harmonic oscillator, and its oscillations are called forced oscillations.
  • Such external periodic force can be represented by
         F(t)=F0cosωft          (31)
    where, F0 is the amplitude of the periodic force and ωf is th e frequency of external force causing oscillations.
  • Differential equation of motion under forced oscillations is
  • In this case particle will neither oscillate with its free undamped frequency nor with damped angular frequency rather it would be forced to oscillate with angular frequency ωf of applied force.
  • When damped oscillator is is set in forced motion, the initial motion is combination of damped oscillation and forced oscillations .
  • After certain amount of time the amplitude of damped oscillations die out or becomec so small that they can be ignoredand only forced oscillationd remainand the motion is thus said to reached steady state. 
  • Solution of equation 32 is
         x=Acos(ωft+φ)               (33)
    where A is the amplituse of oscollation of forced oscillator and φ is the initial phase.
  • In case of forced oscillations both amplitude A and initial phase φ are fixed quantities depending on frequency ωf of applied force.
  • Calculations show sthat amplitude

    and initial phase
         tanφ=-v0/(ωfx0)
    where, x0 is displacement of particle at time t=0, the moment driven force is applied and v0 is the velocity of the particle at time t=0.
  • When ωf is very close to ω, then m(ω2-(ωf)2 would be much less than ωfγ, for any reasonable value of γ, then equation 34 becomes
         A=F0/γωf          (35)
  • Thus the maximum possible amplitude for a given driven frequency is governed by the driving frequency and the damping ,and is never infinity.
  • This phenomenon of increase in amplitude when the driving force is close to natural frequency of oscillator is called RESONANCE.
  • Thus resonance occurs when frequency of applied force becomes equal to natural frequency of the oscillator without damping.

No ratings yet - be the first to rate this.

Add a comment

You're using an AdBlock like software. Disable it to allow submit.

Make a free website with emyspot.com - Report abuse