Basic Science World Blog

Figure (1.4) shows a simple pendulum consisting of a cord of length  l having a particle of mass m attached to its free end and making  an   angle θ  with the vertical.

The forces acting on m are its weight (mg) and the tension acting along the cord. The tangential component of the weight is the restoring force mg sin θ acting on m tending to return it to the equilibrium position.

F= – mg sin θ,

But  sin θ= x/l and by using Newton’s second law   ()    we have:

Or

Which represents SHM with angular frequency

and Period

T=2π/ω =2π             (1.21)

Notice that T is independent of m and depends only on l and g.

Let a body of mass m is attached to a spring of force constant k and free to move over a horizontal frictionless surface. If the body is displaced to the right, Fig.(1.5), the force exerted by the spring on the body is directed to the left and is F = – kx, by using Newton’s second law ()

 

the  solution of this equation must be that of a SHM

x= a sin(ωt+δ)   

with     and period

T=2π           (1.22).

                        To find the resulting motion of a system which moves in the x-direction under the effect of two SHMs of equal angular frequencies but of different amplitudes and phases, we can represent each SHM by its vector and carry out a vector addition.

 

If the displacement of the first motion is given by:

X₁=a₁ cos(ωt+δ₁),                    (1.23)

 

and that of the second by:

X₂=a₂ cos(ωt+δ₂)                      (1.24)

 

Then Fig.(1.6) show that the resulting displacement amplitude R is given by:

 

R²=(a₁+a₂ cosδ)² + (a₂ sinδ)²      (1.25)

 

where δ=δ₂ – δ₁    is constant. The phase constant θ of R is given by:

 

     (1.26)

so the resulting SHM has the displacement

X= R cos(ωt+θ)                     (1.27)

an oscillation of the same frequency  ω but having an amplitude R and phase constant  θ.

When a medium is disturbed by a passage of a wave through it, the particles comprising the medium are caused to vibrate. This vibration will propagate in the medium from particle to another with a certain velocity. The process of the propagation of vibrations in the medium is called a wave.

Examples of the oscillatory motion are: the oscillations of a mass on a spring; the vibrations of a stringed musical instrument; the motion of a pendulum; and the oscillations of the molecules in a solid about their equilibrium position. Also, the electromagnetic waves, light waves, are characterized by oscillating electric and magnetic field vectors.

A particle moving along the X-axis undergoes a SHM if a small displacement  X  from its equilibrium position set up a restoring force  F  which is proportional to X and acting in a direction towards the equilibrium position. This restoring force F may be written as:

 F = – k X      (1.1)

 

Where K, the constant of proportionality, is called the stiffness and the negative sign shows that the force is acting against the direction of increasing displacement. The stiffness k is the restoring force per unit distance and has the dimensions

K= F/x=MT ^-2 . Appling Newton’s second law;   we get:

, where           is the acceleration. This gives

            (1.2)

Where the dimensions of  k=T^-2.,here T is the periodic time, the time necessary for one complete wave to pass any points, and v=1/T frequency, the number of waves which pass any point in a unit time. Introducing the angular frequency as ω=2πν                           

so the periodic time is:

      (1.3)

 

if k/m is written as ω².The equation of SHM becomes:

                 (1.4)

This is a linear a second-order differential equation governing the motion. It is not the equation of motion. To find the equation of motion we have to solve Eq.(1.4) for x.

 

The behavior of a simple harmonic oscillator is expressed in terms of its displacement x, its velocity    , and its acceleration    

Which are obtained due to the solution of the differential equation(1.4):

Integrate this equation yields

                          (1.5)

We can write the last equation in the form

 

Integration with respect to t gives

 

 

where  δ is the constant of integration. Thus

                 (1.7) 

Equation (1.7) is a solution of Eq.(1.5)

The limiting values of     are   ±1 so the motion takes place between the limits   x=±a   and the constant "a" is the maximum value of x, known as the amplitude of displacement.

The quantity       is known as the phase of the motion and  δ is known as the phase constant. The form of Eq.(1.7) is shown in Fig.(1.1)

 

 

 

The period T; is also shown in Fig.(1.1); if we add T to t in Eq.(1.7) the value of x must remain unchanged. Hence,

                  (1.8)

This means that the phase has been increased by exactly   2π  radians. Thus by inspection ωT=2πhence        

  T=2π/ω                  (1.9)

and

 ω=2π/T=2πν          (1.10)

ω  Is known as the angular frequency  of the motion.

Then we can write the equation of motion in more convenient form as:

           (1.11)

The values of the velocity and acceleration of a particle undergoing SHM are given by:

         (1.12)

and

        (1.13)

Thus the acceleration is proportional to the displacement but is in opposite direction, as can be seen from Fig.(1.2).

 

 

 

 

 

Fig(1.2) variation with time of displacement, velocity and acceleration where δ=0

Note that at any specified time the velocity is  π/2    out of phase with the displacement and the acceleration is  π     out of phase with the displacement. Since the sinr and cosine functions oscillate between ±1      , then the maximum value of the velocity and acceleration are given by  ωa      and  ω²a,   respectively.

The fact that the velocity is zero at maximum displacement in SHM and is a maximum at zero displacement illustrates the important concept of an exchange between kinetic and potential energy. It is known that for motions, the total mechanical energy E which is the sum of kinetic energy KE plus the potential energy PE, is conserved.

E= KE + PE=KE_max=PE_max    (1.14)

The kinetic energy at any instant is given by:

Using the relations  v=a ω cos(ωt+δ)   and ω²=k/m   gives:

KE=1/2 m ω²a² cos²(ωt+δ)      (1.15)

The kinetic energy has a maximum value of    1/2 m ω²a²  or 1/2ka²During the motion, the kinetic energy varies between zero and this maximum value, as can be seen from the curve in Fig.(1.3).

The potential energy PE, at any instant is given by:

 PE=1/2kx²     

      =1/2 ka² sin²(ωt+δ)                    (1.16)

The potential energy has a maximum value of  1/2ka²  . During the motion the potential energy varies between zero and this maximum value, Fig.(1.3). The total energy is the sum of KE plus PE:

E=1/2ka²                               (1.17)

 

 

Fig.(1.3): Kinetic energy KE versus displacement for SH oscillator. Note that KE+PE=const.

 

Thus the total energy is constant and has the value  1/2ka²       which it is proportional to the square of the amplitude of the motion. At the equilibrium position, the PE is zero, but  1/2ka²        . At other positions, the kinetic and potential energies have different values but their sum is always a constant equal to 1/2ka².      .

 In order to obtain the velocity at any displacement x, let us use the energy conservation relation:

thus

                                  (1.18)

Or

                                  (1.19)

This relation shows clearly that the speed is a maximum at the equilibrium position x=0 and is zero at maximum displacement x=a.