 # Introduction

On the basis of mechanics, there are two Fundamental equations of classical mechanics to explain a system-

### First,

The total energy of a particle is the sum of kinetic energy mv²/2 and potential energy V .thus, E =mv²/ 2 +V          (1)

Here v is the velocity of the particle and m is mass of the particle. We know that velocity is the function of time (t), so that linear momentum is given by-

P =mv                                       (2)

Therefore,

E= 1/2(P²/m) + V              (3)

Equation (3)can be used in the number of ways-

For example, since momentum is

P=mv

P = m (dx/dt)                    (4)

Where (dx) is the distance traveled in time dt

Above equation, equation (4) is in the form of the differential equation in which x is the function of time t. When we solve Equation (4) it gives the position and momentum of the particle, which is again the function of time (t). Thus position and momentum of the particle may be represented as x(t) and p(t) respectively. And we can also determine the trajectory of a particle by knowing x(t) and p(t).

Let us suppose that a particle is in stationery state. Thus at this State, potential energy V of particle remains zero. Now substitute V =0 in equation (3) we get,

E = p²/2m             (5)

Now, on substituting the values of p from equation (4) into equation (5) we see the that,

E= m²/2m (dx/dt)²

Or E= m/2 (dx/dt)²

or (2E/m)½ = dx/dt

Now, dx/dt = √2E/m            (6)

Hence solution of equation (6) can be expressed in the form of

x(t) = x(0) + (√2E/m).t

Or, x(t) = x(0) + (√2mE/m²).t

x(t) =x(0) + p.(t/m)              (7)

Here p may be written as p(0) i.e p represents initial momentum of particle, therefore

x(t) = x(0) + p(0) t/m

Thus if we know the initial position and momentum of the particle then position and momentum of particle can be assumed.

## Second

Second equation of classical mechanics is the second law of motion of Newton.i.e. force acting on a particle is may be defined as the multiplication of mass of the particle with its acceleration,

F= ma

Where ‘m’ is mass of the particle and ‘a’ is the acceleration of the particle.

Since a=dv/dt

Therefore F = m(dv/dt) = d(mv)/dt =dp/dt

That means force can be defined as the rate of change of momentum.

Thus,  dp/dt= F                  (8)

Since rate of change of momentum is directly proportional to the acceleration, therefore

F= dp/dt  = m.(d²x/dt²)      (9)

It means if we know the force acting on a particle, then it is possible to determine its path and position.

Let F be the force applied on particle for time τ(tau),then, Newton’s equation of motion will be-

(1)dp/dt =F , a constant between t=0  and t=τ

And (2) ±dp/dt =0 (after t= τ)

Solution of equation (1) is :

p(t) = p(0) t + Ft (0,t,τ)            (10)

Momentum of particle in between this time interval will be –

p(τ) = p(0) + F (τ)                      (11)

Solution of equation (1) is given by-

p = constant

Or, after t=τ the momentum of the particle will be p(τ) at any time.

Let us Suppose that particle is in rest initially. therefore, its momentum is p(0)=0 and kinetic energy will be E= p²/2m.

After time τ the value of energy will be

E= F²τ²/2m           (12)

Since F and τ have any value ,so total energy E of moving particle may have any value.

If a particle ihabiving rotatory motion,then its angular momentum (J) is related with its angular velocity (ω) as

J = Iω                      (13)

Where I is moment of inertia. Equation (13) is like (12). To accelerate rotatory motion of particle a twisting force T(torque) is applied on the particle,then equation (13) will became

J= T                          (14)

If torque is applied for time (τ) then energy of particle will be increased by

E = T²τ²/2I           (15)

Where T is called twisting force and I is called moment of inertia of particle.

Equation (15) is similar as equation (12).

#### Vibrational motion

If a particle moves linearly on both side of its mean position and have simple harmonic motion then this motion is called vibrational motion.

In vibrational motion, restoring force is directly proportional to the displacement (x) . I.e.

F ∝x

Or F = -kx           (16)

Where k is force constant. Negative sign indicates that F is acting ioppsoosite direction od x. when F is in left hand side displacement is in right hand side, or vice versa. For this motion, Newton’s equation will be

m(d²x/dt²) = -kx          (17)

Solution of above equation is

x(t) =Asinωt,         ω  =(k/m)¹/₂     (18)

Since momentum is m ,dx/dt

So  p(t) = m(dx/dt)

p(t) = mωAcosωt                (19)

According to sinωt position of particle changes harmonically. Frequency of motion is (ν) = ω/2π And momentum of the particle is minimum when displacement is maximum (x=A) and momentum is maximum when displacement is minimum (x=0).

Total energy of particle

Kinetic energy+ potential energy = KE + PE

Or  E = (p²/2m )+ V

But potential energy is related to F as

F= -dV/dx               (20)

dV/dx= -F =+kx

dV = kxdx

On integration

V= (1/2)kx²               (21)

Thus

E=p²/2m + (1/2)kx²      (22)

Putting the values of x(t) and p(t) from equation (18) and (20) in equation (22) ,we have

E =[ (mω²Acosωt)²/2m ]+ [k(Asinωt)²/2]

E =[ (m²ω²A²cos²ωt)/2m] + kA²sin²ωt/2

E= m²(k/m).(A²cos²ωt/2m)+ 1/2(kA²sin²ωt)

E= [(1/2)A²k(cos²ωt+sin²ωt)

E = kA²/2

If we change the motion particle (translation,rotational or vibrational) then particle may can be excited to some extent.

By classical mechanics we can conclude that-

1) we can measure position as well as momentum of moving particles

2) A moving particle can be  excited upto any extent by applying small force on translatrota ,rotational or vibrational motion.

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