Derivation of Physics Formula

What are the Derivations in Physics?

In Physics, it’s all about a better understanding of the basics. The derivation in physics defines the origination of some mathematical algorithm by understanding any physical phenomena.

On this page, we will learn about the following:

  • Why do we derive Physics formulas?

  • How to derive Physics formulas?

  • How to derive equations in Physics?

  • Derivation of Physics formulas

  • Deriving Physics equations

Why do we Derive Physics Formulas?

Everything we study in Physics has some logic behind and Mathematics gives us the logic to understand the phenomena and when we see the connection between mathematics and physics our understanding increases much more. During applications, students may come across many concepts, problems, and mathematical formulas. With the help of derivations, students use their ability and creativity and good sort of potential to find solutions.

How to Derive Physics Equations?

Mathematical derivations are important in deriving the physical equations because it helps to make us understand where the equation came from, why that is the equation for a particular problem.

How to Derive Physics Formulas?

Physics formulas are derived from observations and experiments.

There are few derivations done below to describe how to derive physics formulas.

Deriving Physics Formulas

Torque on a Bar Magnet

                                         →

A uniform magnetic field B  is represented by parallel lines.

 NS is a bar magnet with the length 2l and strength at each pole is m. 

                                                                                               →     

The magnet is held at an angle Ө with the direction of B.

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                                                          →  

Force on North pole = mB, along B. 

                                                                    →

Force on South pole = mB, opposite to B.

                                                                                                                                          

The forces are equal and they tend to rotate the magnet clockwise so as to align it along 

  →

  B.

Therefore, torque on the bar magnet is given by

て = force x perpendicular distance = mB x NA…(1)

Consider Δ NAS

SinӨ  = NA/ NS  = NA/ 2l =>   NA = 2lSinӨ…(2)

Putting the value of (2) in (1)

て  = mB x 2lSinӨ.  Since M = m x 2l

て  = B x M sinӨ

In vector form, the equation (3) will be written as

 

→     →   →

て  = B x M


                                                                                                                                 →

The direction of the torque  is orthogonal to the plane containing M and B

 

 

Heat Engine

 A heat engine is a device that converts heat energy into mechanical energy without the change in the internal energy of the system.

A heat engine essentially consists of the following components:

  • Source of heat at high temperatures

  • Working substance

  • Sink of heat at a lower temperature

Suppose, Q1 = Amount of heat absorbed by the working substance from the source at T1 K,

Q2 = The amount of heat rejected to the sink at T2 K,

W is the net amount of external work done by the working substance.

Therefore, the net amount of heat absorbed,

dQ = Q1 - Q2

Since the working substance returns to its initial state, the change in its internal energy dU will be zero.

According to the first law of thermodynamics,

dQ = dU + dW

Therefore, dQ = dW …(1)

i.e. the net amount of heat absorbed = external work done by the engine

Q1 - Q2 = W…(2)

Thermal Efficiency (η )

It is defined as the ratio of work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source.

η = The net work done per cycle(W)  / Total amount of heat absorbed per cycle (Q1)

= Q1- Q2 / Q1

η = 1 - Q2 /Q1

 

The Total Energy in S.H.M

Simple harmonic motion is the motion executed by a point mass subjected to a force that is proportional to the displacement of the particle but opposite in sign.

A particle executing S.H.M possess the following: Potential energy and Kinetic energy

Potential energy.

This is on account of the displacement of the particle from its mean position.

Consider a particle of mass m, executing S.H.M with amplitude a and constant angular frequency ω.  Suppose t second after starting from its mean position, the displacement of the particle is y, which is given by

y =  a Sin ωt…(1)

 

The velocity of the particle at instant t,

V = dy /dt = d ( a Sin ωt)/ dt = a  x d(Sin ωt)/dt

V = a ωCosωt ….(2)

 

Acceleration of the particle at this instant,

A= dV/ dt = d( a ωCosωt)/ dt = a ω d(Cosωt)/ dt

A = - a ω2 Sin ωt = - ω2 y….(3)

 

Here, a negative sign indicates that acceleration is always directed from its mean position.

Restoring force, F =mass x acceleration

     = - m x  ω2 y = - hy

Here, m x  ω2 = h =  force constant or spring constant of S.H.M....(4)

 

Total work done on displacing the particle from its mean position to a position of displacement y will be 

W = \[\int_{0}^{y}\]  hy dy = h [y(1 + 1)/ (1 + 1) = h y2 / 2]     

So, the work done appears as the potential energy at a given instant. Thus 

Kp = 1/ 2 h y2   (putting value of ‘h’ from eq (4))

= 1 /2 m  ω2 y (putting the value of ‘y’ from eq(1))

Kp = 1/ 2 m  ω2  a2 Sin2 ωt….(5)


 

Kinetic Energy

This energy is on account of the velocity of the particle.

Kinetic energy at the instant time t is given by

Kk = 1/ 2 m V2 (Putting the value of ‘V’  from eq(2))

 = 1/ 2 m (aωCosωt )2 = 1/ 2 m a2 ω2 (1 - Sin2 ωt)

=1/ 2 h a2( 1 - y2 / a2)

Kk = 1/ 2 m ω2 ( a2 - 1 - y2)


 

Total energy = Kp + Kp 

 =   1 / k y2 + 1/ 2 m ω2(  a2 - 1 - y2) = 1/ 2 h a2 =1/ 2 m ω2 a2

= 1 /2  m  (2 ሀ π)2 a2

T.E. = 2 m ሀ2 π2 a2

 

List of Important Physics Derivations in Class 12

Topic

Formula

Description

Drift speed

Vd = 1/ て =  ½ (eE/m) (て)

Vd is proportional to the electric field E, and to the collision-time て

Biot-Savart law (The magnitude of a magnetic field)

dB = μo / 4π(i d l Sinθ/ r2

Here, θ is the angle 

                 →      →

between dl and r 

Coulomb’s law in vector form

→                 →

F  = kq1q2  r /r3


Here, q1 and q2 are the charges on the particles, r is the separation between 

         →

them, r  is the position vector of the force-experiencing particle

Cyclotron frequency

ሀ = 1 /T = qB/ 2 πm

Ω = 2 πf = qB/ m

Here, f is the cyclotron frequency, T, the orbital period.

The Q factor of a resonant circuit

Q = 1/R ✓L /C


Q is taken as a voltage multiplication

Here, Q is the ratio of voltage developed across the inductance or capacitance at resonance to the impressed voltage, which is applied across R.

Motional or induced EMF

E = -vLB

An emf induced by the motion of the conductor across the magnetic field is a motional electromotive force. This equation is true as long as the velocity, field, and length are mutually perpendicular.

Schrödinger wave equation

i h ძΨ/  ძt = - h2/ 2m  ძ2/  ძx2 + V(x).(x,t)

Here, Ψ = Aei (kx-ωt)

K = wave number = 2π/ λ

Hamiltonian of a system 

T =P.E. + K.E. 

E = hω 

Young’s double slit experiment

Bandwidth, β = (D/d) λ

The distance between (n+1)th and nth order consecutive bright fringes. 

xn+1 – xn =  [(D/d) [(n+1)λ] –  (D/d) [(n)λ]] = (D/d) λ

 

Deriving Physics Equations

Equation of uniformly accelerated motion by calculus method

Consider a body having linear motion with uniform acceleration a.

Let v, and v2 be the velocity of a body at time t1 and t2 respectively.

Velocity-time Relation

Let at an instant time t, the velocity be v, and change in velocity be dv in time interval dt.

Acceleration = a =  dv/ dt => dv = a dt…(1)

Integrating eq(1) with conditions when velocity v1 changes to v2 and time t1 to t2

 \[\int_{v1}^{v2}\] dv  =  a\[\int_{t2}^{t1}\] dt

v2 - v1 = a (t2 - t1)


v2 = v1 + a (t2 - t1) 

 

If initial velocity V1= u, final velocity v2 = v, and time t1 =0  then t2 =t.

 

v  = u +at

First equation of motion

 

Displacement -time relation

v = ds /dt

ds = v dt

ds = (v0 + at) dt                          

\[\int_{s0}^{s}\]ds = \[\int_{0}^{t}\] (v0 + at) dt                                                             

s − s0 = v0t + ½ at2  

 

s = s0 + v0t + ½at

 

List of important physics derivations in class 11

Topic

Formula

Description

Acceleration due to gravity and its variation with depth

gd  = g(1- d/R)

This is an expression for the acceleration due to gravity at the depth below the surface of the earth, and at the center of the earth d = R.

Hooke’s law

Fs = hx

Fs =spring force

h = spring constant

x= string stretch or compression

Stoke’s law

F=6πηrv

vt = 2 a2 (ρ -σ)/ 9η



The value of k for a spherical body = 6π

F= The viscous force on a spherical body falling through a liquid.

vt= Terminal velocity. ρ and σ are mass densities of sphere and fluid resp.

Equation of the path of a projectile


Y= x tanθ −g × x2/

2u2cos2θ

Equation of trajectory

 

Summary

  • Derivation means the action of obtaining something from a source or origin.

  • Through derivation, we find a logical connection between a natural phenomenon and a mathematical description of that phenomenon. In general, this points to an important conclusion about nature itself.

FAQ (Frequently Asked Questions)

Q1: What is the Significance of Dimensional Formula?

Ans: The expressions or formulae in physics tell us how and what of the fundamental quantities that are present in a physical quantity and such expressions are called the dimensional formula of that physical quantity. The dimensional formula helps in deriving units from one system to another.

Q2: Why are Physics Equations Important?

Ans: Equations are actually used by us in daily life. They are the mathematical representation of the things and they are useful to solve our daily life problem. They allow us to explain the past, and gives us the logics of the same via mathematical equations, and predict the future -  as far as the ultimate fate of the Universe.