Derivation of Physics Formula

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 it 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 \[\vec{B}\] 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    \[\vec{B}\] .

(Image will be uploaded soon)

(Image will be uploaded soon)

Force on North pole = mB, along \[\vec{B}\]

Force on South pole = mB, opposite to  \[\vec{B}\]

The forces are equal and they tend to rotate the magnet clockwise so as to align it along \[\vec{B}\].

Therefore, torque on the bar magnet is given by

τ = force✕ perpendicular distance = mB ✕ 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

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 

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

The velocity of the particle at instant t,

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

Acceleration of the particle at this instant,

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

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

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

\[W = \int_{0}^{y}hy\text dy = h[\frac{y^{2}}{2}]^{y}_{0} = \frac{hy^{2}}{2}\]

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

\[K_{p} = \frac{1}{2}hy^{2}\] (Substituting the value of ‘h’ from eq (4))

\[K_{p} = \frac{1}{2}mω^{2}y^{2}\] (Substituting the value of ‘y’ from eq(1))

Kinetic Energy

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

Kinetic energy at the instant time t is given by

\[K_{k} = \frac{1}{2}mV^{2}\] (Substituting the value of ‘V’  from eq(2))

\[K_{k} = \frac{1}{2}m(aωCosωt)^{2} = \frac{1}{2}ma^{2}ω^{2}(1 - Sin^{2}ωt)\]

\[K_{k} = \frac{1}{2}ha^{2}(1 - \frac{y^{2}}{a^{2}})\]

List of Important Physics Derivations in Class 12

Deriving Physics Equations

Equation of uniformly accelerated motion by calculus method

Consider a body having linear motion with uniform acceleration a.

Let v1 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.

\[Acceleration = a = \frac{dv}{dt}\]

dv = adt…..(1)

Integrating eq(1) with conditions when velocity v1 changes to  v2and time t1 to t2

\[\int_{v_{1}}^{v_{2}}dv = a\int_{t_{1}}^{t_{2}}dt\]

\[v_{2 } - v_{1} = a(t_{2} - t_{1})\]

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

Displacement -time relation:

v = \frac{ds}{dt}

ds = v dt

ds = (v_{o} + at)dt

\[\int_{s_{o}}^{s}ds = \int_{o}^{t}(v_{o} + at)dt\]

\[S - S_{o} = v_{o}t + \frac{1}{2}at^{2}\]

List of important physics derivations in class 11

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.