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Dimensional Analysis - Principle of Homogeneity, Applications and ...

We quantify the size and shape of things using dimensional measurement. ... Physics Important Questions ... of the relationship between physical quantities with the help of dimensions and units of measurement is termed as dimensional analysis. ... Using Dimensional Analysis to Check the Correctness of Physical Equation.

For more information, see Dimensional Analysis - Principle of Homogeneity, Applications and ...

Krishna
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According to the Principle of Homogeneity, the dimensions of each term of a dimensional equation on both sides should be the same. This theory allows us to convert the units from one form to another.

Write the of dimensional formulae for the given physical quantities

Displacement (y) units meters

Dimensional formula for displacement $y = [M^0L^1T^0]$

Dimensional formula maximum displacement $a = [M^0L^1T^0]$

time $t = [M^0L^0T^1]$

Time period $T = [M^0L^0T^1]$

Dimensional formula for velocity $v = [M^0L^1T^{-1}]$

A trigonometrical function's argument, i.e. angle, is dimensionless.

a) y = a \sin \frac{2 \pi t}{T}

Dimensional consistency of equations

$[M^0L^1T^0] = [M^0L^1T^0]$

L.H.S = R.H.S

Therefore, the formula given is dimensionally correct.

b) y = a \sin vt

Dimensional consistency of equations

$[M^0L^1T^0] = [M^0L^1T^0] ([M^0L^1T^{-1}] * [M^0L^0T^1])$

$[M^0L^1T^0] = [M^0L^1T^0] ([M^0L^1T^0])$

LHS \neq RHS

Therefore, the formula given is dimensionally incorrect.

c) y = \frac{a}{T} \sin \frac{t}{a}

Dimensional consistency of equations

$[M^0L^1T^0] = \frac{[M^0L^1T^0]}{[M^0L^0T^1]} * \frac{[M^0L^0T^1]}{[M^0L^1T^0]}$

$[M^0L^1T^0] = [M^0L^0T^0]$

LHS \neq RHS

Therefore, the formula given is dimensionally incorrect.

d) y = a \sqrt{2} (\sin \frac{2 \pi t}{T} + \cos \frac{2 \pi t}{T})

Dimensional consistency of equations

A trigonometrical function's argument, i.e. angle, is dimensionless.

y = a \sqrt{2}

$[M^0L^1T^0] = [M^0L^1T^0]$

L.H.S = R.H.S

Hence, the formula given is dimensionally correct.