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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 ...

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.

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Write the of dimensional formulae for the given physical quantities

Displacement (y) units meters

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

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

time [math] t = [M^0L^0T^1] [/math]

Time period [math] T = [M^0L^0T^1] [/math]

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

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

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

Dimensional consistency of equations

[math] [M^0L^1T^0] = [M^0L^1T^0] [/math]

L.H.S = R.H.S

Therefore, the formula given is dimensionally correct.

b) y = a \sin vt

Dimensional consistency of equations

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

[math] [M^0L^1T^0] = [M^0L^1T^0] ([M^0L^1T^0]) [/math]

LHS \neq RHS

Therefore, the formula given is dimensionally incorrect.

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

Dimensional consistency of equations

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

[math] [M^0L^1T^0] = [M^0L^0T^0] [/math]

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}

[math] [M^0L^1T^0] = [M^0L^1T^0] [/math]

L.H.S = R.H.S

Hence, the formula given is dimensionally correct.