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A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic function:
(A) $y = a\sin 2{\pi ^{t/T}}$
(B) $y = a\sin \nu t$
(C) $y = \left( {a/T} \right)\sin t/a$
(D) $y = a/\sqrt 2 \left( {\sin 2\pi /T + \cos 2\pi /T} \right)$

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: This could be simply solved by breaking the total distance into three parts and applying the basic formula of mean speed. The average speed is the distance per time ratio.

Complete step by step answer:
Since all the equations are for the displacements of the particle, they all should have the dimension of the length.
Now, checking all the equations of displacement to be dimensionally correct.
For the first option:
Since the trigonometric functions are dimensionless, $\dfrac{{2\pi t}}{T}$​ will also be dimensionless.
$\dfrac{{2\pi t}}{T} = \dfrac{T}{T} = 1 = \left[ {{M^0}{L^0}{T^0}} \right]$
Since the trigonometric functions are dimensionless, so $vt$ should be dimensionless.
$vt = \left( {L{T^{ - 1}}} \right)\left( T \right) = L = \left[ {{M^0}{L^{ - 1}}{T^{ - 1}}} \right]$
Thus, the given equation is not correct.
Similarly, we check for other options:
The trigonometric functions are dimensionless, $2\pi t$​ will also be dimensionless.
$\dfrac{{2\pi t}}{T} = \dfrac{T}{T} = 1 = \left[ {{M^0}{L^0}{T^0}} \right]$
The given equation is dimensionally correct.
Thus, the formulas in b) and c) are dimensionally wrong.
The correct ones are a) and d).

Additional Information: In engineering and science, dimensional analysis is the analysis of the relationships between different physical quantities by identifying their base quantities (such as length, mass, time, and electric charge) and units of measure (such as miles vs. kilometres, or pounds. It is a method of reducing the number of variables required to describe a given physical situation by making use of the information implied by the units of the physical quantities involved.

Note: It should be always kept in mind that we make use of dimensional analysis for three prominent reasons: To check the consistency of a dimensional equation. To derive the relation between physical quantities in physical phenomena. To change units from one system to another. The main advantage of a dimensional analysis of a problem is that it reduces the number of variables in the problem by combining dimensional variables to form non-dimensional parameters. By far the simplest and most desirable method in the analysis of any fluid problem is that of direct mathematical solution.
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