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# A book with many printing errors contains four different formulas for the displacement y of a particle undergoing certain periodic motion:1) $y = a\sin 2{\pi ^{t/T}}$2) $y = a\sin vt$3) $y = (a/T)\sin t/a$4) $y = (a/\sqrt 2 )(\sin 2\pi /T + \cos 2\pi /T)$(a = maximum displacement of the particle, v = speed of the particle. AT = time period of motion). Rule out the wrong formulas on dimensional grounds.

Last updated date: 15th Sep 2024
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Hint:-Dimensional analysis is being used to check the equation if it is correct or not. If the dimensions in the LHS and RHS are equal then the equation is said to be correct if not then it is not correct.

Complete step-by-step solution:
Step1: Dimensional Analysis of$y = a\sin 2{\pi ^{t/T}}$. Here the dimensions on the LHS must be equal to the dimension on the RHS.
$y = a\sin 2{\pi ^{t/T}}$;
${L^1} = {L^1}$;
Functions that are trigonometric generally are dimensionless, so $\dfrac{{2\pi t}}{T}$will also be a dimensionless quantity.
$\dfrac{{2\pi t}}{T} = \dfrac{T}{T} = {M^0}{L^0}{T^0}$;
Hence the dimensions are correct, so, this formula is dimensionally right.

Step2: Dimensional Analysis of$y = a\sin vt$. Here the dimensions on the LHS must be equal to the dimension on the RHS.
$y = a\sin vt$;
Here the amplitude will have the same dimension as length.
${L^1} = {L^1}$;
Now, ${\text{Distance = velocity}} \times {\text{Time}}$. So$L' = vt$;
$vt = {L^1}{T^{ - 1}}{T^1}$;
$vt = {L^1}{T^{ - 1}}{T^1} = {M^0}{L^1}{T^0}$;
${M^0}{L^1}{T^0} \ne {M^0}{L^0}{T^0}$;
This is not a dimensionally right formula;
Step3: Dimensional Analysis of$y = (a/T)\sin t/a$. Here the dimensions on the LHS must be equal to the dimension on the RHS.
$y = (a/T)\sin t/a$;
$y = (\dfrac{a}{T}) = (\dfrac{{{L^1}}}{{{T^1}}}) = ({L^1}{T^{ - 1}})$;
${L^1} \ne ({L^1}{T^{ - 1}})$;
The formula $y = (a/T)\sin t/a$is not dimensionally correct.
Step4: Dimensional Analysis of$y = (a/\sqrt 2 )(\sin 2\pi /T + \cos 2\pi /T)$. Here the dimensions on the LHS must be equal to the dimension on the RHS.
$y = (a/\sqrt 2 )(\sin 2\pi /T + \cos 2\pi /T)$;
${L^1} = {L^1}$;
$(\dfrac{{2\pi t}}{T}) = (\dfrac{T}{T}) = 1 = {M^0}{L^0}{T^0}$;
The formula $y = (a/\sqrt 2 )(\sin 2\pi /T + \cos 2\pi /T)$is dimensionally correct.

Final Answer: Option “2 and 3” is incorrect.

Note:- Here one has to dimensionally analyze each and every option that is available. Then we have to compare if the dimensions on the LHS and RHS are equal. Every equation that is used in the world of physics has to be dimensionally correct.