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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements3 Marks
Question
A book with many printing errors contains four different formulas for the displacement $y$ of a particle undergoing a certain periodic motion:
  1. $\text{y}=\text{a}\sin2\pi\text{ t/T}$
  2. $\text{y}=\text{a}\sin \text{vt}$
  3. $\text{y}=(\text{a/T})\sin\text{t/a}$
  4. $\text{y}=(\text{a}\sqrt{2})(\sin2\pi\text{t/T}+\cos2\pi\text{t/T})$
$(a =$ maximum displacement of the particle, $v =$ speed of the particle. $T =$ time$-$period of motion$)$. Rule out the wrong formulas on dimensional grounds.
✓

Answer

The displacement $y$ has the dimension of length, therefore, the formula for it should also have the dimension of length. Trigonometric functions are dimensionless and their arguments are also dimensionless. Based on these considerations now check each formula .
  1. $\frac{2\pi\text{t}}{\text{T}}=\frac{\text{T}}{\text{T}}=1=(\text{M}^0\text{L}^0\text{T}^0)\ \ \dots$ dimensionless
  2. $\text{vt}=(\text{LT}^{-1})(\text{T})=\text{L}=\big[\text{M}^0\text{L}^0\text{T}^0\big]\ \ \dots$ not dimensionless
  3. $\frac{\text{t}}{\text{a}}=\frac{\text{T}}{\text{L}}=[\text{L}^{-1}\text{T}^1]\ \ \dots$ not dimensionless
  4. $\frac{2\pi\text{t}}{\text{T}}=\frac{\text{T}}{\text{T}}=1=\big[\text{M}^0\text{L}^0\text{T}^0\big]\ \ \dots$ dimensionless
dimensionally. The formulas in $(ii)$ and $(iii)$ are dimensionally wrong.

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