### Question 2.14

NCERT Class XI Physics

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion (a) y = a sin 2π (t/T) (b) y = a sin vt (c) y = (a/T) sin (t/a) (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

### Solution in Detail

$ \begin{aligned} &\text{write dimensions first }\\ & [a] = [L], [t] = [T], [v] = [LT^{-1}], [T] = [T]\\\\ &\underline{\text{Question (a)}}\\ &y = a \sin 2\pi (t/T)\\ &\text{Step 1: verify sine angle first}\\ &[2\pi(t/T)] = [T]/[T] = \text{dimensionless}\\ &\therefore \sin 2\pi(t/T) \text{verified dimensionless}\\\\ &\text{Step 2: verify RHS and LHS}\\ &[\text{RHS}] = [a \sin 2\pi (t/T)]\\ & = [a] \times [\sin 2\pi (t/T)]\\ & = [L] \times \text{dimensionless}\\ &= [L] \\ &= [y]= \text{LHS verified dimensionally}\\\\ &\underline{\text{Question (b)}}\\ &y = a \sin vt\\ &\text{Step 1: verify sine angle first}\\ &[vt] = [LT^{-1}]\times[T] = [L]\\ &\text{fails, because an angle is dimensionless}\\\\ &\underline{\text{Question (c)}}\\ &y = a/T \sin (t/a)\\ &\text{Step 1: verify sine angle first}\\ &[t/a] = [T]/[L] = [TL^{-1}]\\ &\text{fails, because an angle is dimensionless}\\ \end{aligned} $

### Video Explanation

Please watch this youtube video for a quick explanation of the solution:

### Solution in brief and Physical Concept

LHS of all the expressions is displacement, y, so the dimensions on the LHS are [L]

for dimensional correctness, the dimensions on the RHS must also be [L].

Additionally, the angles in sin and cos should evaluate to dimensionless. But in (b) and (c) the angles fail this test, hence (b) and (c) are ruled out.

As verified above, the angles in (a) and (d) are dimensionless and the RHS also evaluate to [L]. Hence the expressions are dimensionally correct.

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