Ncert Exercises & Solutions

A book with many printing errors contains four different formulas for the displacement \(y\) of a particle undergoing a certain periodic motion. Rule out the wrong formulas on dimensional grounds.
(A) \(y=a\sin\left(\dfrac{2\pi t}{T} \right) \)
(B) \(y=a\sin (vt) \)
(C) \(y=\dfrac{a}{T}\sin\left(\dfrac{t}{a} \right) \)
(D) \(y=(a\sqrt2) \left[\sin\left(\dfrac{2\pi t}{T} \right)+\cos \left(\dfrac{2\pi t}{T} \right) \right] \)
where, \(a\)= maximum displacement of the particle, \(v\) = speed of the particle. \(T\) = time-period of motion.

Choose the correct option from the given ones:

1.	\(A,B\) and \(C\)	2.	\(B, C\) and \(D\)
3.	\(B\) and \(C\)	4.	\(B\) and \(A\)

The argument of a trigonometrical function, i.e. angle is dimensionless. Now using the principle of homogeneity of dimensions.

(a) \frac{2 π t}{T} = \frac{[T]}{[T]} = d i m e n s i o n l e s s

(b) v t = [L T^{- 1}] [T] = [L] = h a v i n g d i m e n s i o n

(c) \frac{t}{a} = \frac{[T]}{[L T^{- 2}]} = [L^{- 1} T^{- 3}] = h a v i n g d i m e n s i o n

(d) \frac{2 π t}{T} = \frac{[T]}{[T]} = d i m e n s i o n l e s s

S o, (b) a n d (c) a r e d i m e n s i o n a l l y w r o n g .

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