(a) $y=a\sin \left( \frac{2\pi t}{T} \right)$

(b) y = a sin vt

Answer :

(a) Here, the dimension of ‘y’ is M⁰ L¹ T⁰

And the dimension of ‘a’ = M⁰ L¹ T⁰

And the dimension of

$\sin \frac{2\pi t}{T}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

The formula is dimensionally valid since the dimensions on both sides are equal.

(b) The dimensions on both sides are not equal, hence it is dimensionally inaccurate.