किसी पुस्तक में, जिसमें छपाई की अनेक त्रुटीयां हैं, आवर्त गति कर रहे किसी कण के विस्थापन के चार भिन्न सूत्र दिए गए हैं 

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

$(b)\;y=a \sin v t$

$(c)\;y=\left(\frac{a}{T}\right) \sin \frac{t}{a}$

$(d)\;y=(a \sqrt{2})\left(\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right)$

$(a=$ कण का अधिकतम विस्थापन, $v=$ कण की चाल, $T=$ गति का आवर्त काल ) । विमीय आधारों पर गलत सूत्रों को निकाल दीजिए |

$(a)$ Correct $\quad y=a \sin \frac{2 \pi t}{T}$

Dimension of $y= M ^{0} L ^{1} T ^{0}$

Dimension of $a= M ^{0} L ^{1} T ^{0}$

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

Dimension of L.H.S $=$ Dimension of R.H.S

Hence, the given formula is dimensionally correct.

$(b)$ Incorrect $y=a \sin v t$

Dimension of $y= M ^{0} L ^{1} T ^{0}$

Dimension of $a= M ^{0} L ^{1} T ^{0}$

Dimension of $v t= M ^{0} L ^{1} T ^{-1} \times M ^{0} L ^{0} T ^{1}= M ^{0} L ^{1} T ^{0}$

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

$(c)$ $\text { Incorrect } \quad y=\left(\frac{a}{T}\right) \sin \left(\frac{t}{a}\right)$

Dimension of $y= M ^{0} L ^{1} T ^{0}$

Dimension of $\frac{a}{T}= M ^{0} L ^{1} T ^{-1}$

Dimension of $\frac{t}{a}= M ^{0} L ^{-1} T ^{1}$

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

$(d)$ Correct $y=(a \sqrt{2})\left(\sin 2 \pi \frac{t}{T}+\cos 2 \pi \frac{t}{T}\right)$

Dimension of $y= M ^{0} L ^{1} T ^{0}$

Dimension of $a= M ^{0} L ^{1} T ^{0}$

Dimension of $\frac{t}{T}= M ^{0} L ^{0} T ^{0}$

since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of $y$ and $a$ are the same. Hence, the given formula is dimensionally correct.