A book with many printing errors contains four different formulas for the displacement $y$ of a particle undergoing a certain periodic motion:
$(a)\;y=a \sin \left(\frac{2 \pi t}{T}\right)$
$(b)\;y=a \sin v t$
$(c)\;y=\left(\frac{a}{T}\right) \sin \frac{t}{a}$
$(d)\;y=(a \sqrt{2})\left(\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right)$
$(a=$ maximum displacement of the particle, $v=$ speed of the particle. $T=$ time-period of motion). Rule out the wrong formulas on dimensional grounds.
A book with many printing errors contains four different formulas for the displacement $y$ of a particle undergoing a certain periodic motion:
$(a)\;y=a \sin \left(\frac{2 \pi t}{T}\right)$
$(b)\;y=a \sin v t$
$(c)\;y=\left(\frac{a}{T}\right) \sin \frac{t}{a}$
$(d)\;y=(a \sqrt{2})\left(\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right)$
$(a=$ maximum displacement of the particle, $v=$ speed of the particle. $T=$ time-period of motion). Rule out the wrong formulas on dimensional grounds.
$(a)$ 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.
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