Question.
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.
solution:
(a) Answer: Correct
$y=a \sin \frac{2 \pi t}{T}$
Dimension of $y=M^{0} L^{1} T^{0}$
Dimension of $a=M^{0} L^{1} T^{0}$
Dimension of $\sin \frac{2 \pi t}{T}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\because$ Dimension of L.H.S = Dimension of R.H.S
Hence, the given formula is dimensionally correct
(b) Answer: 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) Answer: Incorrect
$y=\left(\frac{a}{T}\right) \sin \left(\frac{t}{a}\right)$
Dimension of $y=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
Dimension of $\frac{a}{T}=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}$
Dimension of $\frac{t}{-}=M^{0} L^{-1} T^{1}$
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) Answer: Correct
$y=(a \sqrt{2})\left(\sin 2 \pi \frac{t}{T}+\cos 2 \pi \frac{t}{T}\right)$
Dimension of $y=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
Dimension of $a=M^{0} L^{1} T^{0}$
Dimension of $\frac{t}{T}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
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.
(a) Answer: Correct
$y=a \sin \frac{2 \pi t}{T}$
Dimension of $y=M^{0} L^{1} T^{0}$
Dimension of $a=M^{0} L^{1} T^{0}$
Dimension of $\sin \frac{2 \pi t}{T}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\because$ Dimension of L.H.S = Dimension of R.H.S
Hence, the given formula is dimensionally correct
(b) Answer: 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) Answer: Incorrect
$y=\left(\frac{a}{T}\right) \sin \left(\frac{t}{a}\right)$
Dimension of $y=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
Dimension of $\frac{a}{T}=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}$
Dimension of $\frac{t}{-}=M^{0} L^{-1} T^{1}$
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) Answer: Correct
$y=(a \sqrt{2})\left(\sin 2 \pi \frac{t}{T}+\cos 2 \pi \frac{t}{T}\right)$
Dimension of $y=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
Dimension of $a=M^{0} L^{1} T^{0}$
Dimension of $\frac{t}{T}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
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.