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Question: Answered & Verified by Expert
A book with many printing errors contains four different formulas for the displacement $y$ of a particle undergoing a certain periodic motion:
(a) $y=a \sin \frac{2 \pi t}{T}$
(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.
PhysicsUnits and Dimensions
Solution:
1822 Upvotes Verified Answer
According to dimensional analysis an equation must be dimensionally homogeneous.
(a) $y=a \sin \frac{2 \pi t}{T}$
Here, $[$ L.H.S. $]=[y]=[$ L $]$
and $[$ R.H.S. $]=\left[a \sin \frac{2 \pi t}{T}\right]$
$$
=\left[\mathrm{L} \sin \frac{T}{T}\right]=[\mathrm{L}]
$$
So, it is correct.
(b) $y=a \sin v \mathrm{t} \quad$ Here, $[y]=[\mathrm{L}]$
and $[a \sin v \mathrm{t}]=\left[\mathrm{L} \sin \left(\mathrm{LT}^{-1} \cdot \mathrm{T}\right)\right]$
$=[\mathrm{L} \sin \mathrm{L}]$
So, the equation is wrong.
(c) $y=\left(\frac{a}{T}\right) \sin \frac{t}{a} \quad$ Here, [y] $=[$ L]
and $\left[\left(\frac{a}{\mathrm{~T}}\right) \sin \frac{t}{a}\right]=\left[\frac{\mathrm{L}}{\mathrm{T}} \sin \frac{\mathrm{T}}{\mathrm{L}}\right]$
$$
=\left[\mathrm{LT}^{-1} \sin \mathrm{TL}^{-1}\right]
$$
So, the equation is wrong.
(d) $y=(a \sqrt{2})\left(\sin \frac{2 \pi t}{\mathrm{~T}}+\cos \frac{2 \pi t}{\mathrm{~T}}\right)$
Here, $[y]=[\mathrm{L}],[a \sqrt{2}]=[\mathrm{L}]$
$$
\text { and }\left[\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right]=\left[\sin \frac{T}{T}+\cos \frac{T}{T}\right]
$$
$$
\because[\text { LHS }]=[\text { RHS }]
$$
$=$ dimensionless
So, the equation is correct.

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