Question

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

Answer 1

Since all the equations are for the displacements of the particle, they all should have the dimension of the length.

Now, checking all the equations of displacement to be dimensionally correct.

$\left(a\right)$

Since the tignometric functions are dimensionless, $\frac{2\pi t}{T}$ will also be dimensionless.

$\frac{2\pi t}{T}=\frac{T}{T}=1=\left[M^0{L}^0{T}^0\right]$

Here, $a$ is the amplitude and it will have the dimension of the length. SO, the equation is correct.

$\left(b\right)$

Since the tignometric functions are dimensionless, $vt$ should be dimensionless.

$vt=\left(L{T}^{-1}\right)\left(T\right)=L=\left[M^0{L}^1{T}^0\right]$

Thus, the given equation is not correct.

$\left(c\right)$

The quantity $\frac{t}{a}$ should be dimensionless

$\frac{t}{a}=\frac{T}{L}=\left[M^0{L}^{-1}{T}^1\right]$

Therefore, it is also dimensionally incorrect.

$\left(d\right)$

The tignometric functions are dimensionless, $\frac{2\pi t}{T}$ will also be dimensionless.

$\frac{2\pi t}{T}=\frac{T}{T}=1=\left[M^0{L}^0{T}^0\right]$

The given equation is dimensionally correct.

Thus, the formulas in $\left(b\right)\&\left(c\right)$ are dimensionally wrong.