In equation y-x-2 cos-22 pidfrac-beta y-alpha- - the of x-alpha-beta are m-s-1- and -ms-1-1- respectively- then of y and y- are m- -ms-1- m- -ms-2- m-2- m m-2- -ms-2-

Question

In equation  y-x-2 cos-22 pidfrac-beta y-alpha- - the of  x-alpha-beta  are  m-s-1- and  -ms-1-1-  respectively- then of y and y- are m- -ms-1-  m- -ms-2-   m-2- m  m-2- -ms-2-

Toppr Admin · Accepted Answer

Since argument of cos has to be dimensionless, thus $| 2 π β y^{'} / α | = M^{0} L^{0} T^{0}$
Thus, $| y^{'} | = \frac{| α |}{| β |}$ ( $π$ is dimensionless)
Thus, $| y^{'} | = \frac{s^{- 1}}{(m s^{- 1})^{- 1}} = m s^{- 2}$
Now, since cos is dimensionless, thus $| y | = | x^{2} | = m^{2}$

In equation y=x2cos22πβy′α, if the units of x,α,β are m,s−1and (ms−1)−1 respectively, then units of y and y' arem,ms−1m,ms−2 m2,mm2,ms−2

Since argument of cos has to be dimensionless, thus |2πβy′/α|=M0L0T0Thus, |y′|=|α||β| (π is dimensionless)Thus, |y′|=s−1(ms−1)−1=ms−2Now, since cos is dimensionless, thus |y|=|x2|=m2

In equation $y = x^{2} c o s^{2} 2 π \frac{β y^{'}}{α}$ , if the units of $x, α, β$ are $m, s^{- 1}$ and $({m s}^{- 1})^{- 1}$ respectively, then units of y and y' are
$m, {m s}^{- 1}$
$m, {m s}^{- 2}$
$m^{2}, m$
$m^{2}, {m s}^{- 2}$

Since argument of cos has to be dimensionless, thus $| 2 π β y^{'} / α | = M^{0} L^{0} T^{0}$
Thus, $| y^{'} | = \frac{| α |}{| β |}$ ( $π$ is dimensionless)
Thus, $| y^{'} | = \frac{s^{- 1}}{(m s^{- 1})^{- 1}} = m s^{- 2}$
Now, since cos is dimensionless, thus $| y | = | x^{2} | = m^{2}$