Question

Force f is given in terms of time t and distance x by F = A sin CT + B cos d x .then dimensions of A/B and C / D are given by

Answer 1

Answer:

A/B= no dimensions, C/D= LT^(-1)

Explanation:

obviously...Ct & Dx indicate the angles

Ct = radians

Dx = radians

C = radians/sec

D = radians/metre

since sinCt & sinDx are simply numericals hence A & B indicate Force components

A = Newtons

B = Newtons

Now,

C/D = [radions/sec]/radions/metre]= metre/sec =LT^(-1)

A/B = Newtons/Newtons is a ratio and no dimensions

Answer 2

Answer:

Explanation:

Given the force, $F = A sin CT + B cos D x$

where T is the time and x is the distance.

The given equation have the terms with trigonometric functions, i.e., sin and cos, which are dimensionless.

Therefore, $[CT] =[M^0L^0T^0]$

$\implies [CT^1] ={M^0L^0T^0}$

Hence, the dimensions of C are

$[C] =\dfrac{{M^0L^0T^0}}{T^1} =[M^0L^0T^{-1}]$

Similarly, $[Dx] =[M^0L^0T^0]$

$\implies [DL^1] ={M^0L^0T^0}$

Hence, the dimensions of D are

$[D] =\frac{{M^0L^0T^0}}{L^1} =[M^0L^{-1}T^0]$

Thus,

$\dfrac{C}{D} =\dfrac{[C]}{[D]}$

$=\dfrac{[M^0L^0T^{-1}]}{[M^0L^{-1}T^0]}={[M^0L^1T^{-1}]}$

Now for A/B, since the terms are under addition, applying the principle of homogeneity, dimensions of A and B must be same as force F.

Therefore, since the dimensional formula of force, F is

$[F]}={[M^1L^1T^{-2}]}$

Therefore, $[A]=[M^1L^1T^{-2}]~\&~[B]=[M^1L^1T^{-2}]$

And hence,

$\dfrac{A}{B} =\dfrac{[A]}{[B]}$

$=\dfrac{[M^1L^1T^{-2}]}{[M^1L^1T^{-2}]}={[M^0L^0T^0]}$

Therefore,  $\dfrac{A}{B} ={[M^0L^0T^0]}$ and  $\dfrac{C}{D} =[M^0L^1T^{-1}]$