The position of a particle moving under uniform accelerationis some function of time and the acceleration. Suppose we writethis position as: x=ka^mt^n where k is adimensionless constant. Show by dimensional analysis that thisexpression is satisfied if m=1 and n=2. Thus bydimensional analysis this expression issatisfied.

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adelekeademolu adelekeademolu · Answer 1 · 2020-02-09T02:02:49+01:00

Answer:

For [tex]m=1[/tex] and [tex]n=2[/tex], the expression is satisfied dimensionally.

Explanation:

[tex]x=ka^mt^n[/tex]

[tex]x[/tex] is a position and has a dimension of length, [tex]L[/tex].

[tex]a[/tex] is acceleration which has dimensions of [tex]LT^{-2}[/tex].

[tex]t[/tex] is time with dimension of [tex]T[/tex].

Since [tex]k[/tex] is dimensionless, we do not factor it into the dimensional equation as below:

[tex]L = (LT^{-2})^mT^n[/tex]

Expanding the first term on the right hand side,

[tex]L = L^mT^{-2m}T^n[/tex]

Applying the laws of indices,

[tex]L = L^mT^{-2m+n}[/tex]

The index of each fundamental dimension must be equal on both sides.

For [tex]L[/tex],

[tex]1=m[/tex]

For [tex]T[/tex],

[tex]0=-2m+n[/tex]

But [tex]1=m[/tex]

[tex]0=-2\times1+n[/tex]

[tex]0=-2+n[/tex]

[tex]2=n[/tex]

Thus, the equation is dimensionally satisfied for the given values of [tex]m=1[/tex] and [tex]n=2[/tex].