Physics, asked by manojamway68, 8 months ago

Y=a sin bt
Find dimension of a and b where y=displacement and t=time

Answers

Answered by muneshwarpragya
1

Explanation:

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Answered by talasilavijaya
0

Answer:

a=[M^1L^0T^0] and b=M^0L^0T^{-1}

Explanation:

Given a function, Y=\mbox{a sin bt}

where y is the displacement and t is the time.

Displacement is measured in units of length, therefore,

the dimensions of displacement can be written as [M^1L^0T^0].

And the dimensions of time can be written as [M^0L^0T^1].

Given \mbox{ sin bt} which is nothing but a trigonometric ratio, \theta has no dimensions and therefore bt has no dimensions.

so, bt can be written as

bt=[M^0L^0T^0]

\implies b=\dfrac{[M^0L^0T^0]}{t}

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

Since \mbox{ bt} has no dimensions, therefore y=a

y is the displacement and has the dimensions [M^1L^0T^0].

So, the dimensions of a can be written as  [M^1L^0T^0].

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