The force f is given in terms of time t and displacement X by the equation F= Asin Bx +C sin dt. What is the dimension of D/B?

Here's what I think,

A_Trigonometric_ratio (whatever)

The general form of sin/cos/tan(angle).

So here as in the general form the "whatever" is always an "angle" input which (may sound a bit unfair but) is a dimensionless quantity.

As mentioned in your question,

F= Asin(Bt) + C cos (Dy)

Dimentions of these terms will be same (by homogenity of dimensions).

So what we gotta do here is make Bt and Dy dimensionless for this whole equation to work.

According to your question t is time,

So to make Bt dimensionless,

B must be equal to inverse of time i.e T^(-1)

Using the same process on Dy, gives inverse of length i.e L^(-1)

So,

B= M^(0)L^(0)T^(-1)

D = M^(0)L^(-1)T^(0)

The dimension of D/B then will be M^(0)L^(-1)T^(1)

it is the second option.

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