Question

(a) suppose that the displacement of an object is related to the time according to the expression x=Bt^2. What are the dimensions of B? (b) A displacement is related to the time as x=A sin (2?ft), where A and f are constants. Find the dimensions of A. (hint: A trigonometric function appearing in an equation must be dimensionless.)

Answer 1

a.Dimension of B=
`[LT^(-2)]`

b.Dimension of A=
`[L]`

Explanation:

We are given that

a.Suppose that the displacement of an object is related to time according to the expression

`x=Bt^2`

We have to find the dimension of B

Dimension of time=T

Dimension of displacement =L

`B=(x)/(t^2)`

Substitute the value then we get

Dimension of B=
`(L)/(T^2)=[LT^(-2)]`

b.A displacement is related to the time as

x=A sin(2ft)

Where A and f are constants.

We have to find the dimensions of A.

We know that trigonometric function is dimensionless.

`A=(x)/(sin 2ft)`

Substitute the value then we get

Dimension of A=
`[L]`

Answer 2

A simple way to get the dimensions is just to rearrange the equation.

So in a.) (i'm going to assume t2 is t squared)
rearrange:
x = Bt2 ----> B= x/t2
Now you are told that x is displacement (L) and t is time (T) so sub these in.
B= L/T2 Therefore the dimensions are L/T2.


In b.) following the same steps:
x = A sin(2πft) ----> A = x/sin(2πft)
The hint tells you that sin(2πft) is dimensionless so you can disregard that part.
A = x
A=L