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 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

Answer 2

Answer:

a.Dimension of B=$[LT^{-2}]$

b.Dimension of A=$[L]$

Step-by-step 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

$B=\frac{x}{t^2}$

Substitute the value then we get

Dimension of B=$\frac{L}{T^2}=[LT^{-2}]$

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.

$A=\frac{x}{sin 2ft}$

Substitute the value then we get

Dimension of A=$[L]$