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Asuppose 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.)

Verified Answer :

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

lublana

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=\([L]\)

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