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

Dimension of LHS = \[ M^0L^1T^0 \]
Dimension of RHS must be the same. Let Dimensions of B = \[ M^xL^yT^z \]
Now, Dimensions of RHS = \[ M^xL^yT^{z+2} \]
Compare the powers of M, L, T individually to find the values of x, y, z

Answer 2

Similarly for second part: LHS: \[M^0L^1T^0\] and for the sin function, it is \[M^0L^0T^0 \]
So, If A's dimensions are \[ M^xL^yT^z\] \[x+0 = 0 \Rightarrow x = 0\] \[y+0 = 1 \Rightarrow y = 1\] \[z+0 = 0 \Rightarrow z = 0\]

Answer 3

did this help?

Answer 4

it looks really complicated and well thought out, but i just started learning physics so im having trouble understanding what u just wrote..... :" (oh and it's actually supposed to be { x=A sin (2 pi ft), } instead of x=A sin (2?ft),

Answer 5

So basically if you have something that is a distance on one side of the equation, the other side of the equation must also have distance. Otherwise it doesn't make much sense. So M means mass, L means length, T means time. If you make sure that the result from the left hand side is of the same dimensions as that on the right hand side, it solves your question and makes things sane.

Answer 6

x is distance, which is a length so the dimensions of this are \[ M^0L^1T^0\] which means no mass, one power of length, no time.
\[t^2\] has a square of time. So its dimensions are \[M^0L^0T^2\] which means no length, no mass but two powers of time. I hope now you can understand the solution I provided. Else tell me which part is not clear.

Answer 7

then how about B? (for x=Bt^2.)

Answer 8

now LHS has \[ M^0L^1T^0 \] and RHS has B and \[ M^0L^0T^2 \] but we want RHS to be same as LHS. So, B must be such that when multiplied with the rest of RHS, it should give LHS.
\[M^0L^1T^0 = dim(B) M^0L^0T^2\] dim(B) will be of the type \[ M^xL^yT^z\] and must satisfy the equation. So, x + 0 = 0, y + 0 = 1, z + 2 = 0: this gives us x = 0, y = 1, z = -2

I hope the concept is clear.

Answer 9

i think im starting to slightly understand your explanation....thanks!!! :)

Answer 10

Or simply. x is in units of metres (m). The RHS is is something (B) times seconds squared \[s ^{2}\]. So the B must include a distance (m) multiplied by a quotient of time squared.\[\frac{ 1 }{ s ^{2} }\] So therefore B must be acceleration. \[ms ^{-2}\]