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Homework Help: Question about Dimensions (very basic)

  1. Aug 25, 2008 #1
    1) Which of the equations below is dimension-
    ally correct?
    a. y = (2 m) cos(k x), k = 2 m-1
    b. v = v0 + a x

    my work for:
    a) cos(kx) = 2ym-1
    So this would lead to k = (1/x)cos-1(2ym-1)

    But i dont know whether the COSINE operation on RHS changes m-1 to any other unit or not. So, please help me with that.

    b) in this case i thought that the presence of two unknown quantities with unknown units on the RHS would mean that the dimension of "v" is correct or not. So i would say it cannot be determined

    2) The volume of an object is given as a func-
    tion of time by V = A+(B/t)+Ct4. Determine
    the dimension of the constant C.
    1. L2/T4
    2. L4/T3
    3. L3/T4
    4. L/T
    5. L/T4

    My work:

    V = A+(B/t)+Ct4
    so, V=A+(B+Ct5/t)
    or, C= {[(V-A)t]-B}/t5

    Which i thought would give the dimension for C as L/T4. But i am not sure at all so please help.
    Last edited: Aug 25, 2008
  2. jcsd
  3. Aug 25, 2008 #2


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    Homework Helper

    Sorry I don't understand your use of m in this equation. Are you intending that m is some unit like mass?

    Again what is the units of a? Are you meaning acceleration here? Is V units of velocity? If a is supposed to be x/t2 and v in units of x/t then no the equation is not consistent as the last term is apparently x2/t2 and not x/t.
    Here doesn't each term have to be in units of volume? Doesn't that require the last term Ct4 to be in units of L3?
  4. Aug 25, 2008 #3
    see that is the question. i posted them as is, without ANY modification and thats the same thing i thght. what does "a" and other variables mean and what are their units. nothing is given.
  5. Aug 25, 2008 #4


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    Homework Helper

    That may be true for question 1. But your answer to question 2 is apparently figured incorrectly.
  6. Aug 25, 2008 #5
    apparently the rite answer for the second question was 3 which is L3/T4. apparently the first one is the correct one in the first question. well thanks for helpin out
  7. Aug 26, 2008 #6
    My guess is that in 1a, the use of "m" is intended to be meters. The use of m^-1 in the given value of k seems to indicate this.

    I'd like to point out something that you may have forgotten from trig. Sine/Cosine, etc are dimensionless functions. They have no dimensions. Remember their definitions from triangles.

    Sine(theta) = opposite side / hypotenuse
    Cos(theta) = adjacent side / hypotenuse

    In any ratio like this the numerator (top) units will cancel the denominator (bottom) units. So it doesn't matter if the triangle sides will be measured in feet, meters, inches, thumb widths, etc. The units will always cancel and the answer (a ratio) will be the same for all units.

    So you can ignore the cosine in 1a and use just what's out front. BUT...You DO have to take into account if the quantity inside the () of the cosine come out to radians. Sometimes the convention is not to explicitely type out radians in the units because radians is itself defined as a ratio. So I would check that "kx" comes out with no units in order to be correct for inside the cosine.
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