Results 1 to 5 of 5

Math Help - Dimensional Analysis

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    131

    Dimensional Analysis

    I have got two final questions that are really stumping me. I cant figure how to do these. Any help would be great..!
    Attached Thumbnails Attached Thumbnails Dimensional Analysis-q4bc.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    a)

    Quote Originally Posted by moolimanj View Post
    I have got two final questions that are really stumping me. I cant figure how to do these. Any help would be great..!
    Write out an equation using only the units and what you want to solve for.
    Note: Since the webber constant is "unitless" we can write its units as 1.

    remember that \rho=\frac{m}{V} \\\ V=m^3

    So the equation looks like.

    1=\frac{\frac{m}{V}\left( \frac{m}{s}\right)^2m}{\sigma}

    Solving for sigma and multiplying the numerator we get

    \sigma=\frac{m^4}{Vs^2}=\frac{m^4}{m^3s^2}=\frac{m  }{s^2}

    So sigma must have units distance over time squared.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2007
    Posts
    131
    Thanks - sort of makes sense. Can you help with the second part as well?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2007
    Posts
    131
    For part (c) I get the following (can someone check this please:

    v-ms^-1

    p=Nm^-2

    N= gms^-2 (i.e. Force = mass x acceleration)

    so p = gms^2/m^2 = g/s^2m

    V=m^3

    Density (theta) = g/m^3

    so

    p/density = g/ms^2/g/m^3 which simplifies to m/s^2

    Hence,

    V is proportional to square root of pressure/Density

    Does this sound right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    Posts
    61
    moolimanj, that is indeed correct. The relation you derived indicates that the speed of sound is independent of the volume as can be expected.

    There is a more general theory on this method of working. You were given the dependent parameters, but in more complicated cases you don't know which are important and this is were the theorem of Buckingham comes in the picture. It is a very important theorem in applied physics for looking at parameters which form dimensionless groups that define the problem completely. If you want to read more on them look at the following link.

    Buckingham ? theorem - Wikipedia, the free encyclopedia

    best regards, Coomast
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dimensional Analysis
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 7th 2011, 06:54 PM
  2. dimensional analysis help?
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: August 20th 2010, 09:37 PM
  3. Dimensional analysis check
    Posted in the Algebra Forum
    Replies: 0
    Last Post: January 7th 2010, 07:20 AM
  4. Dimensional Analysis check
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: December 9th 2009, 01:36 AM
  5. Dimensional Analysis
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: December 6th 2007, 07:53 PM

Search Tags


/mathhelpforum @mathhelpforum