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Homework Help: I need some help with pendulum experiment

  1. Feb 23, 2006 #1
    I need some help!!

    i have two things i'm stuck on and i really can't get my head around them.

    i have done a pendulum experiment using a 1m rule hung over a point using an inverted v shaped piece of string, i have calculated the period and have been given this equation to relate period T to perpendicular height h,

    T^2.h=4pi^2/g(h^2 + k^2)

    k is a constant and g is acceleration of free fall.

    i need to draw a suitable graph to find g and k

    The other question is,

    ice of temp -10 celsius is dropped into a glass of water at temp 20 cesius, what is the temperature eventually?

    i have the respective heat capacities and the latent heat of ice, i also know i have to use Q=mc delta t from -10 to 0. i think i have to use Q=ml after that but i don't know how they relate and if i have to goback to the original equation after 0 celsius?

    please please help i just don't seem to be getting anywhere, thanks, Dan.
  2. jcsd
  3. Feb 23, 2006 #2


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    The equation for a simple pendulum period is [tex] T = 2\pi \sqrt{\frac{h}{g}} [/tex].
  4. Feb 23, 2006 #3


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    You want to re-arrange this so that you've got something like
    [tex] T^2 = g\cdot k \cdot h [/tex]
    This will allow you to plot the square of the time period against height, where [itex] kg [/itex]gives the gradient. [itex]k[/itex] can be defined using the equation and [itex]g[/itex] can be found by dividing the gradient by [itex]k[/itex]
  5. Feb 23, 2006 #4
    thanks, that's really helpfull. Any idea about the other question?
  6. Feb 23, 2006 #5
    no matter how hard i try i can't rearrange the given equation to make
    T^2 = g.k.h although it makes sense and works out correctly!

    Also how would i find k using a T^2/h graph?
  7. Feb 23, 2006 #6
    You've got data for T and h and need to find a way to graph to get g and k. Here's some creative graphing logic for you. You've got data for T and h, so they have to be your x and y variables somehow. So solve for h^2:
    [tex]h^2=\frac{4 \pi^2}{g} * \frac{1}{T^2} - k^2 [/tex]
    How does this help? Set your y variable equal to h^2 and your x variable to 1/T^2. That means whatever slope you get from your linear regression you can equate to [tex]4 \pi^2/g[/tex]. Likewise the intercept you will get you can equate to [tex]-k^2[/tex].

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