I need some help with pendulum experiment

  • Thread starter dan greig
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  • #1
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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.
 

Answers and Replies

  • #2
Hootenanny
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The equation for a simple pendulum period is [tex] T = 2\pi \sqrt{\frac{h}{g}} [/tex].
 
  • #3
Hootenanny
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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]
 
  • #4
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thanks, that's really helpfull. Any idea about the other question?
 
  • #5
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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?
 
  • #6
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dan greig said:
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.

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].

-Dan
 

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