I need some help with pendulum experiment

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?

Hootenanny
Staff Emeritus
Gold Member
The equation for a simple pendulum period is $$T = 2\pi \sqrt{\frac{h}{g}}$$.

Hootenanny
Staff Emeritus
Gold Member
You want to re-arrange this so that you've got something like
$$T^2 = g\cdot k \cdot h$$
This will allow you to plot the square of the time period against height, where $kg$gives the gradient. $k$ can be defined using the equation and $g$ can be found by dividing the gradient by $k$

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?

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?

$$h^2=\frac{4 \pi^2}{g} * \frac{1}{T^2} - k^2$$
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 $$4 \pi^2/g$$. Likewise the intercept you will get you can equate to $$-k^2$$.