I need some help with pendulum experiment

Click For Summary

Homework Help Overview

The discussion revolves around a pendulum experiment and a heat transfer problem involving ice and water. The original poster describes conducting a pendulum experiment and seeks assistance with the relationship between the period of the pendulum and its height, as well as the final temperature of ice dropped into water.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation relating the period of the pendulum to height and explore how to rearrange it for graphing purposes. There are attempts to clarify how to derive values for gravitational acceleration and the constant k from the graph. The heat transfer question raises inquiries about the relationship between heat capacities, latent heat, and the equations needed to find the final temperature.

Discussion Status

Some participants have provided insights on how to rearrange the pendulum equation for graphing, suggesting a method to derive g and k from the slope and intercept of the graph. The heat transfer question remains less clear, with ongoing inquiries about the necessary equations and their interrelations.

Contextual Notes

The original poster expresses confusion about the rearrangement of equations and the relationships between variables in both problems, indicating a need for clarification on these concepts.

dan greig
Messages
41
Reaction score
0
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.
 
Physics news on Phys.org
The equation for a simple pendulum period is T = 2\pi \sqrt{\frac{h}{g}}.
 
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 kggives the gradient. k can be defined using the equation and g can be found by dividing the gradient by k
 
thanks, that's really helpfull. Any idea about the other question?
 
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?

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:
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.

-Dan
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
9
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
6K