Physics Experiment: Calculating G from T

In summary: T^2?(you mean g = 4pi2*l / T2 :wink:)In summary, the equation for T is T=2*Pi*sqrt(l/g), where T2 is the result of squaring T and then re-arranging the terms. To work out G, first square T and then divide by 2 pi.
  • #1
donniemateno
45
0
morning ladies n gents

just wondering has anyone studied this physics experiment before? I have to do it as part of my last year at uni as I didnt do physics at college ( chose mechanics much more interesting!)

anyway its apparently been the same exam for the last 4/5 years but the lecturers give you no help in advance and just give you this equation:

T = 2 * Pi * sqrt(l / g)

T = period (in seconds)
Pi = 3.14...
sqrt = root of ...
l = length of pendulum in m
g = acceleration in m/s^2 (9.81)

I am wondering if anyone has done this experiment but working out G rather than t? if so what would the formula be rearranged? I have found 2 through some googling :

1. g = l/(t/tpi)^2
2. g = 4pi^2L/T^2

Sorry if this is in the wrong section please move if it is
 
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  • #2
good morning, donniemateno! :wink:
donniemateno said:
T = 2 * Pi * sqrt(l / g)

square this equation, then re-arrange it …

what do you get? :smile:
 
  • #3
Quote by donniemateno
T = 2 * Pi * sqrt(l / g)
square this equation, then re-arrange it …
what do you get?

g = sqrt (2*pi*l )/ T?
 
  • #4
hi donniemateno! :smile:

why aren't there more squares there? :confused:
 
  • #5
g = sqrt (2*pi*l )/ T^2

I think

Im really not very good at rearranging formulas

I understand the concept of taking one from one side to the other through division but get lost when sqreroots are involved
 
  • #6
a square roots becomes a ^2 so does that mean:

g =(2*pi*l)^2/ T^2
 
  • #7
ok, then do it one step at a time :wink:

first, square T = 2 *π* √(l / g) without any rearranging …

what do you get? :smile:
 
  • #8
first, square T = 2 *π* √(l / g) without any rearranging

I'm going to say

T^2=2*pi*(l/g)?
 
  • #9
T2 correct :smile:

(l/g) correct :smile:
 
  • #10
is the next step to divide by 2 pi? if so would that be :

T^2 / 2*pi = (l/g)?
 
  • #11
whoa!

go back … your 2 *π was not correct …

you needed to square it! :wink:
 
  • #12
would pi become 4 pi?
 
  • #13
yup! :smile:

T = 2 *π* √(l / g)

becomes

T2 = 4 *π2* (l / g) :wink:

(you see how you have to square each term separately?)

ok, now get g on the left and T2 on the right :smile:
 
  • #14
so next if i divide by g to get

T2 / g = (4 *π2* (l / g)) / g

then divide by t^2 to move it over giving me

g = (4pi*l)/T^2
 
  • #15
donniemateno said:
so next if i divide by g to get

T2 / g = (4 *π2* (l / g)) / g

divide by g?? :confused:
 
  • #16
would it become

t^2*g = 4pi^2*l

then

g = 4pi*l / T^2?
 
  • #17
(you mean g = 4pi2*l / T2 :wink:)

yes! :smile:
 

What is the purpose of calculating g from T in a physics experiment?

The purpose of calculating g from T is to determine the acceleration due to gravity, which is a fundamental constant in physics. It is necessary in many experiments and can also help verify the accuracy of theoretical models.

What equipment is needed to perform this experiment?

To calculate g from T, you will need a pendulum, a stopwatch or timer, a measuring device (such as a ruler or protractor), and a calculator. A stable support (such as a table or stand) and a string or wire to suspend the pendulum are also necessary.

What are the steps to perform this experiment?

The steps to perform this experiment are as follows:

  1. Set up the pendulum by suspending it from a stable support with a string or wire.
  2. Measure the length of the pendulum (from the point of suspension to the center of mass).
  3. Release the pendulum and use a stopwatch or timer to measure the time it takes for one full swing (from one side to the other and back).
  4. Repeat this process multiple times and calculate the average time for one full swing.
  5. Use the equation g = 4π²L/T², where g is the acceleration due to gravity, L is the length of the pendulum, and T is the average time for one swing, to calculate g.

What are some sources of error in this experiment?

Some sources of error in this experiment include air resistance, the accuracy of the timing device, and the precision of the length measurement. Other factors such as temperature, humidity, and the angle of release can also affect the accuracy of the results.

How can the results of this experiment be used in real-world applications?

The results of this experiment can be used in various real-world applications, such as engineering and construction, navigation, and space exploration. Knowing the acceleration due to gravity is crucial in designing structures and vehicles to withstand specific forces and in accurately determining locations and trajectories. It is also essential in understanding the behavior of objects in space and on other planets.

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