anyone studied physics and done the pendulum test?

[donniemateno]

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 a stupid topic to post I am just aware a lot of people here are and have studied and wondering if anyone might think piece of piss and lend some help/advice

cheers in advance
____________________
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[weasley]

Rather than google the answer, it is a simple arithmetic rearrangement so have a go. This is maths, not physics - you must have done basic algebra to do mechanics?

Write out the equation in full, then simply do a step-by-step series of operations to isolate g. For example, to start with you know you want to get the "2.pi" from the right to the left, so divide both sides by 2.pi. Keep on going until g is on its own.

By the way, if you're interested, that equation is an approximation of the period of a pendulum, which deviates from the truth more as the amplitude of the swing increases. This is for interest only and not pertinent to your equation solution.
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[donniemateno]

my mechanics diploma involved no maths! or english really.... just take this off show me well done put it back together. ah well done you can change brakes discs and pads now!

so from what you said would it be ?

G= squareroot f L*T/2*pi
____________________
Current bike: Honda CBR 600 F3
Previous bikes: Honda XR 125L5 , BMW 800 ST, Honda CB500

[MattJ]

Just done it in my head quickly so not had time to write it down but I think it's

G = l/(t/2*pi)^2

[donniemateno]

thanks matt thats the same as an equation above but I dont understand where the squareroots disppears! will have to do some reading
____________________
Current bike: Honda CBR 600 F3
Previous bikes: Honda XR 125L5 , BMW 800 ST, Honda CB500

[Lord Percy]

Hello, I did this test quite recently! (Or a version of it)

If you want to calculate g:

Set up the pendulum in any form you like.
Measure the length of the string/cable/rope/whatever.
Raise the pendulum with your arm, ready to start it swinging. Don't raise it too high - it needs to swing smoothly.
Let go of the pendulum
Watch it do one full oscillation (a full journey from side to side to it ends up back where it started)
After it has completed the first oscillation, start a timer.
Count ten more oscillations with the timer running. (I repeat this is ten FULL cycles of it going from side to side),
left, right, 1
left, right, 2
left, right, 3
left, right, 4
left, right, 5
left, right, 6
left, right, 7
left, right, 8
left, right, 9
left, right, 10

Like that.

Stop the timer immediately after the tenth oscillation.
Record the time.
Divide that time by 10, to get the time per single oscillation.

Repeat that several times, then work out the average of all your results. This is all for accuracy.

So now you will have the following information.

T = (calculated using the process given above) - in seconds
g = ?
pi = pi
l = (length - measured with as much accuracy as possible) - in metres (eg 0.017m)

So, you're told that T = 2pisqrt(l/g)

Here's how to make g the subject:

T = 2pisqrt(l/g)
T/2pi=sqrt(l/g)
To get rid of the square root, you just square both sides. i.e. multiply them with themselves. It's a balanced equation so both terms will still be equal if you do that. (Consider: sqrt4=2 If you multiply both sides with themselves, you're left with 4=4)
(T^2)/(4pi^2)=l/g
g=l/(T^2)/(4pi^2)

g=4pi^2L/T^2

Now you can plug your measurements into the formula to find g.

For better accuracy, try repeating the experiment with different lengths.

And... Mechanics much more interesting? I BEG TO DIFFER! You just wait till I can quantum mechanic you to kingdom come, nyah-haaah!

[MattJ]

As he said, once you get the square root of l/g alone, you square both sides. This gets rid of the square root so it's just l/g but the other side is now squared as a whole.

[stinkwheel]

G and g are two different things. It would be worth remembering that, do not mix them up. If they use g, do not put G. G is the gravitational constant. g is acceleration due to gravity.

In this case though, if you are solving the pendulum equation with regard to g in metres per second per second, you are actually calculating G in newtons per kilogram. You will be docked marks if you don't use the correct units.

Frankly I'm stunned you've got to final year at university in an engineering degree unable to do basic algebra. What you've asked is nothing to do with either the pendulum experiment or physics but about how to manipulate an equation.

Ok. So if my rusty 17 year old algebra skills still serve.
https://cdn.bikechatforums.com/files/codecogseqn.gif
So square both sides because nobody likes a square root.
https://cdn.bikechatforums.com/files/codecogseqn_1_.gif

Get rid of the fraction because I fucking hate fractions. Multiply both sides by g:
https://cdn.bikechatforums.com/files/codecogseqn_2_.gif

But put in a fraction because you need to know g. Divide both sides by T squared:
https://cdn.bikechatforums.com/files/codecogseqn_3_.gif

Someone should check that though because I can't maths but this is GCSE level maths.
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[donniemateno]

wow saw thanks for going to all that trouble with the post. it has broken it down for me well

I think in the experiment i have to do i have to use 3 different weights and 3 different lengths and see how this affects the results

im still not sure how you end up with the final equation using 2L? i can understand it with the 2 pi becoming 4 pi.

im guessing as you have done 2pi * 2 which is 4 pi 1L becomes 2L?

and quantom mechanics esh....ill pawn ur ass at....changing a tyre?
____________________
Current bike: Honda CBR 600 F3
Previous bikes: Honda XR 125L5 , BMW 800 ST, Honda CB500

[MattJ]

The reason I've put G in my equation not g is because I wrote the post on my phone just after I had got up and it auto capitalised it.

[donniemateno]

im shit at algebra full stop

normally before an exam ill revise like this. so when i go in ill remember the method uses to adjust the equation to my needs.

i wont know why or how....that kind sir is how i got to the final year..
____________________
Current bike: Honda CBR 600 F3
Previous bikes: Honda XR 125L5 , BMW 800 ST, Honda CB500

[MattJ]

I think your way is correct Stinkwheel but I think it could be simpler than you've done it. You squared both sides straight away, which I didn't do. I shifted the 2pi over to the other side first so the square root of l/g was alone and then squared both sides. That way the left hand side has brackets around all of it but only one squared symbol.

Then I just swapped the entire left hand side (in brackets) with g to isolate g. Looks nicer in my opinion as there is only one squared symbole and you havent had to alter anything (e.g. you squared 2 to make 4 because you squared things individually rather than squaring the whole lot.

Happy for someone to show me why I'm wrong but I think that's right.

[el_oso]

this is really a maths problem, not a physics one. pretty simple algebra , stinkwheel with only his GCSE maths got it correct.
If you are in your third year at uni and can't do something like this in what I'm guessing is a fairly technical discipline, then how the hell did you pass your other years? unless of course you just had a brain fart (we all do)
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[Lord Percy]

I tested the formula on the dangly light switch in my toilet.

I timed 10 oscillations on my phone to get an average for T, estimated the length of the cord, plugged the numbers into the formula and got g as 10.7 or something. I then realised, since g is actually known, that I could rearrange the formula to make 'l' the subject and get an accurate measurement of the length of my toilet light switch cord WITHOUT ACTUALLY MEASURING IT! 0.78m, thank you science

[Lord Percy]

[stinkwheel]

[treeno]

Maybe this report I did for a module earlier this year will help?

https://www.filedropper.com/pendulum