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A pendulum clock in X city is 2.5 minutes behind actual time. Another clock with same mass and length is 2.5 minutes ahead of actual time in Y city. Gravity is exactly 9.8m/s[sup]2[/sup] in Y city. Determine the gravity in X city.

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The formulas for this problem are found under Simple Harmonic Motion, Simple Pendulum and Torsion pendulum.

They're the same gravity, but one clock is set back and one ahead, or so it seems.

I hope I don't have to do this stuff

uh.... 9.8m/s2?

I dislike these plug and chug type of equations with all those extra variables tied to them.

Either this.

Maybe...

14.8m/s[sup]2[/sup]

or...

4.8m/s[sup]2[/sup]

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Just guesstimates. I am only a completely non-physics student.

How long has each clock been running?

That was my answer, because of no such variable being presented to us, we could not determine the differences in periods of the pendulums.

I know that by Calculus I could've determined an answer like X m/s[sup]2[/sup] per each second that the pendulum has been running. Possibly using an integral I could've compiled the formulas and then get the primitive formula that will tell me my infered answer.

Anyone think that such a thing is possible via Calculus? I only had about 30 minutes to do that problem, as I've done the other 9 in about 25 minutes and it would've taken me too much time to do it.

thats the freefall accerlartion of earth as if it was in a vaccum

how is the y varible effecting the answer. there is no definate relaionship other that it is time and they are both effected by gravity.

and ne way, correct me if im wrong, but earths gravity is the same where ever you are on the ground at 0 sea level elevation.

disclaimer : im not in a calcalus class or physics so dont get mad if im wrong lol

oops... forgot the squared was acceleration =_=

The earth is an elliptical shpere, therefore, gravity is different on the equator than the poles. Altitude affects it the same way as the earth not being a perfect shpere does.
The farther away you are from earth's central gravitational point, the less force is exerted on you. The diffrence is so small that you won't really notice it until are 30k+ feet above sea level.

Answer: 1.8 m/s^2

How I got it:

The equation I used for this was: L = T^2/(4PI^2)*g

T is the period of the pendulum swing, g is gravity, and L is the length of the pendulum. Since the pendulums in both cities are the same length, it doesn't matter what the exact length is.

So one clock is 150 seconds (2.5 minutes x 60 seconds) ahead of time and the other one is 150 seconds behind time. A normal pendulum's Period as we all know is 1 second per cycle. To make a clock run faster, you need to make the Period less than 1 second since the pendulum has to swing faster, and to make a clock run slower, you need to make the Period greater than 1 second since the pendulum has to swing slower.

Now here's the part where I dunno how to explain, but it's just how I got the answer. You divide 60 by 150 and you get 0.4. So for city Y's pendulum's T, it would be 1 - 0.4 = .6 sec. For the other one, it would be 1 + 0.4 = 1.4 sec.

For the pendulum in city Y, here's the equation:

L = (0.6^2)/(4PI^2)*(9.8)
L = 0.08937

Now for city X equation:

0.08937 = (1.4^2)/(4PI^2)*g
g = 1.8 m/s^2


gg

Well.. simple logic tells me that's wrong.

I haven't actually done these kind of problems but 1.8 seems too low (or high if it's positive ;O) for gravity if gravity is 9.8 m/s^2 at one section of the planet.

Your simple logic will change once you've done problems like this

I'm taking AP physics B, so unless I got wrong information, then I'm pretty sure I should be correct since that equation came out of my textbook.

T = 2pi (L/g)^(1/2)
T[sup]2[/sup]/4pi[sup]2[/sup] = L/g

T[sup]2[/sup]g/4pi[sup]2[/sup] = L

The equation is correct, but I think that you have implemented it wrong. The approach you used had logic, but if you analize all your variables, you wil notice that something doesn't seem quite right.

The period of the clock may be 1 second per cycle, but your approach is correct only if the periods of the pendulum's are 0.6 and 1.4. The problem that none of us have been able to explain is that the periods could be 0.97 and 1.05 (Example values) and the gravity would be around 9.78m/s[sup]2[/sup] (Example value).

Your answer is correct, but if and only if the periods are really the values which you gave them.

TOO MUCH ADVANCED MATH!

holy fack...i hope i never have to read that again... u sound so smart

SEN is where geniuses come together

Yea, that was my biggest problem.

Beer, is this a real question or something for school? Or was it just made up?

You could have said T = 2pi sqrt(L/g)
Easier to read in my opinion.
Whatever, I dont like math that much. Ask bob, he got a B in calc at nyu, but he's in high school.

It was in a Physics test that I took on Friday. Teacher still hasn't given us test results...



I got A in a wierd Calculus that my teacher is giving. It's a full course of Calculus I with some Calculus II and Diferential Equations mixed in.

Right now, I can do anything related to derivatives, differentials, simple differential equations, most integrals and some integrals by changing variables.