Using Log base 10 to find gravity

How can i use log base 10 to plot a graph and find the gravity value for this question.

I am thinking take the log of both sides and plot the log (T) on y axis and log (L) on the x axis and using the gradient to find value of gravity. But taking the log of T gives me a negative number so i was thinking of taking log of T^2. I dunno for sure.

Hoping some of you people who are more comfortable with logs would be able to explain better.

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Reply 1

Zacken

Original post by Jyashi

Plot a graph of

T^2

(square of time period) against

\ell

(length), this will be a straight line.

The gradient will be

\frac{4\pi^2}{g}

from which it should be easy to find

g

Reply 2

Jyashi

Original post by Zacken

Plot a graph of

T^2

(square of time period) against

\ell

(length), this will be a straight line.

The gradient will be

\frac{4\pi^2}{g}

from which it should be easy to find

g

I have already done that once. But i need to use log base 10 to tackle this question.

Reply 3

Zacken

Original post by Jyashi

I have already done that once. But i need to use log base 10 to tackle this question.

Fair enough. Take the logarithm of both sides:

\log T = \log 2\pi + \frac{1}{2} \log \frac{\ell}{g} \Rightarrow \log T = \frac{1}{2} \log \ell + \log \frac{2\pi}{\sqrt{g}}

So plot

\log T

against

\log \ell

, it should have gradient

\frac{1}{2}

and then investigate the y-axis intercept.

Reply 4

Jyashi

Original post by Zacken

Fair enough. Take the logarithm of both sides:

\log T = \log 2\pi + \frac{1}{2} \log \frac{\ell}{g} \Rightarrow \log T = \frac{1}{2} \log \ell + \log \frac{2\pi}{\sqrt{g}}

So plot

\log T

against

\log \ell

, it should have gradient

\frac{1}{2}

and then investigate the y-axis intercept.

Yes but when i take log T i get a negative number. Is it ok for time to have a negative number? The same thing happens with log L. So i was thinking of doing log (T/2pi). Dunno if it will work.

Reply 5

Zacken

Original post by Jyashi

Yes but when i take log T i get a negative number. Is it ok for time to have a negative number? The same thing happens with log L. So i was thinking of doing log (T/2pi). Dunno if it will work.

Time is not negative, the logarithm of time is. And that's fine, nothing wrong with that. :dontknow:

Reply 6

Jyashi

Original post by Zacken

Time is not negative, the logarithm of time is. And that's fine, nothing wrong with that. :dontknow:

But i will plot the negative values on the negative y and x axis right?

Reply 7

Zacken

Original post by Jyashi

But i will plot the negative values on the negative y and x axis right?

Yeps.

Reply 8

Jyashi

Original post by Zacken

Yeps.

Thanks ill plot this and post back.

Reply 9

Jyashi

Original post by Zacken

Yeps.

Ok so i get this graph where the intercept looks like 0.302.

I see that 1÷2 is the gradient and the intercept represents the 2pi/sq g. So does this mean that i should reformulate g = sq (2pi/0.302)?

I see that 1÷2

Reply 10

Zacken

Original post by Jyashi

The intercept looks negative, no? And the intecept requires log (2pi/rt(g)). So you'll need to take the exponential.

Reply 11

Jyashi

Original post by Zacken

The intercept looks negative, no? And the intecept requires log (2pi/rt(g)). So you'll need to take the exponential.

Ah yes its -0.2 but i just read the equation of line and shut my brain after that lol.

So i take e^-0.2 = 2pi/sq g ?

Reply 12

Zacken

Original post by Jyashi

Ah yes its -0.2 but i just read the equation of line and shut my brain after that lol.

So i take e^-0.2 = 2pi/sq g ?

Yep! Although something has gonewrong somewhere because that's a horrible estimate of g, I'm not sure what though. Hopefully someone whose brain isn't completely fried will jump in. :redface:

Reply 13

Jyashi

Original post by Zacken

Yep! Although something has gonewrong somewhere because that's a horrible estimate of g, I'm not sure what though. Hopefully someone whose brain isn't completely fried will jump in. :redface:

Something has gone wrong somewhere. I see grahically the intercept as negative but when i feed in the x and y values to work out c algebraically the intercept always comes out to 0.3

Hopefully an advanced alien civilization can recover this forum thousands of years in the future and find out the answer to this paradox which was clearly the limit of human intelligence.

Reply 14

Student403

Original post by Jyashi

The intercept should be 0.3. 0.3 is correct. Check you have not plotted your data using the incorrect column

Ensure you are using the correct units

Reply 15

Jyashi

Original post by Student403

The intercept should be 0.3. 0.3 is correct. Check you have not plotted your data using the incorrect column

Ensure you are using the correct units

I cant see an error. I have plotted log (L) on x axis and log (T) on the y axis and they are correct to the right decimal points.
This is why we need the aliens.

Reply 16

Student403

Ignore my above post

Your intercept of 0.302 is correct. @Zacken watch out :smile:

Intercept of -0.2 is where x = approx (-1), not where x = 0

See the x axis labels

Anyway the second reason you were not getting the correct value of g is because of the base.

Now that you have 0.302 as your intercept, you need to be taking 10 to the 0.302, not e to the 0.302

So with 10^0,302 = 2pi/sqrtg

You can solve to get g correct to 2 S.F.

Reply 17

Zacken

Original post by Student403

Now that you have 0.302 as your intercept, you need to be taking 10 to the 0.302, not e to the 0.302

So with 10^0,302 = 2pi/sqrtg

You can solve to get g correct to 2 S.F.

Bloody hell, literally just realised I was using the wrong base!! Argh, too much pure maths for me. Thank you!! :biggrin:

Reply 18

Jyashi

Original post by Jyashi

I cant see an error. I have plotted log (L) on x axis and log (T) on the y axis and they are correct to the right decimal points.
This is why we need the aliens.

Got it the graph is also showing the right intercept. Hint: look at the x axis