Jyashi
Badges: 2
Rep:
?
#1
Report Thread starter 4 years ago
#1
Name:  Screenshot_2016-03-01-18-33-03.jpg
Views: 296
Size:  207.1 KB

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.
0
reply
Zacken
Badges: 22
Rep:
?
#2
Report 4 years ago
#2
(Original post by Jyashi)
Name:  Screenshot_2016-03-01-18-33-03.jpg
Views: 296
Size:  207.1 KB

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.
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.
0
reply
Jyashi
Badges: 2
Rep:
?
#3
Report Thread starter 4 years ago
#3
(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.
0
reply
Zacken
Badges: 22
Rep:
?
#4
Report 4 years ago
#4
(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.
0
reply
Jyashi
Badges: 2
Rep:
?
#5
Report Thread starter 4 years ago
#5
(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.
0
reply
Zacken
Badges: 22
Rep:
?
#6
Report 4 years ago
#6
(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:
0
reply
Jyashi
Badges: 2
Rep:
?
#7
Report Thread starter 4 years ago
#7
(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?
0
reply
Zacken
Badges: 22
Rep:
?
#8
Report 4 years ago
#8
(Original post by Jyashi)
But i will plot the negative values on the negative y and x axis right?
Yeps.
0
reply
Jyashi
Badges: 2
Rep:
?
#9
Report Thread starter 4 years ago
#9
(Original post by Zacken)
Yeps.
Thanks ill plot this and post back.
0
reply
Jyashi
Badges: 2
Rep:
?
#10
Report Thread starter 4 years ago
#10
(Original post by Zacken)
Yeps.
Name:  Screenshot_2016-03-01-20-02-11.jpg
Views: 215
Size:  501.9 KB

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
0
reply
Zacken
Badges: 22
Rep:
?
#11
Report 4 years ago
#11
(Original post by Jyashi)
Name:  Screenshot_2016-03-01-20-02-11.jpg
Views: 215
Size:  501.9 KB

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
The intercept looks negative, no? And the intecept requires log (2pi/rt(g)). So you'll need to take the exponential.
0
reply
Jyashi
Badges: 2
Rep:
?
#12
Report Thread starter 4 years ago
#12
(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 ?
0
reply
Zacken
Badges: 22
Rep:
?
#13
Report 4 years ago
#13
(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.
1
reply
Jyashi
Badges: 2
Rep:
?
#14
Report Thread starter 4 years ago
#14
(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.
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.
0
reply
Student403
Badges: 19
Rep:
?
#15
Report 4 years ago
#15
(Original post by Jyashi)
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.
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
0
reply
Jyashi
Badges: 2
Rep:
?
#16
Report Thread starter 4 years ago
#16
(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.
0
reply
Student403
Badges: 19
Rep:
?
#17
Report 4 years ago
#17
Ignore my above post


Your intercept of 0.302 is correct. Zacken watch out 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.
1
reply
Zacken
Badges: 22
Rep:
?
#18
Report 4 years ago
#18
(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!!
0
reply
Jyashi
Badges: 2
Rep:
?
#19
Report Thread starter 4 years ago
#19
(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
0
reply
Student403
Badges: 19
Rep:
?
#20
Report 4 years ago
#20
(Original post by Zacken)
Bloody hell, literally just realised I was using the wrong base!! Argh, too much pure maths for me. Thank you!!
All good I was stumped for 5 minutes as well xD
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (560)
34.06%
Yes, I like the idea of applying to uni after I received my grades (PQA) (677)
41.18%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (332)
20.19%
I think there is a better option than the ones suggested (let us know in the thread!) (75)
4.56%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise