# Using Log base 10 to find gravity

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#1

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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4 years ago
#2
(Original post by Jyashi)

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 (square of time period) against (length), this will be a straight line.

The gradient will be from which it should be easy to find .
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#3
(Original post by Zacken)
Plot a graph of (square of time period) against (length), this will be a straight line.

The gradient will be from which it should be easy to find .
I have already done that once. But i need to use log base 10 to tackle this question.
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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:

So plot against , it should have gradient and then investigate the y-axis intercept.
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#5
(Original post by Zacken)
Fair enough. Take the logarithm of both sides:

So plot against , it should have gradient 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.
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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.
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#7
(Original post by Zacken)
Time is not negative, the logarithm of time is. And that's fine, nothing wrong with that.
But i will plot the negative values on the negative y and x axis right?
0
4 years ago
#8
(Original post by Jyashi)
But i will plot the negative values on the negative y and x axis right?
Yeps.
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#9
(Original post by Zacken)
Yeps.
Thanks ill plot this and post back.
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#10
(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
0
4 years ago
#11
(Original post by Jyashi)

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.
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#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 ?
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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
#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
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
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#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
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
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
#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
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
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