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# Where did I go wrong on this pendulums question? watch

1. So I was doing this question and here is my working out

time period (P) is 86401/86400 . Normal period is 1. Therefore, we can infer that
EDIT: Picture doesn't seem to work but I calculate "L" by rearranging it and I get 0.253...... metres

So we now need to find to delta L
$P' = 2\pi \sqrt{\frac{L+ \Delta L }{g}} = \frac{86401}{86400} \\ \\ \\ L+ \Delta L = \frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} } \\ \\ \\ \therefore L=\frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} }-0.253 ... = 5.863...\textup{x}10^{-6}$
EDIT2: ^^^^ Delta L ^^^^
which when divided by that constant, alpha, does NOT lead to the right answer!

Any ideas?
2. Okay fine fine no one is coming forward ... how are you SUPPOSED to do it then? Maybe then we can see if I am wrong by comparison?
3. Tagged the most helpful guy I know ... plz help
4. (Original post by DrSebWilkes)
Tagged the most helpful guy I know ... plz help
It would be useful if you posted the preamble (assuming there is one) and the beginning of the question, since there is some theory that I'm not familiar with.

However, having plugged numbers into my calculator, I disagree with your numbers.

Typing by my calculator.

However, without the previous bits of the question, I wouldn't know whether or not this is right.
5. (Original post by DrSebWilkes)
Tagged the most helpful guy I know ... plz help
(Original post by K-Man_PhysCheM)
It would be useful if you posted the preamble (assuming there is one) and the beginning of the question, since there is some theory that I'm not familiar with.

However, having plugged numbers into my calculator, I disagree with your numbers.

Typing by my calculator.

However, without the previous bits of the question, I wouldn't know whether or not this is right.
Oh, and either way, we're looking for a temperature, not a length...? In which case, I don't see how your working has much relevance...
6. (Original post by K-Man_PhysCheM)
It would be useful if you posted the preamble (assuming there is one) and the beginning of the question, since there is some theory that I'm not familiar with.

However, having plugged numbers into my calculator, I disagree with your numbers.

Typing by my calculator.

However, without the previous bits of the question, I wouldn't know whether or not this is right.
There was, but it effectively just gave a more complicated version of the pendulum formula that isn't needed for this bit.

$\dpi{120} \textup{The pendulum Length for 24 hours:} \\ L=\frac{1 second\cdot10}{4\pi ^{2}}=0.253302.....$

So basically, what I did was find the "new" time period, and then found out the length of that new time period, and subtracted away from it the old "orignal" length; this delta L is divided by the constant in the question to give the maximum delta Temperature.
7. (Original post by DrSebWilkes)
So I was doing this question and here is my working out

time period (P) is 86401/86400 . Normal period is 1. Therefore, we can infer that
EDIT: Picture doesn't seem to work but I calculate "L" by rearranging it and I get 0.253...... metres

So we now need to find to delta L
$P' = 2\pi \sqrt{\frac{L+ \Delta L }{g}} = \frac{86401}{86400} \\ \\ \\ L+ \Delta L = \frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} } \\ \\ \\ \therefore L=\frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} }-0.253 ... = 5.863...\textup{x}10^{-6}$
EDIT2: ^^^^ Delta L ^^^^
which when divided by that constant, alpha, does NOT lead to the right answer!

Any ideas?
Okay, to actually answer the question now (part d).

Let be the time period, since will be used for temperature.

If the temperature is increased by , then the length will increase by , since we are told is the fractional increase in length.

For an expanded pendulum at a higher temperature:

Now,

After some algebra, you can write in terms of .

Now, for it to be accurate to within one second in one day, the ratio must be equal to the ratio of one second to one day.

This will then allow you to solve for using your previous result
8. (Original post by DrSebWilkes)
There was, but it effectively just gave a more complicated version of the pendulum formula that isn't needed for this bit.

$\dpi{120} \textup{The pendulum Length for 24 hours:} \\ L=\frac{1 second\cdot10}{4\pi ^{2}}=0.253302.....$

So basically, what I did was find the "new" time period, and then found out the length of that new time period, and subtracted away from it the old "orignal" length; this delta L is divided by the constant in the question to give the maximum delta Temperature.
Okay, well as you probably know, I don't like numbers. I've worked it through now (I think; I don't actually have the official answers!) and given the start of the workings in my previous post.
9. (Original post by K-Man_PhysCheM)
Okay, well as you probably know, I don't like numbers. I've worked it through now (I think; I don't actually have the official answers!) and given the start of the workings in my previous post.
Yeah no that's pretty similar to how the "mark-scheme" did it but this was a calculator paper so I used it.

I don't see anything wrong with how you do it, but I don't see how it differs either? You L+kL could just be equal to some constant "C".

so we have time period P which equals "stuff" * Root C and P which is stuff * Root L, and in both cases it's easy to re-arrange each function to find C (or L) in terms of P

And so if we have a value for both C and L, then the difference in length is just C-L ...

Tell me I'm not going crazy!
10. (Original post by DrSebWilkes)
Yeah no that's pretty similar to how the "mark-scheme" did it but this was a calculator paper so I used it.

I don't see anything wrong with how you do it, but I don't see how it differs either? You L+kL could just be equal to some constant "C".

so we have time period P which equals "stuff" * Root C and P which is stuff * Root L, and in both cases it's easy to re-arrange each function to find C (or L) in terms of P

And so if we have a value for both C and L, then the difference in length is just C-L ...

Tell me I'm not going crazy!
How do you get your values for C and L?
11. (Original post by K-Man_PhysCheM)
How do you get your values for C and L?
Well "L" is mentioned above, and for "C", exactly the same method but with P=86401/86400 (if thought your clock was right and counted 86400 counts in a day, it'd still be accurate to the required time)
12. (Original post by DrSebWilkes)
Well "L" is mentioned above, and for "C", exactly the same method but with P=86401/86400 (if thought your clock was right and counted 86400 counts in a day, it'd still be accurate to the required time)
Okay...

So your method for finding C is invalid, as far as I can see...
13. (Original post by K-Man_PhysCheM)
Okay...

So your method for finding C is invalid, as far as I can see...
But P = 1 so = P

And if that's what P is equal to then I should be able to find "C" just like I found L
14. (Original post by DrSebWilkes)
But P = 1 so = P

And if that's what P is equal to then I should be able to find "C" just like I found L
Sorry, I didn't clock it before, but why do you suppose P=1?

Pretty much every pendulum clock I've seen has a period of 2 seconds... (it ticks every time the bob swings to one side).
15. (Original post by K-Man_PhysCheM)
Sorry, I didn't clockit before, but why do you suppose P=1?

Pretty much every pendulum clock I've seen has a period of 2 seconds... (it ticks every time the bob swings to one side).
Hmm that's a remarkably good question . Well I said that in 24 hours there are 86400 seconds and for the purposes of this, there will be 86400 periods in a day; so if that's true, if the same number of periods were counted but we still want to be accurate, what's the maximum value of P that would allow this.

Still ... now you bring this to my attention I do see that your method seems to be less assumption-y which is good. I think the take home message, regardless, is that the fewer assumptions the better and in this case. Hmm I think, I guess, this was just going to be one of those questions that wasn't inherently compatible with me :|
16. (Original post by DrSebWilkes)
So I was doing this question and here is my working out

time period (P) is 86401/86400 . Normal period is 1. Therefore, we can infer that
EDIT: Picture doesn't seem to work but I calculate "L" by rearranging it and I get 0.253...... metres

So we now need to find to delta L
$P' = 2\pi \sqrt{\frac{L+ \Delta L }{g}} = \frac{86401}{86400} \\ \\ \\ L+ \Delta L = \frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} } \\ \\ \\ \therefore L=\frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} }-0.253 ... = 5.863...\textup{x}10^{-6}$
EDIT2: ^^^^ Delta L ^^^^
which when divided by that constant, alpha, does NOT lead to the right answer!

Any ideas?
I am not sure what have you done to get the "wrong answer". I can get the same answer based on your method and the method proposed by K-Man_PhysCheM.
You may want to check your algebra.
17. (Original post by Eimmanuel)
I am not sure what have you done to get the "wrong answer". I can get the same answer based on your method and the method proposed by K-Man_PhysCheM.
You may want to check your algebra.
Hmm? Well I'm glad on one level. So just to double check I am not totally crazy and there might be some method in my much-madness? I'll have another go tomorrow but thanks for letting me know (sorry away / late!)
18. (Original post by Eimmanuel)
I am not sure what have you done to get the "wrong answer". I can get the same answer based on your method and the method proposed by K-Man_PhysCheM.
You may want to check your algebra.
I forgot about this and now I just remembered! So I put into wolfram alpha, and it spat out the exact same answer as what I wrote above showing it's wrong. So what on earth am I doing wrong then?!
19. (Original post by DrSebWilkes)
So I was doing this question and here is my working out

time period (P`) is 86401/86400 . Normal period is 1. Therefore, we can infer that
EDIT: Picture doesn't seem to work but I calculate "L" by rearranging it and I get 0.253...... metres

So we now need to find to delta L
$P' = 2\pi \sqrt{\frac{L+ \Delta L }{g}} = \frac{86401}{86400} \\ \\ \\ L+ \Delta L = \frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} } \\ \\ \\ \therefore L=\frac{7465132801\cdot 10}{7464960000\cdot 4\pi^{2} }-0.253 ... = 5.863...\textup{x}10^{-6}$
EDIT2: ^^^^ Delta L ^^^^
which when divided by that constant, alpha, does NOT lead to the right answer!

Any ideas?
(Original post by DrSebWilkes)
I forgot about this and now I just remembered! So I put into wolfram alpha, and it spat out the exact same answer as what I wrote above showing it's wrong. So what on earth am I doing wrong then?!

As I mentioned I did not know what had you done to get the wrong answer.

Based on your method, L is already known ( and T = 1 s)

If you manipulate the above expression correctly, you will get

From this, you should get δT as 1.2 K.
20. (Original post by Eimmanuel)
<snip>
I think ... it's because I was effectively doing L+"x", and solving for x, rather than "L+kL" ... but never the less, I'm still somewhat puzzled why that wouldn't work.

Thanks for the help so far! (You have restored some confidence I had lost :O )

EDIT: I can confirm that if I solve it as "L+kL" I get the right answer ....

and when I divide my "x" by L, I get the right answer ... euughh ... so that means the "x" isn't an absolute change but a ratio? Well thanks for pointing me in the right direction.

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