# HNC electrical Engineering Maths Dimensional analysis

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#1
I'm after a few tips I'm a mature student 47 doing my HNC/HND my question is:
You are testing a decorative clock, to possibly be manufactured by your Consumer
Electronics division, and attach a mass (m) to a string of length (𝑙𝑙) to form a simple
pendulum. Assuming that the acceleration due to gravity (g) of the earth may have an
influence on the period (t) of the pendulum swing, use dimensional analysis to find a
formula for t which could possibly involve m, 𝑙𝑙, and g.

t=constant x M(A) l(B) g(C)

Please could you give me some examples I'm not after the answer i just wat to understand it..and any help on writing answers on here would be great as I'm very rusty with maths but getting there.
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6 months ago
#2
(Original post by shoneyman2001)
I'm after a few tips I'm a mature student 47 doing my HNC/HND my question is:
You are testing a decorative clock, to possibly be manufactured by your Consumer
Electronics division, and attach a mass (m) to a string of length (𝑙𝑙) to form a simple
pendulum. Assuming that the acceleration due to gravity (g) of the earth may have an
influence on the period (t) of the pendulum swing, use dimensional analysis to find a
formula for t which could possibly involve m, 𝑙𝑙, and g.

t=constant x M(A) l(B) g(C)

Please could you give me some examples I'm not after the answer i just wat to understand it..and any help on writing answers on here would be great as I'm very rusty with maths but getting there.
Put the units in your formula. Do the units on the right simplify to a unit of time? They need to.
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#3
(Original post by Plücker)
Put the units in your formula. Do the units on the right simplify to a unit of time? They need to.
1
#4

This is my working out
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#5
(Original post by shoneyman2001)

This is my working out
Can you please tell me if i am wrong or where i am going wrong
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6 months ago
#6
(Original post by shoneyman2001)
Can you please tell me if i am wrong or where i am going wrong
Your formula is correct but I would leave out leaving where k is a constant.
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#7
Thank you for all your help
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#8
Yes i understand it now, so no need for m^0
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4 weeks ago
#9
Hi All,

So I'm having some difficulty with this Question. I know I'm incorrect and instead of just looking through the answers on the web I thought I would ask you guys at the bottom of this thread. I'm probably doing something really stupid but I'd rather know.

Can someone identify the stupid for me? as I said I know it's wrong, and I could just take an answer but I would be cheating myself.

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4 weeks ago
#10
When you write
B + C = 0
you should have got
B = -C = ...
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4 weeks ago
#11
(Original post by mqb2766)
When you write
B + C = 0
you should have got
B = -C = ...
I still think I'm being daft here. can you elaborate please?
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4 weeks ago
#12
(Original post by $Tiaan82$)
I still think I'm being daft here. can you elaborate please?
Not much to be honest.

You know C (well you solve for it on the following line) and you have written
[L]: B+C = 0
So that means its easy to solve for B. What do you think B is?
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#13
(Original post by mqb2766)
Not much to be honest.

You know C (well you solve for it on the following line) and you have written
[L]: B+C = 0
So that means its easy to solve for B. What do you think B is?
I've got B as 1/2 and C as -1/2. Therefore B-C=0
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4 weeks ago
#14
So I've been thinking about this and I believe I know a little more; I believe the problem may be down to the L^B+C confusing me... Where I'm treating L as a single component when it's actually two (L length and LT acceleration) so I end up with the following;

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4 weeks ago
#15
(Original post by $Tiaan82$)
So I've been thinking about this and I believe I know a little more; I believe the problem may be down to the L^B+C confusing me... Where I'm treating L as a single component when it's actually two (L length and LT acceleration) so I end up with the following;

L^0 = 1
and not sure why the T goes on the denominator.
You seem to be making some basic errors.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#16
(Original post by mqb2766)
L^0 = 1
You seem to be making some basic errors.
I agree, that is why I've asked for help. L^0=1 yes, but is that my problem?
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4 weeks ago
#17
(Original post by $Tiaan82$)
I agree, that is why I've asked for help. L^0=1 yes, but is that my problem?
Youve solved for A, B and C, so you've answered the question.
Not sure what problem you mean.
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4 weeks ago
#18
(Original post by mqb2766)
Youve solved for A, B and C, so you've answered the question.
Not sure what problem you mean.
I solved A B & C but it was after that is where I got confused; I was adding L^1/2+(-1/2) which was adding to 0 and giving me L^0 and my equation was just t=t....

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4 weeks ago
#19
(Original post by $Tiaan82$)
I solved A B & C but it was after that is where I got confused; I was adding L^1/2+(-1/2) which was adding to 0 and giving me L^0 and my equation was just t=t....

Which means its dimensionally correct. If the two sides were different, I'd be worried.

Edit - note here you're comparing SI units in that equation - they have to match. When you set up the equations for A, B and C, you're constraining the problem to say time = time. However, the basic problem is that you don't have time on the right hand side, rather you have mass, length and acceleration. The values of A, B and C tell you how the mass, length and acceleration "mix" together to give time. In this case, no mass is necessary (A=0) and you take the 1/sqrt(acceleration) to get time (C=-1/2), but you then have to then cancel the sqrt(length) on the denominator, so B=1/2.
Last edited by mqb2766; 4 weeks ago
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