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HNC electrical Engineering Maths Dimensional analysis

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.

My answer:

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.
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.

My answer:

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.
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.

Thanks i will reply with an answer
IMG_0085.jpg

This is my working out
Original post by shoneyman2001
IMG_0085.jpg

This is my working out

Can you please tell me if i am wrong or where i am going wrong
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 m0m^0 leaving T=klgT=k\sqrt{\frac{l}{g}} where k is a constant.
Thank you for all your help
Original post by Plücker
Your formula is correct but I would leave out m0m^0 leaving T=klgT=k\sqrt{\frac{l}{g}} where k is a constant.

Yes i understand it now, so no need for m^0
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.

Screenshot 2021-09-21 080904.png
Reply 9
When you write
B + C = 0
you should have got
B = -C = ...
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?
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?
(edited 2 years ago)
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
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;

Screenshot 2021-09-21 080904.png
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;

Screenshot 2021-09-21 080904.png

L^0 = 1
and not sure why the T goes on the denominator.
You seem to be making some basic errors.
(edited 2 years ago)
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?
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.
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....

Screenshot 2021-09-21 080904.png
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....

Screenshot 2021-09-21 080904.png

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.
(edited 2 years ago)

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