Hi,
Say I have a particle performing SHM between two points A and B
AMCB
Where M is the middle, would the time it takes to get from C to B = to the time it takes to get from B to C? Will this also apply to any two points you pick?
Thnaks

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 15012010 22:29

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 15012010 22:32
Yes.

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 15012010 22:32
(Original post by 2710)
Hi,
Say I have a particle performing SHM between two points A and B
AMCB
Where M is the middle, would the time it takes to get from C to B = to the time it takes to get from B to C? Will this also apply to any two points you pick?
ThnaksLast edited by Unbounded; 15012010 at 22:36. 
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 15012010 22:42
OK can you please tell me where I have gone wrong then:
APMQB
A to B is 8m
Period of Osc = 12s
P and Q are 3 and 6 m from A respectively.
Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.
x = A cos wt
For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:
therefore
Which is the wrong answer.
Any help appreciated 
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 15012010 22:54
(Original post by 2710)
OK can you please tell me where I have gone wrong then:
APMQB
A to B is 8m
Period of Osc = 12s
P and Q are 3 and 6 m from A respectively.
Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.
x = A cos wt
For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:
therefore
Which is the wrong answer.
Any help appreciated
coz x is the displacement from MLast edited by peaches6994; 15012010 at 22:56. 
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 15012010 22:58
(Original post by 2710)
OK can you please tell me where I have gone wrong then:
APMQB
A to B is 8m
Period of Osc = 12s
P and Q are 3 and 6 m from A respectively.
Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.
x = A cos wt
For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:
therefore
Which is the wrong answer.
Any help appreciated
then minus them. try that and see? 
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 15012010 23:00
No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre. The way you did it is the way the MS did, which is what I dont want to do. Besides, its 1.
So can anyone see the flaw in my method? 
peaches6994
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 15012010 23:10
are you sure? I think x is fixed, no matter what formula you use.

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 16012010 01:26
(Original post by 2710)
No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre. The way you did it is the way the MS did, which is what I dont want to do. Besides, its 1.
So can anyone see the flaw in my method?
Also Q is 2m away from M so I think x should be 2. so time taken to travel from Q to M ,2=4coswtLast edited by rbnphlp; 16012010 at 01:52. 
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 16012010 07:14
the answer should be 5.38 given you calculations are correct

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 16012010 07:17
(Original post by 2710)
OK can you please tell me where I have gone wrong then:
APMQB
A to B is 8m
Period of Osc = 12s
P and Q are 3 and 6 m from A respectively.
Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.
x = A cos wt
For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:
therefore
Which is the wrong answer.
Any help appreciated 
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 16012010 07:25
(Original post by rbnphlp)
are you familiar with the considering the SHm as a circular motion and finding the time , I find it much more logical and easy to follow
Also Q is 2m away from M so I think x should be 2. so time taken to travel from Q to M ,2=4coswt
OP .... the above is an important post and I would suggest that you consider the method rbnphlp suggests. 
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 16012010 10:33
(Original post by 2710)
No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre. 
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 16012010 12:30
Ah, so I tkae distance from the centre, BUT time is the one which varies right?
With the cos formula I take the time from the end. Ok, Let me try that.
(Original post by ananda2010)
the answer should be 5.38 given you calculations are correct 
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 16012010 13:00
First of all, thanks to everyone who helped. I appreciate your help with other solutions, but I just wanted to find the flaw in my working using my method. And yes, I am familiar with the circular motion thing, but I just wanted to use this method
OK, I've seem to have got my answer, but I'm a tad confused:
3m1m]2m2m
APMQB
But this is the weird part. I am taking my time from A. And I put into the formula x=1. The time it gives me is over 3 seconds (the time it takes to get to the middle), when it is clear that the time it takes to get to P should be less than 3. Ie, the formula takes negative distances to be on the other side of where you take your time to be. Is this meant to be? I always though negative meant (in this case) left of M.
Anyways:
Same here, ie, positive x =2 and it takes it to be on the left hand side of M, ie 2m from M
So basically, my formula flips my whole diagram right to left..kinda >__> which I found hard to get in my head.
Besides that 3.482 = 1.48, which is the correct answer.
I just need someone to confirm my thinking for x distances, I hope you understand what I mean
THANKS help appreciated
EDIT: Ah, I know an easier way to help you understand me.
For example, say the question asked me the time it took to get from A to P.
So I use this:
Which is obviously the wrong answer. I have to do 3.48  3, since it is on the wrong side. Do you understand what I'm getting at?Last edited by 2710; 16012010 at 13:06. 
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 16012010 15:20
Bump ? T__T

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 16012010 20:29
Anyone ? I really need to know this

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 16012010 20:50
The position of P will be 1, not 1, as it is 3 units away from A.
Edit: Whichever end you have chosen as the starting point (if you are using the cosine) determines the positive direction.Last edited by ghostwalker; 16012010 at 21:00. 
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 16012010 21:01
(Original post by ghostwalker)
Notice that using the cosine, at time t=0 the displacement is 4, which is the position of A relative to the centre. Note: it is positive.
The position of P will be 1, not 1, as it is 3 units away from A.
Edit: Whichever end you have chosen as the starting point (if you are using the cosine) determines the positive direction.
Ok let me think about this. So if I had to find A to Q, do I put x as negative 2?
Thanks
EDIT: Nevermind it works! Thanks ghostwalkerLast edited by 2710; 16012010 at 21:06.
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