how would you go about this question and why?

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 23022018 13:59

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 23022018 14:49
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tiny hobbit
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 23022018 15:00
(Original post by Maths&physics)
how would you go about this question and why?
In this question the strings are identical and so the equilibrium position is at the centre of AB. 
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 23022018 15:30
(Original post by tiny hobbit)
You need to consider when the particle is a distance x from the equilibrium position. Use F = ma and substitute Hooke's law into it for each string. This should simplify to (x double dot) = kx where k should turn out to be positive. You have then proved that SHM is taking place and should say so.
In this question the strings are identical and so the equilibrium position is at the centre of AB.
how can T  4g = ma
and also T  4g = T? 
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 23022018 15:53
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says 4 x 9. underneath.
They are in the process of subtracting 4g from each side.
I don't think that it is set out very well.
I would complete the T  4g = m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards. 
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 23022018 16:09
(Original post by tiny hobbit)
The first is true.
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says 4 x 9. underneath.
They are in the process of subtracting 4g from each side.
I don't think that it is set out very well.
I would complete the T  4g = m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards.
its quite confusing!Last edited by Maths&physics; 23022018 at 16:20. 
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 23022018 16:31
(Original post by tiny hobbit)
The first is true.
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says 4 x 9. underneath.
They are in the process of subtracting 4g from each side.
I don't think that it is set out very well.
I would complete the T  4g = m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards.
leading to x(double dot) =  (omega^2)x ???
why is this the case? why should force/tension/acceleration have to be opposite to displacement?
can you please explain it because I dont get it? thanksLast edited by Maths&physics; 23022018 at 16:43. 
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 23022018 16:46
(Original post by Maths&physics)
so, to prove SHM in this case: force/tension/acceleration has to be opposite to displacement?
leading to x(double dot) =  (omega^2)x ???
why is this the case? why should force/tension/acceleration has to be opposite to displacement?
can you please explain it because I dont get it? thanks
If u differentiate twice wrt to t u will get an equation for acceleration.
You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.
The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM.Posted on the TSR App. Download from Apple or Google Play 
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 23022018 16:56
(Original post by Shaanv)
Are u familiar with the equation x=Acos(wt).
If u differentiate twice wrt to t u will get an equation for acceleration.
You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.
The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM. 
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 23022018 17:18
(Original post by Shaanv)
Are u familiar with the equation x=Acos(wt).
If u differentiate twice wrt to t u will get an equation for acceleration.
You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.
The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM.
and the second one (part b, ii), where he is finding the period but multiples it by the period found earlier?
thanks 
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 23022018 18:10
(Original post by Maths&physics)
whats the difference between the first period (part b), where he uses T = 2pi/w
and the second one (part b, ii), where he is finding the period but multiples it by the period found earlier?
thanks 
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 23022018 23:08
(Original post by tiny hobbit)
At the top of the motion in the second version, the string goes slack. What you shown is the proportion of the SHM oscillation that takes place. Now you need to calculate the time for the top part of the motion, when the particle just moves under gravity, using suvat with a=9.8 
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 24022018 10:51
(Original post by Maths&physics)
in the second screenshot, hes using a different equation to work out the period. what equation is that, whats the difference and whats the algebraic form? thanks
Do you get a running commentary with these videos that explain what is going on? 
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 27022018 15:37
(Original post by tiny hobbit)
He's finding what proportion of the whole period occurs when the string is stretched. The circle business is an alternative to using either x = a cos wt or x = a sin wt.
Do you get a running commentary with these videos that explain what is going on?Last edited by Maths&physics; 27022018 at 15:39. 
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 27022018 16:06
(Original post by Maths&physics)
whats the algebraic formula for that method?
yes, but he doesn't explain everything and he didn't explain that.
It's not one I use, I go with the x=asin/cos wt 
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 27022018 16:16
(Original post by tiny hobbit)
I suppose that it is something like (angle at centre of circle for the part that's required)/360 x period
It's not one I use, I go with the x=asin/cos wt
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