# M3: S.H.MWatch

#1
in trying to do a: ive chosen a point (x) which I will use to prove SHM. its below the natural length and below E. at this point, would tension be facing up towards A below x, and down towards B above x?
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1 year ago
#2
(Original post by Maths&physics)
in trying to do a: ive chosen a point (x) which I will use to prove SHM. its below the natural length and below E. at this point, would tension be facing up towards A below x, and down towards B above x?
I don’t understand your references to “below x” and “above x” but for SHM centred on E, you would expect the net force on P to act downwards when P is above E and upwards when P is below E. Tension will always be acting upwards as we are told the string does not go slack.
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#3
(Original post by old_engineer)
I don’t understand your references to “below x” and “above x” but for SHM centred on E, you would expect the net force on P to act downwards when P is above E and upwards when P is below E. Tension will always be acting upwards as we are told the string does not go slack.
brilliant. thats makes sense but how do I prove SHM?
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1 year ago
#4
(Original post by Maths&physics)
brilliant. thats makes sense but how do I prove SHM?
You have to show that the net force acting on P, and hence its acceleration, is proportional to and opposite in sign to its displacement from the centre of oscillation.
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#5
(Original post by old_engineer)
You have to show that the net force acting on P, and hence its acceleration, is proportional to and opposite in sign to its displacement from the centre of oscillation.
because its been pulled down 0.4 meters below the point of equilibrium (A), does that mean it will go 0.4 above A?
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#6
(Original post by old_engineer)
You have to show that the net force acting on P, and hence its acceleration, is proportional to and opposite in sign to its displacement from the centre of oscillation.
so, ive done point x below the point of equilibrium, with acceleration always acting towards the point of equilibrium?

with the following equation:

therefore:

but does the particle go 0.4 above point of equilibrium at first?
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1 year ago
#7
(Original post by Maths&physics)
because its been pulled down 0.4 meters below the point of equilibrium (A), does that mean it will go 0.4 above A?
The point of equilibrium is E not A but you are right that, for SHM, the motion is symmetrical either side of E. However, the twist with this question is that the upward excursion of P will be big enough for the string to go slack, at which point SHM ceases and P becomes a projectile.
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#8
(Original post by old_engineer)
The point of equilibrium is E not A but you are right that, for SHM, the motion is symmetrical either side of E. However, the twist with this question is that the upward excursion of P will be big enough for the string to go slack, at which point SHM ceases and P becomes a projectile.
sorry, I meant point E. cool. did you check my working in the following post?
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1 year ago
#9
(Original post by Maths&physics)
sorry, I meant point E. cool. did you check my working in the following post?
Your value of w matches the period defined in part (a) so I assume the preceding working is correct.
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#10
(Original post by old_engineer)
Your value of w matches the period defined in part (a) so I assume the preceding working is correct.
according to the mark scheme, its mg - T , not T - mg (like me). I think thats because the particle has been realised?

mg will always be acting down, and because its beyond its natural length, T will always be acting up. so, im not sure why mg is positive and T negative in the mark scheme?
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1 year ago
#11
(Original post by Maths&physics)
so, im not sure why mg is positive and T negative in the mark scheme?
I would guess they have taken downwards as the positive direction. You can equally take upwards as the positive direction as long as you working is clear and consistent on this point.
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#12
(Original post by old_engineer)
Your value of w matches the period defined in part (a) so I assume the preceding working is correct.
and for part b, to work out max acceleration: which equation would we use?
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#13
(Original post by old_engineer)
I would guess they have taken downwards as the positive direction. You can equally take upwards as the positive direction as long as you working is clear and consistent on this point.
but isn't acceleration always towards the point of equilibrium?
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1 year ago
#14
(Original post by Maths&physics)
but isn't acceleration always towards the point of equilibrium?
Yes, but if you take downwards as positive then for a positive (downwards) displacement, the net force acting on P will be negative (upwards), towards the centre of SHM.
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1 year ago
#15
(Original post by Maths&physics)
and for part b, to work out max acceleration: which equation would we use?
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#16
(Original post by old_engineer)
because w = 7, you would expect x to be as small as possible for the greatest value (least negative), but in the mark scheme: w is taken as positive and therefore is is as large as possible (0.4) in this case.
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1 year ago
#17
(Original post by Maths&physics)
because w = 7, you would expect x to be as small as possible for the greatest value (least negative), but in the mark scheme: w is taken as positive and therefore is is as large as possible (0.4) in this case.
The question asks for the greatest MAGNITUDE of acceleration, which will be w^2 multiplied by the greatest magnitude of displacement.
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#18
(Original post by old_engineer)
The question asks for the greatest MAGNITUDE of acceleration, which will be w^2 multiplied by the greatest magnitude of displacement.
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#19
(Original post by old_engineer)
The question asks for the greatest MAGNITUDE of acceleration, which will be w^2 multiplied by the greatest magnitude of displacement.
for part d:

1)time from B to its natural length (where t = 0)

2)then the time it takes traveling up from where t = 0, and then down again until string under tension again.

3)and then, time from string becoming under tension again, until string back to b again (will it even return to b again)?
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1 year ago
#20
(Original post by Maths&physics)
so, ive done point x below the point of equilibrium, with acceleration always acting towards the point of equilibrium?

with the following equation:

therefore:

but does the particle go 0.4 above point of equilibrium at first?
(Original post by old_engineer)
I would guess they have taken downwards as the positive direction. You can equally take upwards as the positive direction as long as you working is clear and consistent on this point.
x is increasing downwards so x double dot is downwards. You must therefore use sum of forces = m x double dot downwards so that the minus sign emerges from your working.
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