# M3: SHMWatch

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#1
shouldn't where I've put a small red circle be the equilibrium position and not immediately below the natural length?
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2 months ago
#2
The center would be where the forces balance so the resulting force on the body is zero.
Probably need to see the question, but I'm sure you can sort it.
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#3
(Original post by mqb2766)
The center would be where the forces balance so the resulting force on the body is zero.
Probably need to see the question, but I'm sure you can sort it.
exactly, where I have put the red circle is where the forces are 0, or rather the equilibrium position but they've drawn the equilibrium position at the end of of the strings natural length. so, the diagram is wrong?
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#4
(Original post by mqb2766)
The center would be where the forces balance so the resulting force on the body is zero.
Probably need to see the question, but I'm sure you can sort it.
I've drawn an arrow to where I think the equilibrium point should be while the diagram has it where the natural length ends?
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2 months ago
#5
My reading is that the natural length is l, the extension to the equilibrium position is e, and the point shown is past that so the acceleration would be upwards?
(Original post by Maths&physics)
I've drawn an arrow to where I think the equilibrium point should be while the diagram has it where the natural length ends?
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#6
(Original post by mqb2766)
My reading is that the natural length is l, the extension to the equilibrium position is e, and the point shown is past that so the acceleration would be upwards?
I then e stands for extension. anyway, where ive drawn the arrow, is that where the equilibrium position should be?
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2 months ago
#7
I believe so. The question says it's a distance x below the equilibrium position and that is x below your red arrow.
(Original post by Maths&physics)
I then e stands for extension. anyway, where ive drawn the arrow, is that where the equilibrium position should be?
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#8
(Original post by mqb2766)
I believe so. The question says it's a distance x below the equilibrium position and that is x below your red arrow.
how could I do part d using the conservation of energy?
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2 months ago
#9
You could do, but you could also just write down the cos() term which represents the position using c) then differentiate. The answer should be a simple combination of amplitude and frequency.
(Original post by Maths&physics)
how could I do part d using the conservation of energy?
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#10
(Original post by mqb2766)
You could do, but you could also just write down the cos() term which represents the position using c) then differentiate. The answer should be a simple combination of amplitude and frequency.
sorry, I meant part e
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2 months ago
#11
Similar, find the time it returns to the position, then sub that value into the velocity sin() expression.
... Or do conservation of energy.
(Original post by Maths&physics)
sorry, I meant part e
Last edited by mqb2766; 2 months ago
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#12
(Original post by mqb2766)
Similar, find the time it returns to the position, then sub that value into the velocity sin() expression.
... Or do conservation of energy.
for a string: it's extension at the equilibrium point is x, but I pull it (x + 2) beyond that point, so its 2x + 2 beyond it's natural length, what would it's amplitude b?

same question for a spring?
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2 months ago
#13
For SHM, itsi relative to the equilibrium position. For a single spring on a horizontal table, the natural spring length is the Equilibrium position. For a single spring hanging down, the equilibrium pos is where gravity and the spring balance. The displacement and amplitude are measured with respect to the equilibrium.

Similar for a string, but be careful about the motion if it goes slack.
(Original post by Maths&physics)
for a string: it's extension at the equilibrium point is x, but I pull it (x + 2) beyond that point, so its 2x + 2 beyond it's natural length, what would it's amplitude b?

same question for a spring?
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#14
(Original post by mqb2766)
For SHM, itsi relative to the equilibrium position. For a single spring on a horizontal table, the natural spring length is the Equilibrium position. For a single spring hanging down, the equilibrium pos is where gravity and the spring balance. The displacement and amplitude are measured with respect to the equilibrium.

Similar for a string, but be careful about the motion if it goes slack.
horizontally: so however much we stretch the spring back, that is amplitude?

the strings amplitude would be natural length + extension if it's attached to a fix point (fixed point being centre of amp)

vertically: I understand spring and strings amp is when T = mg.

for this question d (attached): I calculated the velocity from A to when it comes to its natural length (which the mark scheme also did) and then I took the period from when the string becomes taut to k, to when it passes it's natural length again, which the mark scheme also did. however, the string becomes slack before then (when T = mg). however, I don't understand which values I should chose for amplitude and extension in this case.

mqb2766 it's this question, part d. I can see why they taken into account the extended point of equilibrium from the calculation - as the new place it will become slack? I just don't understand why they did it twice.

DFranklin this is the question here.
Last edited by Maths&physics; 2 months ago
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2 months ago
#15
I'm losing it a bit with all your different cases. I'm presuming that in each case you've identified the equilibrium position for all forces acting on the body and x is measured with respect to that point and the motion is SHM. Then
x = a*cos(w*t)
Where a is the amplitude and it may be a sin or some shift. The amplitude is the max displacement. If it's realease (zero velocity) then that will be the amplitude, but other initial conditions are possible. It's when the velocity is zero or force and acceleration are max magnitude.

Original post by Maths&physics)
horizontally: so however much we stretch the spring back, that is amplitude?

the strings amplitude would be natural length + extension if it's attached to a fix point (fixed point being centre of amp)

vertically: I understand spring and strings amp is when T = mg.

for this question d (attached): I calculated the velocity from A to when it comes to its natural length (which the mark scheme also did) and then I took the period from when the string becomes taught to k, to when it passes it's natural length again, which the mark scheme also did. however, the string becomes slack before then (when T = mg). however, I don't understand which values I should chose for amplitude and extension in this case.
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#16
(Original post by mqb2766)
I'm losing it a bit with all your different cases. I'm presuming that in each case you've identified the equilibrium position for all forces acting on the body and x is measured with respect to that point and the motion is SHM. Then
x = a*cos(w*t)
Where a is the amplitude and it may be a sin or some shift. The amplitude is the max displacement. If it's realease (zero velocity) then that will be the amplitude, but other initial conditions are possible. It's when the velocity is zero or force and acceleration are max magnitude.

Original post by Maths&physics)
is the string only taut when T > mg?
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2 months ago
#17
It's taut, T>0, when it exceeds it's natural length.
Your condition would break half the SHM cycle.
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#18
(Original post by mqb2766)
It's taut, T>0, when it exceeds it's natural length.
Your condition would break half the SHM cycle.
why in d have they excluded the time when the string passes equilibrium point on it's way back up, and they've multiplied it twice?
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2 months ago
#19
Do you have the quextion, other bits of the answer?
(Original post by Maths&physics)
why in d have they excluded the time when the string passes equilibrium point on it's way back up, and they've multiplied it twice?
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#20
(Original post by mqb2766)
Do you have the quextion, other bits of the answer?
here are parts a, b, and c.

isn't it intuitive that they've taken away the new extended point of equilibrium from the calculation? I just don't understand why they did it twice.
Last edited by Maths&physics; 2 months ago
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