M3: SHM

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.

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?

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?

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?

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

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?

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 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.

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)

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?

Do you have the quextion, other bits of the answer?