M3: S.H.M

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.

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.

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?

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?

Your value of w matches the period defined in part (a) so I assume the preceding working is correct.

so, im not sure why mg is positive and T negative in the mark scheme?

Your value of w matches the period defined in part (a) so I assume the preceding working is correct.

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?

and for part b, to work out max acceleration: which equation would we use?

You already have (x doubledot) = -w^2.x at your disposal.

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.

The question asks for the greatest MAGNITUDE of acceleration, which will be w^2 multiplied by the greatest magnitude of displacement.

so, ive done point x below the point of equilibrium, with acceleration always acting towards the point of equilibrium?

with the following equation:

$T - mg = 98(0.2 + x) - 2(9.8)$

$T - mg = 98x$

$F = 98x$

$ma = 98x$

$2(x double dot) = 98x$

$(x double dot) = 49x$

$(x double dot) = -(w^2)x$

therefore: $w = 7$

but does the particle go 0.4 above point of equilibrium at first?

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.