M3: S.H.M

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)?

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.

That looks about right, assuming t denotes tension (rather than time). It’s reasonable to assume no energy loss, so P will return to B. Also (1) and (3) should be the same, by symmetry.

Thanks. Can you write that out for me please? Thanks

mg -T = m (X double dot)

This is almost the same as you had, just different order for the first two terms. As you found, the constant bits cancel out when you use Hooke's Law and this time you will come out with a minus sign as needed for SHM.

cool, so the direction for x double dot (acceleration) is based on the direction of x? but I thought acceleration was always in direction of the equilibrium?

Yes, the acceleration is towards the equilibrium position, i.e. in the opposite direction to the displacement. That's why there is a minus sign in the equation.

ah, ok . what if we're moving towards the equilibrium?

If you are where x is positive, the acceleration is towards the equilibrium position so x double dot is negative.

If you are where x is negative, the acceleration towards the equilibrium position is in the positive direction, so x and x double dot still have opposite signs.

when would x be negative? it wouldn't be negative in the example in the question given?

x is the displacement from the equilibrium position, i.e. E in this example. As we have chosen to take down as positive, as soon as the particle goes above E, x is negative.

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.

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

so, I understand these 2 equations: $x = a.sin(wt)$ also $x = a.cos(wt)$

but what is the third? is it $x = a.sin(wt + a)$ or $x = a.sin(wt + pi/2)$. I think its the first but can you explain the conditions for me please. thanks

You should choose whichever sin or cos option matches your starting conditions. The particle in this question starts its SHM with displacement at its maximum value, and I think you settled on downwards as the positive direction, in which case x = a.cos(wt) is the one to use.

im asking in general and not about the above question. however, what is the third equation and what are its condition - like is what x and t both = 0.

You would use the third equation in situations where the SHM is part way through a cycle at t = 0.

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?