A-levels Simple Harmonic Motion Help

In the sinusoidal shm calculation, x is the displacement from the equilibrium which occurs at a height of 1.15.
So when x<0.55, the height, h, from the floor is >0.6 as
h = A-x
Where A=1.15.

A sketch sometimes helps.

I have sketched the graph here :
https://www.thestudentroom.co.uk/attachment.php?d=1610438069&attachmentid=991018

Now I have a question, when bottom of the barrier goes over the equilibrium position of the diagram of the question, lets say it is at point A now (as in my sketch), the value of h would still be > 0.6m, which indicates that the contestant can still pass. The same applies for the barrier over A, right?

So why not use the value of x (over the equilibrium position) in calculation as well?

That's correct and the answer does include that region.
The t they calculate is ~0.7 and they double it and subtract from T=4.5.
Note the floor is at the top of your curve, if we treat downwards as positive as they do in the ms.

But I thought that they doubled it because the barrier goes down and goes up that displacement.

Sorry, I still dont understand why did they substracted the time x>0.6m

Im confused now

Hi !

I need help regarding this

Question (Click to see the photo): https://www.thestudentroom.co.uk/attachment.php?d=1610430559&attachmentid=991002

Mark scheme: https://www.thestudentroom.co.uk/attachment.php?d=1610430559&attachmentid=991004

1.15 m is the amplitude of the barrier found by (2.3/2).
It means the equilibrium position is at the center, not at the ground.

And 0.55m is displacement from the equilibrium position (1.15- 0.6).

Why should I use 0.55 m instead of 0.60 m to calculate the time he should not get under the barrier?

I am just trying to say that, when the barrier is within 0.6m of the ground, the contestant will not be able to get under the barrier.

I have an exam soon so please help me ASAP! :-)

I have sketched the graph here :
https://www.thestudentroom.co.uk/attachment.php?d=1610438069&attachmentid=991018

Now I have a question, when bottom of the barrier goes over the equilibrium position of the diagram of the question, lets say it is at point A now (as in my sketch), the value of h would still be > 0.6m, which indicates that the contestant can still pass. The same applies for the barrier over A, right?

So why not use the value of x (over the equilibrium position in question diagram) in calculation as well?