Need help with A Level Simple Harmonic Motion question

Hi, I am struggling with this Simple Harmonic Motion practice question, and looking at the mark scheme didn't help. Could anyone explain further please? I have posted the question in a reply below, it wouldn't let me add it to the main post

(edited 1 year ago)

Reply 1

College student2

this is the question.

So for part iii) I thought you were only meant to mark one point of acceleration that was both the greatest and in the direction of positive displacement? But I think I may have misinterpreted it because the mark scheme says “peak indicated (1 mark) trough indicated (1 mark)”. If they are two separate answers, which is the one for greatest acceleration and which is for acceleration in the direction of positive displacement??

For iv) I tried to use a = v^2/r, and I put in (0.03/0.0375)^2/0.03, to try and use the point I had marked on the graph (I marked the minimum point on the first trough) but got it wrong, and I’m not sure what I did wrong and can’t find a way to get the right answer (which is apparently 4.7ms^-2)

Reply 2

Skiwi

Original post by College student2

this is the question.
So for part iii) I thought you were only meant to mark one point of acceleration that was both the greatest and in the direction of positive displacement? But I think I may have misinterpreted it because the mark scheme says “peak indicated (1 mark) trough indicated (1 mark)”. If they are two separate answers, which is the one for greatest acceleration and which is for acceleration in the direction of positive displacement??
For iv) I tried to use a = v^2/r, and I put in (0.03/0.0375)^2/0.03, to try and use the point I had marked on the graph (I marked the minimum point on the first trough) but got it wrong, and I’m not sure what I did wrong and can’t find a way to get the right answer (which is apparently 4.7ms^-2)

for iv) from what i can see you're finding the velocity using displacement/time. Why can't you do this when the particle is undergoing shm?
Instead try using a=-omega^2r ( which is the same as what you had since v=wr).

Reply 3

username6691833

Original post by College student2

this is the question.
So for part iii) I thought you were only meant to mark one point of acceleration that was both the greatest and in the direction of positive displacement? But I think I may have misinterpreted it because the mark scheme says “peak indicated (1 mark) trough indicated (1 mark)”. If they are two separate answers, which is the one for greatest acceleration and which is for acceleration in the direction of positive displacement??
For iv) I tried to use a = v^2/r, and I put in (0.03/0.0375)^2/0.03, to try and use the point I had marked on the graph (I marked the minimum point on the first trough) but got it wrong, and I’m not sure what I did wrong and can’t find a way to get the right answer (which is apparently 4.7ms^-2)

could you please post the complete question...the first part too?

Reply 4

College student2

Original post by Skiwi

Thank you, I was able to reach the right answer using that equation!

So is it that what I was originally doing wouldn't work for because the displacement is constantly changing during SHM?

Reply 5

username6691833

Btwfor part iii) you did not misinterpret the question. The peak really just means the extreme (point of max displacement in any direction), so your point will be on any of the negative extremes i.e. trough

Reply 6

mqb2766

Original post by College student2

this is the question.
So for part iii) I thought you were only meant to mark one point of acceleration that was both the greatest and in the direction of positive displacement? But I think I may have misinterpreted it because the mark scheme says “peak indicated (1 mark) trough indicated (1 mark)”. If they are two separate answers, which is the one for greatest acceleration and which is for acceleration in the direction of positive displacement??
For iv) I tried to use a = v^2/r, and I put in (0.03/0.0375)^2/0.03, to try and use the point I had marked on the graph (I marked the minimum point on the first trough) but got it wrong, and I’m not sure what I did wrong and can’t find a way to get the right answer (which is apparently 4.7ms^-2)

The two marks are for [1] acceleration is maximum (magnitude) at a peak or trough and [1] its positive at a trough. So 1 mark if you got a peak or 2 marks if you got a trough.

iv) see Skiwi's comment about -w^2x, though note v^2/r is a circular motion formula for the magnitude of the radial acceleration, not shm.

(edited 1 year ago)

Reply 7

username6691833

For (iv) if you want to use (v^2)/r tho, you can use the formula v=wcos(wt)

(edited 1 year ago)

Reply 8

College student2

Original post by mqb2766

The two marks are for [1] acceleration is maximum (magnitude) at a peak or trough and [1] its positive at a trough. So 1 mark if you got a peak or 2 marks if you got a trough.
iv) see Skiwi's comment about -w^2x, though note v^2/r is a circular motion formula for the magnitude of the radial acceleration, not shm.