M3:SHM strings Watch

Maths&physics
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I've spent almost 2 days on this question and I still dont get it. someone has tried to explain it to me but I still do t get it.

part d,

The particle drops free fall from A to the end of its natural length, where tension begins, as well as SHM. it reached K, and begins to move back up again and it becomes slack when it reaches it's natural length 0.6.

it's amplitude is greater than its extension, but I dont know what effect that has - it should project the particle higher than it's natural length and become slack, and no longer be in SHM.

I know the time from A to 0.6 using SUVAT.

I can work out the period using w from earlier but this particle doesn't complete full shm as it surpasses it's natural length. therefore, I need to subtract a value(s) but I dont know what and why.

and please, no rude comments. I genuinely need help.
Last edited by Maths&physics; 2 months ago
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shouldn't the time taken away from the complete oscillation be 0.444 (amplitude) - 0.147 (extension) = 0.297? because if there was a complete oscillation, the particle would go 0.444 above the equilibrium, but the string natural length ends 0.147 after the equilibrium.
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old_engineer
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(Original post by Maths&physics)
shouldn't the time taken away from the complete oscillation be 0.444 (amplitude) - 0.147 (extension) = 0.297? because if there was a complete oscillation, the particle would go 0.444 above the equilibrium, but the string natural length ends 0.147 after the equilibrium.
What was the original question?
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(Original post by old_engineer)
What was the original question?
sorry, I thought that I had attached it. part d.....
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That's what I think is going on but in the answer, they've subtracted 0.147 twice and not 0.297.... :confused:
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(Original post by old_engineer)
What was the original question?
that's what they've done
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If the motion of the particle was full SHM it would look like the left hand graph, in which the horizontal axis is set at the equilibrium position. However, the string will be slack whenever the particle is more than 0.147m above the equilibrium position. So, SHM applies between A and B as marked on the right hand graph. To find the time from A to B you can either determine the time between the two instances per cycle when the displacement is 0.147 upwards, or you can start with the duration of a complete cycle and subtract the bits at both ends when SHM does not apply.
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(Original post by old_engineer)
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If the motion of the particle was full SHM it would look like the left hand graph, in which the horizontal axis is set at the equilibrium position. However, the string will be slack whenever the particle is more than 0.147m above the equilibrium position. So, SHM applies between A and B as marked on the right hand graph. To find the time from A to B you can either determine the time between the two instances per cycle when the displacement is 0.147 upwards, or you can start with the duration of a complete cycle and subtract the bits at both ends when SHM does not apply.
absolutely amazing explanation! I get it now! thank you!
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