HELP With last question on oscillation please

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MrToodles4
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#1
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I need help with the very last question on this please: 4cii

"The cone experiences a mean damping force of 0.25N. Calculate the average power
needed to be supplied to the cone to keep it oscillating with a constant amplitude."

I've attached the rest of the question in case you need context.

http://pmt.physicsandmathstutor.com/...s%202%20QP.pdf

What I did was work out energy from part i. I then worked out time by doing 1/2400 = 4.167*10^-4 secs. So P = e/t and so I subbed in the two values I had.

I don't understand why the markscheme did 1/4 of a cycle??

http://pmt.physicsandmathstutor.com/...A-level/Topic-
Qs/OCR-A/5-Newtonian-World-and-Astrophysics/5.3-Oscillations/Set-M/Oscillations%202%20MS.pdf
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phys981
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They've taken quarter of a cycle because that's what a displacement equal to the amplitude is, so they've taken T/4 to match.

Think of your graph, to go from 0 to max displacement (or the reverse) is quarter of a full wave / full cycle.

So they've used P = Work/time = (force x distance) / time
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Eimmanuel
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(Original post by MrToodles4)
I need help with the very last question on this please: 4cii

"The cone experiences a mean damping force of 0.25N. Calculate the average power
needed to be supplied to the cone to keep it oscillating with a constant amplitude."

I've attached the rest of the question in case you need context.

http://pmt.physicsandmathstutor.com/...s%202%20QP.pdf

What I did was work out energy from part i. I then worked out time by doing 1/2400 = 4.167*10^-4 secs. So P = e/t and so I subbed in the two values I had.

I don't understand why the markscheme did 1/4 of a cycle??

http://pmt.physicsandmathstutor.com/...A-level/Topic-
Qs/OCR-A/5-Newtonian-World-and-Astrophysics/5.3-Oscillations/Set-M/Oscillations%202%20MS.pdf

You may want to show how you work out the energy, as it may or may not be correct depending on how you work it out.


(Original post by MrToodles4)
… I don't understand why the markscheme did 1/4 of a cycle?? …


If the cone undergoes sinusoidal oscillation and starts from x = 0 at t = 0, it would move to x = A at t = ¼T. We can only deduce the exact distance moved by the cone every ¼ of the cycle or period.

But you don’t need really need to use ¼ of the cycle. If you use one period, then you need to work out the distance moves in one period – which is 4A.

Power =  \dfrac{{{F}_{d}}\times 4A}{T}=\dfrac{{{F}_{d}}\times A}{\frac{T}{4}}
where Fd is mean damping force.
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MrToodles4
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(Original post by phys981)
They've taken quarter of a cycle because that's what a displacement equal to the amplitude is, so they've taken T/4 to match.

Think of your graph, to go from 0 to max displacement (or the reverse) is quarter of a full wave / full cycle.

So they've used P = Work/time = (force x distance) / time
Thank you so much really appreciate this I would rep if I could
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MrToodles4
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[QUOTE=Eimmanuel;76223236]You may want to show how you work out the energy, as it may or may not be correct depending on how you work it out.






If the cone undergoes sinusoidal oscillation and starts from x = 0 at t = 0, it would move to x = A at t = ¼T. We can only deduce the exact distance moved by the cone every ¼ of the cycle or period.

Why is this?
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Eimmanuel
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(Original post by MrToodles4)

If the cone undergoes sinusoidal oscillation and starts from x = 0 at t = 0, it would move to x = A at t = ¼T. We can only deduce the exact distance moved by the cone every ¼ of the cycle or period.

Why is this?

Maybe I should rewrite statement as

We can only deduce the exact distance moved by the cone every ¼ of the cycle or period without going through the usage of trigonometric ratio.

In SHM, you should know that if the particle starts from x = 0 at t = 0, the particle will reach x = +A or −A at t = ¼T. If you want to know where the particle is at any time, you need to know how the particle moves. This means that you need to know trigonometric ratio that is used to describe the motion of the particle.

However, regardless of the trigonometric ratio, if the initial condition is x = 0 at t = 0, then we can know when the particle reaches the extreme position.

at t = ¼T, particle will reach x = +A or −A

at t = ½T, particle will reach x = 0.

at t = ¾T, particle will reach x = −A or +A

at t = T, particle will reach x = 0

Hope it makes sense.
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