Physics HL/SL Help

[HiggsHunter]

That's a nice colour. I recommend she tries a really fiery red. To spice the wardrobe up a little bit.

Excellent taste, Award Winning Boss!

[DeBrogliez]

From the same graph that DropBoite attached, how do we calculate the maximum acceleration?

[HiggsHunter]

From the same graph that DropBoite attached, how do we calculate the maximum acceleration?

The maximum acceleration would be 1.5 x 10ˆ-10 x (2pi/T)ˆ2, where T is the period.

But I have not understood why I am a factor of 100 out with the period!

Do you have the full question?

[DeBrogliez]

From the same graph that DropBoite attached, how do we calculate the maximum acceleration?

The maximum acceleration would be 1.5 x 10ˆ-10 x (2pi/T)ˆ2, where T is the period.

But I have not understood why I am a factor of 100 out with the period!

Do you have the full question?

Yes I do!

And you are right about the period

It is 1.1*10^-14

I see what you have done about the acceleration

You multiplied w^2 by the amplitude right?

I did it that way, but I also tried to find it using the second derivative but I got a different answer!

Shouldn't it be the same?

[HiggsHunter]

Yes I do!

And you are right about the period

It is 1.1*10^-14

I see what you have done about the acceleration

You multiplied w^2 by the amplitude right?

I did it that way, but I also tried to find it using the second derivative but I got a different answer!

Shouldn't it be the same?

Great! Can you post the question, or let me know where to find it on the web?

Yes, the acceleration is the second derivative, (-wˆ2) x amplitude x cos(wt), which has maximum value wˆ2 x amplitude.

[DeBrogliez]

I will attach the question, because it is very long!

But I'm not sure if it is allowed or not

And I don't know why the second derivative method is not working with me!

I have been trying a lot, it just doesn't seem to work

SHM.pdf

Edited April 25, 2012 by DeBrogliez

[HiggsHunter]

I will attach the question, because it is very long!

But I'm not sure if it is allowed or not

And I don't know why the second derivative method is not working with me!

I have been trying a lot, it just doesn't seem to work

Thanks a lot for the question! What exam is it from?

The graph of displacement is: amplitude x cos(wt)

So the velocity (1st derivative) is: -w x amplitude x sin(wt)

And the acceleration (2nd derivative) is: -wˆ2 x amplitude x cos(wt)

So the maximum acceleration is: (wˆ2) x amplitude

[DeBrogliez]

I think its from 2009

Thank you!

[HiggsHunter]

Now that DeBrogliez has posted the whole question, I see that actually you were asked to estimate the maximum acceleration using a previously calculated value for the force per unit displacement between the hydrogen and carbon atoms, k.

That value is k = m x wˆ2 = 556 N mˆ-1. (They give approx 560 in the question)

So using this approach, the maximum acceleration (which occurs when x = 1.5 x 10ˆ-10) is 556 x 1.5 x 10ˆ-10 x 1/(1.7 x 10ˆ-27) = 4.9 x 10ˆ19 m sˆ-2.

This is the same as the result that you get by using 1.5 x 10ˆ-10 x (2pi/T)ˆ2 with period T = 1.1 x 10ˆ-14.

I think it's pretty careless of the IBO if the markscheme shows T = 1.1 x 10ˆ-12, but I presume that the examiners would have caught this slip when correcting the papers, and that no student would have been penalized incorrectly.

[DropBoite]

Hi! Thanks for clearing that up! *whew*

I've 2 question on SHM I hope you wouldn't mind answering!

1) In the attached image oscillation refer to diagram 2 and draw a V against t curve. I drew a cosine curve which is wrong, my question is more of a confirmation, but is this because in diagram v the body is at rest and the velocity still has to 'start' and thus a sine curve because it starts from zero?

2) Another tiny question is regarding the same wooden blocks. prior information is that the ratio of its displacement at t=1.75 s and t= 0.35 s is 0.76 and thus we are asked to tell what is the ratio of their energies! What can that be?! As it's a 1 mark question it doesn't even expect much working! the correct answer is: 0.57 or 0.58. A graph for this question is attached called ques 2. Thanks in advance!

[Drake Glau]

Well, I think I can help with number 2

E_k=0.5mw²(x₀²-x²) <-from your data booklet. There is also another kinetic energy equations but keep in mind what would happen when you use them, the max one has x naught by itself, which would zero the whole thing so you can't use that.

w=1.4, you can find this from looking at the graph.

x₀=0, also from the graph.

m is a constant, so we'll treat it like one. It's a ratio so it will magically disappear

Now the ratio you stated is t_1.75/t_0.35 to get the displacement ratio. So we use the same ratio for our energy equation

E_k(1.75)/E_k(0.35)

Lucky for you, everything cancels except for the x² for the two times...

x when t=0.35 is 3.6 so x²=12.96

x when t=1.75 is 2.8 so x²=7.84

7.84/12.96=0.6

Close, but I kinda simply glanced at t=0.35, so it might be a little off

Edited April 26, 2012 by Drake Glau

[DropBoite]

Thanks drake will rep wen using laptop.

@Higgs- remeber tye question about how Lplanet/L sun is written -as a fractiin and nit multiplicatiin- well leading on from that equatiin is this logically corrct im relatively sure but just checking:

LAnt = d^2Ant x bAnt x L Sun? mathematically it seems corrct and thats how it is in the marking scheme but im just looking for an explanation.

[Peanut Butter Jelly]

Can someone explain section 5 (electronics) in option F - communications?

What is an op-amp, basically...

[HiggsHunter]

@Higgs- remeber tye question about how Lplanet/L sun is written -as a fractiin and nit multiplicatiin- well leading on from that equatiin is this logically corrct im relatively sure but just checking:

LAnt = d^2Ant x bAnt x L Sun? mathematically it seems corrct and thats how it is in the marking scheme but im just looking for an explanation.

Page 13 in your data booklet gives b = L/(4pi x dˆ2),

i.e. L = (4pi x dˆ2) x b

So Lant/Lsun = (dant/dsun)ˆ2 x bant/bsun

i.e. Lant = ((dant/dsun)ˆ2 x bant/bsun) x Lsun

The distance of Antares from Earth is given as 3.9 x 10ˆ7 AU, so (dant/dsun)ˆ2 = (3.9 x 10ˆ7)ˆ2

And you are given bant/bsun = 4.3 x 10ˆ-11

Hence Lant = ((3.9 x 10ˆ7)ˆ2 x 4.3 x 10ˆ-11) x Lsun = (6.54 x 10ˆ4) x Lsun

[HiggsHunter]

Can someone explain section 5 (electronics) in option F - communications?

What is an op-amp, basically...

The term op-amp comes from analogue computer days, where high-gain DC amplifiers with stable characteristics were interconnected in various ways to effect different mathematical operations. They are very versatile building blocks for all kinds of electronic circuits.

Because op-amps have differential inputs they can be wired to be inverting or non-inverting and (except in a few cases like a comparator), they are used with negative feedback so that the overall gain is controlled by the values of the feedback components, not by the open-loop gain of the amplifier itself.

The op-amps in IB exam questions are usually considered to be ideal, having infinite open-loop gain, infinite input impedance and zero output impedance. However, when solving problems you have to watch that the maximum and minimum output voltages saturate at the upper and lower power supply voltages.

[DropBoite]

@Higgs- remeber tye question about how Lplanet/L sun is written -as a fractiin and nit multiplicatiin- well leading on from that equatiin is this logically corrct im relatively sure but just checking:

LAnt = d^2Ant x bAnt x L Sun? mathematically it seems corrct and thats how it is in the marking scheme but im just looking for an explanation.

Page 13 in your data booklet gives b = L/(4pi x dˆ2),

i.e. L = (4pi x dˆ2) x b

So Lant/Lsun = (dant/dsun)ˆ2 x bant/bsun

i.e. Lant = ((dant/dsun)ˆ2 x bant/bsun) x Lsun

The distance of Antares from Earth is given as 3.9 x 10ˆ7 AU, so (dant/dsun)ˆ2 = (3.9 x 10ˆ7)ˆ2

And you are given bant/bsun = 4.3 x 10ˆ-11

Hence Lant = ((3.9 x 10ˆ7)ˆ2 x 4.3 x 10ˆ-11) x Lsun = (6.54 x 10ˆ4) x Lsun

I'm sorry but that overcomplicated things for me, I'm genuinely afraid of maths . So in a nutshell does this support the what I was checking? Also it intrigues me how dant/dsun simply becomes d ant in your calculations, thought I think I was right, yes, Lsun is multiplied on the other side (I think aloud a lot! )

A new question, you must be tired by now , : How is the ratio attached obtainable when comparing dplant/dsun- basically everything will be in ratios, what is the math behind the seemingly cross multiplied equation or some such.Secondly, 4 pi cancel out right?

[HiggsHunter]

1) In the attached image oscillation refer to diagram 2 and draw a V against t curve. I drew a cosine curve which is wrong, my question is more of a confirmation, but is this because in diagram v the body is at rest and the velocity still has to 'start' and thus a sine curve because it starts from zero?

Yes, if diagram 2 represents the situation when the partially submerged object is at its lowest point then its velocity starts from zero and will follow a damped sine curve.

[HiggsHunter]

I'm sorry but that overcomplicated things for me, I'm genuinely afraid of maths . So in a nutshell does this support the what I was checking? Also it intrigues me how dant/dsun simply becomes d ant in your calculations, thought I think I was right, yes, Lsun is multiplied on the other side (I think aloud a lot! )

Yes, you are correct, one AU is (approximately) equal to dsun, so the ratio dant/dsun is (3.9 x 10ˆ7).

You are also given bant/bsun as a ratio (4.3 x 10ˆ-11), ready to plug into the formula.

[Drake Glau]

Is this question from and old exam? I remember my astro paper3 having a ratio that involved some heavy distances in it and luminosity as well O.o

My problem was always that the d i calculated was in meters and made my calculator overflow and then i ran out of time before figuring out to use a ratio to convert stuff out of meters and into better units >.>

[HiggsHunter]

Is this question from and old exam? I remember my astro paper3 having a ratio that involved some heavy distances in it and luminosity as well O.o

My problem was always that the d i calculated was in meters and made my calculator overflow and then i ran out of time before figuring out to use a ratio to convert stuff out of meters and into better units >.>

You could use scientific notation but, as in the old slide rule days, the simplest way is just to use your calculator to get the significant figures and then multiply by the appropriate power of 10 separately.