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Thread: Cambridge HSC MX1 Textbook Marathon/Q&A

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by davidgoes4wce View Post














    Just had a bit of trouble with part (a)
    I have nothing else to do on a Monday night and have no social life, so I decided to answer my own question.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.

    do i form an equation of: x double dot = -n^2 (x - 2) ??

    which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n

    or is the equation just 4 = -n^2 x 2

    but then how do i get rid of the negative to get a value for n.

    yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by hedgehog_7 View Post
    The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.

    do i form an equation of: x double dot = -n^2 (x - 2) ??

    which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n

    or is the equation just 4 = -n^2 x 2

    but then how do i get rid of the negative to get a value for n.

    yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!
    We can set the mean position to be x = 0.

    So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

    (If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)

    We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

    Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)

    So -4 = -n^2 * 4, so n = 1.

    So v^2 = 25 - x^2.

    Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
    Last edited by InteGrand; 19 Apr 2016 at 6:34 PM. Reason: Typo
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    We can set the mean position to be x = 0.

    So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

    (If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)

    We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

    Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)

    So -4 = -n^2 * 4, so n = 1.

    So v^2 = 25 - x^2.

    Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
    question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since

    x double dot = - n^2 x

    is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by hedgehog_7 View Post
    question, why is x double dot equal to -4 and not just 4?
    As I said, in simple harmonic motion, the acceleration and displacement from centre of motion are always opposite signs (this is because the acceleration is always trying to push the object back to its centre of motion, as you can visualise by imagining a pendulum for instance). So when x = 4 (displacement from centre of motion), the acceleration must be negative.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by hedgehog_7 View Post
    question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since

    x double dot = - n^2 x

    is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?
    If you take x to be negative, the acceleration you plug in needs to be positive.

    And when they said the acceleration is 4 m s-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    If you take x to be negative, the acceleration you plug in needs to be positive.

    And when they said the acceleration is 4 m s-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.
    oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by hedgehog_7 View Post
    oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?
    Yeah. You only need to take one of them to find n.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    How common is Volumes in the HSC Exams? I rarely see them o.O

    Halp Please

    http://i.imgur.com/kzIHB2w.jpg

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    ew imperial

    yuck
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by si2136 View Post
    How common is Volumes in the HSC Exams? I rarely see them o.O

    Halp Please

    http://i.imgur.com/kzIHB2w.jpg
    Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.
    Yes, the volume. And are feet and foot the same (dumb question)?

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by si2136 View Post
    Yes, the volume. And are feet and foot the same (dumb question)?
    Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

    This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

    We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

    This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

    We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).
    Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by si2136 View Post
    Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge
    I doubt they'd ask it in the HSC without providing a conversion rate though, so don't worry.
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    We can set the mean position to be x = 0.

    So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

    (If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)

    We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

    Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)

    So -4 = -n^2 * 4, so n = 1.

    So v^2 = 25 - x^2.

    Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
    Is there something wrong with the solution for that question because it shows the speed to be
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by davidgoes4wce View Post
    Is there something wrong with the solution for that question because it shows the speed to be
    Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by InteGrand View Post
    Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).
    I fixed up your mistake and redid the solution and was able to get the speeds.

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Hey guys,

    Sorry I'm a bit new to using this. :P

    Could I get some help with Exercise 10I question 10???

    There are two round tables, one oak and one mahogany, each with five seats. In how many
    ways may a group of ten people be seated?

    Thanks

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    There are (5-1)! ways to arrange the 5 people in the first table then another (5-1)! ways to arrange in the other table. But for the first table, any 5 of the 10 people can be chosen and the remaining 5 chosen for the 2nd table.
    So the answer is 10C5*4!*4! = 145152
    Last edited by jathu123; 22 Apr 2016 at 10:51 AM.
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Cambridge EX 10 D Q 19



    Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

    Am I missing something?
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by davidgoes4wce View Post
    Cambridge EX 10 D Q 19



    Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

    Am I missing something?

    Last edited by porcupinetree; 23 Apr 2016 at 1:04 PM.
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    Bachelor of Science (Advanced Mathematics) @ USYD

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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Can understand it a bit better now, my thinking initially was 4/36 but can see why its 4/11, since the total possible combinations with a 3 are: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3) and (6,3).

    Only 4 of those have combined value above 8 or more.
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Ex 3A Q 9 (h)








    Should the solution be between the range?
    8 < t < 16, 32 < t < 40, 54 < t< 60 ?

    The books answer is:

    8 < t < 16
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    Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

    Quote Originally Posted by davidgoes4wce View Post
    Ex 3A Q 9 (h)








    Should the solution be between the range?
    8 < t < 16, 32 < t < 40, 54 < t< 60 ?

    The books answer is:

    8 < t < 16
    Since they only wanted you to consider when you could use the graph AND the table of values...
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