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STEP Maths I, II, III 1996 Solutions

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Good Job People Just one more question to do. Wont be able to make a new thread for the 1995 solutions as ill be at school all day.

However ive got the answer to STEP I Question 2 written down, most of Question 3(Probably get this out today) and most of Question 4. Oh i was able to do the first out of two parts on question 6 too.

Reply 61

insparato

First two parts of STEP III Question 1

cosh^4x - sinh^4x = 1/16(e^x+e^-x)^4 -1/16(e^x-e^-x)^4

= 1/16(e^x+e^-x)^2)^2

=1/16(e^2x+2+e^-2x)^2

=1/16(e^4x+2e^2x+1+2e^2x+4+2e^-2x + 1 + 2e^-2x + e^-4x)
=1/16(e^4x+4e^2x+4e^-2x+e^-4x + 6)
-1/16(e^x-e^-x)^2)^2
-1/16(e^2x-2+e^-2x)^2
-1/16(e^4x-2e^2x+1-2e^2x+4-2e^-2x+1-2e^-2x+e^-4x)
-1/16(e^4x-4e^2x-4e^-2x+6+e^-4x)

1/16(e^4x+4e^2x+4e^-2x+e^-4x + 6) - 1/16(e^4x-4e^2x-4e^-2x+6+e^-4x)

1/2e^2x+1/2e^-2x = 1/2(e^2x+e^-2x) = cosh 2x

1/16(e^4x+4e^2x+4e^-2x+e^-4x + 6) + 1/16(e^4x-4e^2x-4e^-2x+6+e^-4x)

1/8e^4x+12/16+1/8e^-4x = 1/4cosh4x + 3/4

For the next part im thinking

Adding

cosh^4x-sinh^4x = cosh2x

cosh^4x+sinh^4x = 1/4cosh4x + 3/4

to get

2cosh^4x = 1/4cosh4x + cosh2x + 3/4

so

cosh^4x = 1/8cosh4x + 1/2cosh2x + 3/8

Try to generalise this...

Or something like that however i really dont have time to investigate it further for the moment. School ruins everything :frown:

Reply 62

DFranklin

dvs

Is it just me or is III/1 unbearably long?

I had a look at it: it doesn't seem that long to me, but it's a complete pain in the neck notation wise; I was scribbling on 2 inch square post it notes and it just got ridiculous (couldn't fit one equation on a note) so I got fed up and stopped!

Furthering my "what are they thinking with some these questions?" moaning: with a question like this it seems completely obvious how you'd go about answering it; I "know" how to get the a_n etc. The only problem is grinding through the notation without slipups. It just doesn't seem to test mathematical intuition at all.

As you show with the groups question, there are some disproportionately short questions in other areas.

Edit: Putting money where my mouth is, it does come out in well under a side with some careful choice of notation (and if you simply state the equivalent solution for sinh^2m x which I think is reasonable enough here). The special csae for the constant term is a pain though. LaTeXing this in readable format is going to be more pain than I think I can bear, however...

Reply 63

DFranklin

Step III, Q1.
cosh x = (e^x+e^-x)/2, sinh x = (e^x - e^-x)/2. To cut down on notational pain, write p = e^x, q=e^-x, C = cosh x, S =sinh x.

16C^4 = (p+q)^4 = p^4+4p^2+6+4q^2+q^4 = 2 \cosh 4x + 8 \cosh 2x + 6

16S^4 = (p-q)^4 = p^4-4p^2+6-4q^2+q^4 = 2 \cosh 4x - 8 \cosh 2x + 6

Thus

C^4 - S^4 = \cosh 2x, C^4+S^4 = \frac{1}{4}\cosh 4x + \frac{3}{4}

.

Now

2^n C^n = (p+q)^n = p^n + \binom{n}{1}p^{n-2} + ... + \binom{n}{n-1}q^{n-2}+q^n

.

What we'd like to do is pair off terms starting at the end and working to the middle. If n is odd this works fine, but if n is even, there's a single middle term. Write n = 2k+1 (n odd), n = 2k (n even). Then in each case, there are k terms we can pair off, but in the case of n even, there's then a constant middle term

\binom {n}{k}

unaccounted for.

Thus

Unparseable latex formula:

2^n C^n = \Big[ \sum_{i=0}^{k-1} \binom{n}{i} (p^{n-2i}+q^{n-2i})\Big] + R_n,\quad \text{where }R_n = \left\{ \begin{array}{ll}[br]\binom{n}{n/2} & \textrm{if $n$ even}\\[br]0 & \textrm{if $n$ odd}\end{array} \right

Thus

C^n = 2^{1-n} \Big[\sum_{i=0}^{k-1} \binom{n}{i}\cosh (n-2i)x \Big] + R_n/2^n

.

So

a_0 = R_n / 2^n, \quad a_{n-2i} = \binom{n}{i}, \quad a_{n-2i-1} = 0 (i=0,...,k-1)

.

The corresponding terms

b_n

for

S^n

will be

b_{n-2i} = (-1)^i a_{n-2i}, b_{n-2i-1} = 0

(where the relation for the n-2i term goes right down to the constant term, if applicable - in other words if n = 2k,

b_0 = (-1)^k a_0

).

Thus

C^{2m}-S^{2m} = 2^{1-2m}\Big[\sum_{i=0}^{m-1} (1-(-1)^i)\binom{2m}{i}\cosh (2m-2i)x \Big] + (1-(-1)^i)\binom{2m}{m}2^{-2m}

C^{2m}+S^{2m} = 2^{1-2m}\Big[\sum_{i=0}^{m-1} (1+(-1)^i)\binom{2m}{i}\cosh (2m-2i)x \Big] + (1+(-1)^i)\binom{2m}{m}2^{-2m}

.

Comments:

Using (1+(-1)^i) feels a little ugly, but if you want an explicit formula I'm not sure you're going to get anything better.

Again, there is really nothing to this question, other than the notational pain. It is straightforward to get to the point where you could write down

C^{2m}-S^{2m}

for a given m. It is much harder to get an explicit formula, but this is not for any deep reasons. The difficulty is in finding a way of writing down that: "terms from the end pair", "the middle term, if there is one, is a special case", and then "adding the two series for

C^{2m}-S^{2m}

leaves us with a series formed by taking twice every alternate term in the series for

C^{2m}

". I don't even know what form the examiners were expecting for the final answer here.

You may gather I am not a fan of questions like this.

Reply 64

DFranklin

In contrast, III, Q9 is really nice:

Draw a diagram of a quarter circle, measure angle

\theta

so that

\theta

= 0 is a vertical angle. In other words, the height of the particle about the center of the sphere will be

a \cos \theta

.

Conservation of energy gives

v^2 = 2ga(1-\cos \theta).

. Now the radial acceleration due to gravity is

g \cos \theta

, and the particle leaves the sphere when this acceleration is lower than the "centripetal" acceleration

v^2/a

. So the particle leaves the plane when

\cos \theta = 2(1 - \cos \theta) \implies \cos \theta = 2/3

. At this point,

v^2 = aC \implies v = \sqrt{\frac{2ga}{3}}

.

Write C for

\cos \theta

, S for

\sin \theta

. Then

S=\frac{\sqrt{5}}{3}

. When the particle leaves the plane, it's height above the ground is aC and it has travelled aS horizontally.

At this point, it has velocity components

v_y = vS = \frac{\sqrt{5}}{3}\sqrt{\frac{2ga}{3}} = \sqrt{\frac{10ga}{27}},\quad v_x = vC = \frac{2}{3} \sqrt{\frac{2ga}{3}} = \sqrt{\frac{8ga}{27}}

(where

v_y

is measured downwards).

The particle still has aC = 2a/3 to fall. The time t it takes to fall this far satisfies

tv_y+gt^2/2 = 2a/3

, and so we have

\frac{1}{2}gt^2 + t V_y - \frac{2a}{3} = 0

.

Then

t = \frac{-V_y + \sqrt{v_y^2+4ag/3}}{g}

(taking the +ve root as t > 0).

Unparseable latex formula:

\text{so } t = \frac{1}{g}\Big[ \sqrt{\frac{10ga}{27}+\frac{4ga}{3}} - \sqrt{\frac{10ga}{27}}\Big]\\[br]=\sqrt{\frac{2a}{27g}}(\sqrt{5+18}-\sqrt{5}\Big)=\sqrt{\frac{2a}{27g}}(\sqrt{23}-\sqrt{5})

Then the horizontal distance travelled once the particle leaves the sphere is just

tV_x = \sqrt{\frac{8ga}{27}}\sqrt{\frac{2a}{27g}}(\sqrt{23}-\sqrt{5}) = 4a \frac{\sqrt{23}-\sqrt{5}}{27}

.

And the total distance is

aS+tVx = a\frac{\sqrt{5}}{3} + 4a \frac{\sqrt{23}-\sqrt{5}}{27} = a \frac{9\sqrt{5}}{27} + a \frac{4\sqrt{23}-4\sqrt{5}}{27} = a (5\sqrt{5}+4\sqrt{23})/27.

Reply 65

dvs

DFranklin

You may gather I am not a fan of questions like this.

I'm not a fan either. This question was simply mean. It's the first question and looks very approachable, but once you start crunching it out it'll get painful and you will end up wasting precious time tidying everything up.

Reply 66

insparato

dvs

Yeah as i first thought when i did the first two parts :biggrin:

. Looking at the answer to the last two parts eek! *runs away*

Reply 67

datr

STEP I Q13
Start of by looking at the first few possibilities

n=2\:\:\: P=\frac{1}{2} = \frac{1}{2(2-2)!}

n=3\:\:\: P=\frac{1}{2}\times\frac{2}{3} = \frac{1}{3} = \frac{1}{3(3-1)!}

n=4\:\:\: P=\frac{1}{2}\times\frac{1}{3}\times\frac{3}{4} = \frac{1}{2 \times 4} = \frac{1}{4(4-2)!}

n=5\:\:\: P=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4} \times \frac{4}{5} = \frac{1}{5\times 3\times 2} = \frac{1}{5(5-2)!}

n=k\:\:\: P = \frac{1}{k(k-2)!}

Use xp(x) to find the expectation.

\displaystyle E(N) = \sum_{r=2}^\infty \frac{1}{(r-2)!} = \sum_{r=0}^\infty \frac{1}{r!}

This can be evaluated either by comparing with the Taylor series of e^1 or the general term of the Poisson distribution (there may be a better way but I haven't done it yet if there is).

E(N) = e^1

\displaystyle Var(N) = \sum_{r=2}^\infty \frac{r}{(r-2)!} -E(X)^2

\displaystyle\sum_{r=2}^\infty \frac{r}{(r-2)!} = 2 + \sum_{r=3}^\infty \frac{r - 2 + 2}{(r-2)!} = 2 + \sum_{r=3}^\infty \frac{1}{(r-3)!} + \frac{2}{(r-2)!}

From before:

\displaystyle 2 + \sum_{r=0}^\infty \frac{1}{r!} + 2\sum_{r=1}^\infty \frac{1}{r!} = 2 + e^1 + 2(e^1 - 1) = 3e^1

Var(N) = 3e^1 - e^2 = e^1(3-e^1)

Does anyone know a better way of handling these infinite series? This is the only way I could think to do it given what I've done at a level and it seems rather cumbersome.

Reply 68

datr

STEP II Q12

P(X_j < \lambda) = \lambda

P(X_1 \cap X_2 \cap ... \cap X_n < \lambda) = \lambda^n

Unparseable latex formula:

F(x) = \begin{cases}[br] 0 & x < 0 \\[br] x^n & 0 \le x \le 1 \\[br] 1 & x > 1[br] \end{cases}

Unparseable latex formula:

f(x) = F'(x) = \begin{cases}[br] nx^{n-1} & 0 \le x \le 1 \\[br] 0 & Otherwise[br] \end{cases}

\displaystyle E(Y) = \int_{-\infty}^\infty xf(x) \,dx = \int_0^1 nx^n \,dx = \left[\frac{nx^{n+1}}{n+1}\right]_0^1 = \frac{n}{n+1}

\displaystyle E(Y^2) = \int_{-\infty}^\infty x^2f(x) \,dx = \int_0^1 = nx^{n+1} \,dx = \left[\frac{nx^{n+2}}{n+2}\right]_0^1 = \frac{n}{n+2}

Var(Y) = E(Y^2) - E(Y)^2 = \frac{n}{n+2} - \frac{n^2}{(n+1)^2} = \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} = \frac{n(n^2 + 2n + 1 - n^2 - 2n)}{(n+1)^2(n+2)}

Var(Y) = \frac{n}{(n+1)^2(n+2)}

ii) T = Life Span of Neon Light

P(T \le t) = 1 - e^{-\frac{t}{\lambda}}

P(T \ge t) = e^{-\frac{t}{\lambda}}

Let U be the smallest of T₁, T₂, ..., T_n.

U > k => T_j > k for 1 <= j <= n

P(U \ge t) = \left(e^{-\frac{t}{\lambda}}\right)^n = e^{-\frac{nt}{\lambda}}

F(t) = P(U \le t) = 1 - e^{-\frac{nt}{\lambda}}

f(t) = F'(t) = \frac{ne^{-\frac{nt}{\lambda}}}{\lambda}

\displaystyle E(U) = \int_0^\infty \frac{nte^{-\frac{nt}{\lambda}}}{\lambda} \,dt = \left[-te^{-\frac{nt}{\lambda}}\right]_0^\infty + \int_0^\infty e^{-\frac{nt}{\lambda}} \,dt = \left[\frac{-\lambda e^{-\frac{nt}{\lambda}}}{n}\right]_0^\infty

E(U) = \frac{\lambda}{n}

I think that's right but I'm not 100% confident.

Reply 69

DFranklin

If it reassures you, I think these last 2 answers are correct, and I don't think there's anything better you could have done with summing the series in the first question either.

Reply 70

DFranklin

Step III, 1996, Q13.

This is basically "bookwork" from the Part IA tripos, but it's kind of cute.

I won't bother with the E(aX+b), Var(aX+b) stuff as that is totally standard stuff.

The first part of the question is a famous inequality by Tchebyshev (many spellings!). Write E for E(X), V for Var(X).

V = \sum p_i (x_i-E)^2.

Then

V \ge \sum_{i\in S} p_i (x_i-E)^2

, where

S = \{i : |x_i-E| \ge \lambda\}

. In other words, if we only sum the terms where

|x_i-E| \ge \lambda

, we can't end up with more than the original sum over all terms (since each term is positive).

Now

|x_i-E| \ge \lambda \implies (x_i-E)^2 \le \lambda^2

. So

Unparseable latex formula:

[br]V &=& \sum p_i (x_i-E)^2 \ge \sum_{i\in S} p_i (x_i-E)^2 \ge \sum_{i\in S} p_i \lambda^2

Unparseable latex formula:

= \lambda^2 \sum_{i\in S} p_i = = \lambda^2 \bbold{P}(|X-E|\ge\lambda^2)

.

Dividing by

\lambda^2

gives the first result.

For the 2nd part, we want an overestimate for V, rather than an underestimate.

So if

|x_i-E|\ge\lambda

, we'll substitute

k^2

for

(x_i - E)^2

(since

|x_i-E|\le k

for all i). And if

|x_i-E|\le \lambda

we'll substitute

\lambda^2

for

(x_i - E)^2

. So we have:

\displaystyle V = \sum p_i (x_i-E)^2 = \sum_{i\in S} p_i (x_i-E)^2 + \sum_{i \not \in S} p_i (x_i-E)^2

\le \sum_{i\in S} p_i k^2 + \sum_{i \not \in S} p_i \lambda^2

= Pk^2+(1-P)\lambda^2

, where

Unparseable latex formula:

P = \bbold{P}(|X-E|\ge\lambda^2)

Unparseable latex formula:

\displaystyle[br]\text{So } V\le Pk^2+(1-P)\lambda^2 = \lambda^2 + P(k^2-\lambda^2)\\[br]\implies P \ge \frac{V-\lambda^2}{k^2-\lambda^2}\text{ as required.}

Reply 71

DFranklin

STEP III, 1996, Q10 - another very nice mechanics question. This one really does need a diagram though (attached).

Most of the forces should be self explanatory, but I'll add a few words. Firstly, I'll refer to the top log as L1 and the bottom log as L2. All contact forces are resolved into components parallel and perpendicular to the slope. R is the reaction between the two cylinders. F is the frictional force between the two cylinders, F acts clockwise on L1, anticlockwise on L2. A is the point of contact of L1 with the slope, B is the point of contact of L2 with the slope.

Taking moments about A:

W_1 \sin \alpha - R - F = 0

.
Taking moments about B:

W_2 \sin \alpha + R - F = 0

.

Adding and subtracting gives:

2R = (W_1-W_2)\sin \alpha,\quad 2F = (W_1+W_2)\sin \alpha

.

Now

R\ge 0

, and in fact, since

W_1 \sin \alpha > 0

and

F \le \mu R, \quad R\neq0 \text{ and so }R>0

. Thus

W_1 > W_2

, which is (i).

Again, using

F \le \mu R

, we have

\mu \ge \frac{F}{R} = \frac{W1+W2}{W1-W2}

, which is (ii).

For (iii), take moments about the centers of L1, L2 to see that

F_1 = F_2 = F

.

Now resolve for L1 perpendicular to the slope:

W_1 \cos \alpha = R_1 + F \implies R_1 = W_1 \cos \alpha - F

Now

F_1 \le \mu_1 R_1

, and so since

F_1 = F

we have:

Unparseable latex formula:

\displaystyle[br]\mu_1 \ge \frac{F}{R_1} = \left(\frac{W_1 \cos \alpha - F}{F}\right)^{-1}= \left(\frac{W_1 \cos \alpha}{F} - 1\right)^{-1}\\[br]=\left(\frac{2W_1 \cos \alpha}{(W_1+W_2)\sin \alpha} - 1\right)^{-1} = \left(\frac{2W_1 \cot \alpha}{W_1+W_2} - 1\right)^{-1}

.

Similarly resolving for L2 gives

W_2 \cos \alpha = R_2 - F \implies R_1 = W_2 \cos \alpha + F

and the same calculations to the above give

\displaystyle \mu_2 \ge \left(\frac{2W_2 \cot \alpha}{W_1+W_2} + 1\right)^{-1}

Comment: The difficult thing with this question is drawing a sensible diagram and working out where to get started. But once you get going, there are lots of useful relationships around. I'm particularly impressed by the fact the question doesn't need any advanced knowledge: all I've done is resolve forces, take moments, and use the basic definition for friction.

SIII_1996_Q10.JPG11.7KB

Reply 72

evariste_3086

DFranklin

W_1 \sin \alpha - R - F = 0

.
Taking moments about B:

W_2 \sin \alpha + R - F = 0

.

Adding and subtracting gives:

2R = (W_1-W_2)\sin \alpha,\quad 2F = (W_1+W_2)\sin \alpha

.

Now

R\ge 0

, and in fact, since

W_1 \sin \alpha > 0

and

F \le \mu R, \quad R\neq0 \text{ and so }R>0

. Thus

W_1 > W_2

, which is (i).

Again, using

F \le \mu R

, we have

\mu \ge \frac{F}{R} = \frac{W1+W2}{W1-W2}

, which is (ii).

For (iii), take moments about the centers of L1, L2 to see that

F_1 = F_2 = F

.

Now resolve for L1 perpendicular to the slope:

W_1 \cos \alpha = R_1 + F \implies R_1 = W_1 \cos \alpha - F

Now

F_1 \le \mu_1 R_1

, and so since

F_1 = F

we have:

Unparseable latex formula:

.

Similarly resolving for L2 gives

W_2 \cos \alpha = R_2 - F \implies R_1 = W_2 \cos \alpha + F

and the same calculations to the above give

\displaystyle \mu_2 \ge \left(\frac{2W_2 \cot \alpha}{W_1+W_2} + 1\right)^{-1}

Nice work. As you mention you done some of this stuff 20 years ago i just wondered what you do, apart from visit forums, to keep your maths level up? I want to but find it a real struggle ploughing through books just for personal reasons! I need some exams at the end of it to motivate.

edit: should say university level, the A-level stuff i can still manage (on a good day :smile:

) but find it harder to do the uni stuff.

Reply 73

DFranklin

evariste

Nice work. As you mention you done some of this stuff 20 years ago i just wondered what you do, apart from visit forums, to keep your maths level up?

Not a lot, really - that's why I'm doing this, I guess. I'm somewhat a frustrated academic - always expected to go into lecturing but it didn't happen. My real job is in computer graphics; some math involved there but not a whole lot. It pays a lot better than academia though!

edit: should say university level, the A-level stuff i can still manage (on a good day :smile:

) but find it harder to do the uni stuff.

It's similar here. There's some stuff I remember well from university, particularly analysis, which was always my best subject. Other stuff is more variable. These STEP mechanics questions really don't require much "knowledge", so I find them surprisingly accessible considering I've done no mechanics since A-level (other than a few bits of simulation for graphics work).

But going through the STEP pure, it's funny what "blank" spots I've found in my memory. All the "sin A sin B = ..." formulae, for example, which I've not had occasion to use more-or-less since A level. You're definitely expected to know them for STEP!

Reply 74

ad absurdum

I've got most of the mechanics done, I'll try and type some more up tomorrow. For now:

STEP I Q11

Let t₁ be the time at which the particle has travelled a horizontal distance c at which time the height is h, let t₂ be the time at which the particle has travelled a horizontal distance c+d at which time the height is zero. Then:

vt_1\cos\alpha = c

vt_1\sin\alpha - \frac{1}{2}gt_1^2=h

t_1=\frac{c}{v\cos\alpha}

. Substitute into the second equation giving

c\tan\alpha - \frac{gc^2}{2v^2\cos^2\alpha}=h

. Rearrange to get

\frac{g}{2v^2\cos^2\alpha} = \frac{c\tan\alpha - h}{c^2}

- call this equation (1).

And also

vt_2\cos\alpha=c+d

vt_2\sin\alpha = \frac{1}{2}gt^2

. So

t_2 = \frac{c+d}{v\cos\alpha}

. Substitute into the second equation giving

(c+d)\tan\alpha = \frac{g(c+d)^2}{2v^2\cos^2\alpha}

. Rearrange to get

\frac{g}{2v^2\cos^2\alpha} = \frac{\tan\alpha}{c+d}

- call this equation (2).

Combine (1) and (2) to get:

\frac{\tan\alpha}{c+d} = \frac{c\tan\alpha - h}{c^2}

. Rearrange to get

\tan\alpha = \frac{h(c+d)}{cd}

, QED.

Go back to equation (1), and substitute in what we just found for tan(alpha). This gives:

\frac{h(c+d)}{d} - \frac{gc^2\sec^2\alpha}{2v^2} = h

v^2 = \frac{gc^2d\sec^2\alpha}{2h(c+d)-hd}

v^2 = \frac{gc^2d(1+\tan^2\alpha)}{2h(c+d)-hd}

Substitute in tan(alpha) again, to get:

v^2 = \frac{gcd}{2h} + \frac{gh(c+d)^2}{2cd}

v^2 = \frac{g}{2}(\frac{cd}{h} + \frac{(c+d)^2h}{cd})

,QED.

Reply 75

DFranklin

STEP III,Q11. Having spent ages doing a diagram, I realise I've written

\alpha

for

\theta

in it and I can't be bothered to change it now (sorry). The diagram is pretty essential for this question; there are a lot of relevant angles to be marked. So many, in fact, that I introduced

\beta=\pi/2 - \alpha

just to get things to fit. Anyhow, on with the show, so to speak...

First thing to notice (or in my case, dimly recall from geometry that you did at GCSE and never looked at since!), is that

\angle ABC

is a right angle. So

AB = 2a \cos \theta, BC = 2a \sin \theta

(N.B. this is the geometrical equivalent of the

\sqrt{(1 + \cos 2\theta)^2+\sin^2 2\theta} = 2 \cos \theta, \quad \sqrt{(1 - \cos 2\theta)^2+\sin^2 2\theta} = 2 \sin \theta

identities that pop up a lot in STEP).

So the tensions

T_{AB}, T_{BC}

in the strings are given by

T_{AB} = \lambda (2 \cos\theta - 1), T_{BC} = \lambda (2 \sin\theta - 1)

.

Let O be the center of the circle, and put

\phi = \angle AOB

. Then

\phi = \pi - 2\theta

.

Resolving along the circumference of the circle (i.e. resolve along the tangent at B), in the direction of increasing

\phi

(as marked in the diagram), we get a net force of:

T_{BC}\cos \theta - T_{AB} \sin\theta + mg \sin \phi

=\lambda(\sin \theta - \cos \theta) + mg \sin 2\theta

Thus

ma\ddot{\phi} = \lambda(\sin \theta - \cos \theta) + mg \sin 2\theta

. Since

\phi = \pi - 2\theta

we have,

\ddot{\phi} = -2\ddot{\theta}

, giving

2ma\ddot{\theta} = \lambda(\cos \theta - \sin \theta) - mg \sin 2\theta

as required.

From here, it's all pretty much plain sailing:

\sin \theta = \frac{3}{5}, \cos \theta = \frac{4}{5} \implies \sin 2\theta = 2 \frac{3}{5}\frac{4}{5} = \frac{24}{25}

. At equilibrium,

\ddot{\theta} = 0

, so we must have:

\lambda(\frac{4}{5}-\frac{3}{5}) = \frac{24}{25} mg \implies 5 \lambda = 24 mg

. Substituting this value back in, we have:

10ma\ddot{\theta} = 24mg(\cos \theta - \sin \theta) - 5mg \sin 2\theta

\implies 5a\ddot{\theta} = 12g(\cos \theta - \sin \theta) - 5g \sin \theta \cos \theta

We want to examine the behaviour for small variations about the equilibrium point, so set

\alpha = \sin^{-1}\frac{3}{5}

and write

\theta = \alpha + \delta

, where

\delta

is small.

Then we have

5a\ddot{\delta} = 12g(\cos (\alpha+ \delta) - \sin (\alpha+ \delta)) - 5g \sin (\alpha+ \delta) \cos (\alpha+ \delta)

Now up to terms of order

\delta,

we have

\sin \delta \approx \delta, \cos \delta \approx 1

. So

\sin(\alpha+\delta) \approx \sin \alpha \cos \delta + \cos \alpha \sin \delta = \frac{3+4\delta}{5}

\cos(\alpha+\delta) \approx \cos \alpha \cos \delta - \sin \alpha \sin \delta = \frac{4-3\delta}{5}

And then

\sin(\alpha+\delta)\cos(\alpha+\delta) \approx \frac{(3+4\delta)(4-3\delta)}{25} \approx \frac{12+7\delta}{25}

Putting all this together, we have:

\frac{5a}{g} \ddot{\delta} = \frac{12}{5}(4-3\delta-3-4\delta) - \frac{1}{5}(12+7\delta)

\implies \frac{25a}{g} \ddot{\delta} = 12(1-7\delta) - (12+7\delta) = -91 \delta

\implies \ddot{\delta} = -\frac{91g}{25a} \delta

Thus the oscillations have period

2\pi \sqrt{\frac{25a}{91g}}

.

Comment: This question looks a lot worse than it is. I found the most difficult part was keeping track of all the angles when resolving etc. The 2nd half looks long when written out but really didn't take very long in practice.

S3_1996_Q11.png10.7KB

Reply 76

DFranklin

STEP III, 1996, Q14.

Expected number of punctures =

\alpha T

.

p(repairing puncture at time

t + \delta t

) = p(repairing puncture at time t and didn't finish repairing it by time

t + \delta t

) + p(not repairing puncture at time t and got a puncture by time

t + \delta t

) =

p(t)(1-\beta \delta t) + (1-p(t))\alpha \delta t

.

So

p(t+\delta t) = p(t) + (\alpha (1-p(t)) - \beta p(t)) \delta t

\implies \frac{dp}{dt} = \alpha(1-p) - \beta p =\alpha -(\alpha+\beta)p

Put

p(t) = \frac{\alpha}{\alpha+\beta}(1-e^{-(\alpha+\beta)t})

.
Then

p'(t) = \frac{\alpha}{\alpha+\beta}(\alpha+\beta)e^{-(\alpha+\beta)t}

= (\alpha+\beta)\frac{\alpha}{\alpha+\beta}((e^{-(\alpha+\beta)t}-1)+1)

-(\alpha+\beta)p(t)+\alpha

.

So the given function p(t) satisfies the differential equation. The initial condition is p(0) = 0 (since the bike starts in working order), and we see the given function satisfies this as well (as e^0 = 1).

Expected time spent mending punctures

E_T = \int_0^T p(t) dt

= \frac{\alpha}{\alpha+\beta} \int_0^T 1 - e^{-(\alpha+\beta)t} dt

= \frac{\alpha}{\alpha+\beta} \left[ t + \frac{e^{-(\alpha+\beta)t}}{\alpha+\beta}\right]_0^T

= \frac{\alpha}{\alpha+\beta} (T + \frac{e^{-(\alpha+\beta)T} - 1}{\alpha+\beta})

When

(\alpha+\beta)T

is small,

e^{-(\alpha+\beta)} \approx (1 - (\alpha+\beta)T+ (\alpha+\beta)^2 \frac{T^2}{2})

(+ terms in

((\alpha+\beta)T)^3

and higher).

So

E_T / T \approx \frac{1}{T} \frac{\alpha}{\alpha+\beta}(T+\frac{1- (\alpha+\beta)T+ (\alpha+\beta)^2 T^2/2 -1}{\alpha+\beta})

= \frac{1}{T} \frac{\alpha}{\alpha+\beta} (\frac{(\alpha+\beta)^2T^2/2}{\alpha+\beta}) = \frac{\alpha T}{2}

as required.

For the case when

T(\alpha+\beta)

is large, put

S = T(\alpha+\beta)

E_T/T = \frac{\alpha}{\alpha+\beta} (1 + \frac{e^{-(\alpha+\beta)T} - 1}{T(\alpha+\beta)})

= \frac{\alpha}{\alpha+\beta}(1 + \frac{e^{-S}-1}{S}

Then since

\frac{e^{-S}}{S} \to 0, \frac{1}{S} \to 0 \text{ as } S \to \infty

, we have

E_T/T \approx \frac{\alpha}{\alpha+\beta}

when S (

= T(\alpha+\beta)

) is large.

Reply 77

dr_98_98

STEP I Q9

The first part just uses conservation of energy, and if we call the point at which the rope becomes taught B (which is actually the at the point of natural length) then

PE at A = KE at B

mgl = \frac{1}{2}mv^2

v= \sqrt (2gl)

For the second part, if we call the lowest point she reaches C, and if she only just avoids hitting the ground, that means her kinetic energy at C must be zero, so with these lengths

AC = h

BC = h-l

and by the conservation of energy:

PE at A = Elastic Potential Energy at C

mgh = \frac{mg}{2kl}(h-l)^2

2klh= (h-l)^2

h^2 -2hl(k+1) +l^2 = 0

Solving this for h gives

h = l(k+1) \pm l\sqrt(k^2+2k)

But considering that

h>l

h = l(k+1) + l\sqrt(k^2+2k)

Finally, the maximum speed occurs when

\frac{dv}{dt} = 0

. Suppose this occurs at a point D

By Newton Two, which says

mg - T = ma

We must have

T=mg

If we call the extension at D

x_1

then

mg = \frac{mgx_1}{kl}

x_1 = kl

So length

AD = kl + l = l(k+1)

Using conservation of energy again:

PE at A = KE at D + Elastic Potential Energy at D

mgl(k+1) = \frac{1}{2}mv^2 + \frac{mgl^2k^2}{2hl}

v^2 = 2gl(1+k) - gkl

v^2 = (k+2)gl

Reply 78

speedy_s

Speleo

STEP II Question 5

u = z + 1/z
z^4 + z^3 + z^2 + z + 1 = 0
z^2 + z + 1 + 1/z + 1/z^2 = 0
1 + u + u^2 - 2 = 0
u^2 + u - 1 = 0
u = [-1 +- rt5]/2
z^2 + z[1 -+ rt5]/2 + 1 = 0
z = {[-1 +- rt5]/2 +- sqrt[[-1 +- rt5]^2/4 - 4]}/2
z = -(1/4) +- rt(5)/4 +- (1/2)sqrt[-(5/2) -+ (1/2)rt5]
z =
-(1/4) + rt(5)/4 + (1/2)sqrt[-(5/2) - (1/2)rt5]
-(1/4) + rt(5)/4 - (1/2)sqrt[-(5/2) - (1/2)rt5]
-(1/4) - rt(5)/4 + (1/2)sqrt[-(5/2) + (1/2)rt5]
-(1/4) - rt(5)/4 - (1/2)sqrt[-(5/2) + (1/2)rt5]

z^4 + z^3 + z^2 + z + 1 = 0
z^4 + z^3 + z^2 + z = -1

z^5 + z^4 + z^3 + z^2 + z = 0
z^5 - 1 = 0

z = cis(2pi/5), cis(4pi/5), cis(6pi/5), cis(8pi/5)
cos(2pi/5) = Re(z) = [-1 + rt5]/4 (look at cos graph for sign)
sin(2pi/5) = Im(z) =(1/2)sqrt.../i = (1/2)sqrt[(5/2) + (1/2)rt5] = (1/4)sqrt(10 + 2rt5) (see graph to see which is which)