Physics Reference

Monday, May 26, 2014

Physics 9702 Notes | Worked Solutions for Past Papers

Physics 9702 (A-Level) Notes & Worked Solutions for Past Exam Papers

This page contains links to notes for the different chapters of the 9702 Physics syllabus (AS and A-Level) as well as worked solutions (explanations + details) for past exam papers. You may clear any of your doubts at the respective pages.

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More notes and worked solutions will be added soon. Be sure to check again frequently.

Some solutions are temporarily displayed here until solutions for the complete paper is available later.

Physics 9702 (A-Level) Guide

Physics 9702 (A-Level) Notes

Graphs:
Physics Graphs Basics 1 – Coordinates of Points | 2 - Nature of Gradients |

Units - Base (Fundamental) and Derived | Homogeneity
Uncertainty - no. of significant figures (Check Solution 1022)

Adding vectors graphically (Check Solution 703)

Types of Forces - Gravitational, Electric, Frictional, Viscous drag and Upthrust
Weightlessness (Check Solution 989)

Springs in series and in parallel (Check solution 925)

Circular motion in a vertical plane
Derive escape velocity (Check Solution 338)

Kepler's 3rd law of Gravitation (Check Solution 948)

Concept of 'Internal Energy' + examples (Check Solution 588)

Diffraction of water water in ripple tank (Check Solution 858)
How to draw waves with a phase difference? (Check Solution 318)

Ideal diode (Check Solution 675)

Quantisation of Charge (Check Solution 727)

Specific charge (Check Solution 971)

Binary numbers (Check Solution 555)
ADC and DAC: Reproduction of an input signal (Check Solution 667)

Physics 9702 Summary notes (by Zubair Junjunia, Emir Demirkhan & Saif Asmi)
AS:
1. Physical Quantities and Units | 2. Measurement Techniques | 3. Kinematics | 4. Dynamics | 5. Forces |

Complete A-Level Physics 9702 notes for syllabus 2015 (Credit: Neha of India)

Physics 9702 (A-Level) Worked Solutions for Past Papers

Question:

can you tell me if i am reading a volume on a cylinder.....

what error may occur?

and what are the possible improvements?

Explanation:

The measurements to be made are that of the diameter and the length of the cylinder using a vernier caliper (for both).

So, it is obvious that you should know how to handle and record measurements from a vernier caliper. Also, handle the caliper with care and take the necessary precautions (do not squeeze the object, insert object perpendicularly (not diagonally), …)

A possible error that may arise is zero error. This can be prevented by first bringing the jaws of the caliper together, before taking any measurements, and verify whether the zero mark of the vernier scale coincide with the zero mark of the main scale. If it does then you may record measurements directly. If not, then apply the appropriate correction before recording the measurements.

Measurements of the diameter should be taken at different positions (at least 3) along the cylinder and an averaged value can be calculated. The same is done for the length (you can take 2 or 3 measurements).

Question:

Proper definition for a projectile?

Explanation:

An object upon which the only force that acts on it is gravity (that is, other forces such as air resistance is negligible) is called a projectile.

In 1-dimenion motion, an example of a projectile is an object thrown vertically upward (neglecting air resistance). This object would have a constant downward acceleration. The path of this object is along a straight line.

In 2-dimensions, a projectile can be an object thrown at an angle to the horizontal with a constant velocity. This object would have a constant horizontal component of velocity, while its vertical component would be subject to a constant downward acceleration (due to gravity). Again, air resistance is negligible. The path of this object is a trajectory.

Sunday, May 25, 2014

Complex Analysis: #28 Stirling`s Formula

Complex Analysis: #28 Stirling`s Formula

The functional equation for the gamma function, Γ(z + 1) = zΓ(z), together with the observation that Γ(1) = 1, shows that for all n ∈ ℕ, we have Γ(n + 1) = n!. In the analysis lecture, I gave a very simple approximation to the gamma function. Namely, for large integers n, we have n! ≈ nⁿe⁻ⁿ. But this is really a very inexact approximation, which we were able to derive with hardly any thought at all. In this section, we will be looking at the “genuine” Stirling formula. It is

Complex Analysis: #28 Stirling`s Formula equation pic 1

where P₁(t) = t − [t] − 1/2, and [t] is the largest integer which is less than or equal to t. Of course we take the principle value of the logarithm here. The fact that, as long as z is not a negative real number, the integral on the right-hand side of the equation goes to zero as |z| → ∞, shows that

Complex Analysis: #28 Stirling`s Formula equation pic 2

(at least if we stay away from the negative real numbers). In fact we will show that for a given ﬁxed δ between 0 and π, we have that P₁(z) falls uniformly to 0, for z = re^iθ and |θ| ≤ π − δ.

Lemma 1 (Euler’s Summation formula)
Let f : ℝ → ℝ be continuously differentiable (in the sense o real analysis). Then

Complex Analysis: #28 Stirling`s Formula equation pic 3

Taking the sum from k = 1 to n then gives the result.

Similarly we have

Lemma 2
Assume z is not a negative real number or zero. Let P₂(t) = t_∗(t_∗ − 1)/2, where t_∗ = t − [t] for t ∈ ℝ. Then we have

Complex Analysis: #28 Stirling`s Formula equation pic 4

The function Φ which is so defined is analytic in ℂ (minus the negative reals and 0).

Proof
Using partial integration, we have

Complex Analysis: #28 Stirling`s Formula equation pic 5

where K is a ﬁnite number. (When z is a positive real number, the integral obviously also convergesThus we can use Weierstrass’ convergence theorem — on the integrals between successive integer— to conclude that Φ is an analytic function.

Lemma 3
We have

Complex Analysis: #28 Stirling`s Formula equation pic 6

Proof
This follows from the previous lemma, since then K = 0 and π/2y → 0, when y → ∞.

So now that we have looked at these lemmas, let us begin proving Stirling’s formula. It involves the standard, rather awkward problem of the logarithms. So to avoid this, let us ﬁrst take the case that z is just a positive real number. That is, we take z = x + i · 0, and we set about examining log Γ (x). Obviously Γ (x) > 0, so we can use the simple, real logarithm function. The antiderivative of log(x) is x log(x) − x. Therefore, using lemma 1, we see that for n ∈ ℕ we have

Complex Analysis: #28 Stirling`s Formula equation pic 7

is simply some constant real number, let’s call it C. But then we see that the left-hand side of the equation, namely log Γ (x) is just the restriction to the positive real numbers of an analytic function which is defined in the region consisting of C with the negative real numbers and 0 removed. Similarly the right-hand side, namely xlog x − (log x)/2 − x − Φ(x) + C is the restriction to the positive real numbers of another analytic function which is defined in the same region. Since both of these functions coincide on the positive real numbers, they must be same analytic function.

So the only remaining problem is to ﬁnd out what the value of C is. Writing

Γ(1 − z) = −zΓ(−z),

we have

Complex Analysis: #28 Stirling`s Formula equation pic 8

This is true for all y > 0, no matter how large. But according to lemma 3, we have lim_y→∞Φ(iy) = 0. For the rest, we should remember that for a complex number of the form re^iθ, we have

Complex Analysis: #28 Stirling`s Formula equation pic 9

Saturday, May 24, 2014

Complex Analysis: #27 Gamma Function II

Complex Analysis: #27 The Gamma Function II

But what is the relationship with the formula

Complex Analysis: #27 The Gamma Function II equation pic 1

for x a real number greater than 1, which we used in the analysis lectures? Going from real to complex numbers, let us write z = x + iy. Then we have

Therefore — as we saw in Analysis I — the integral defining the gamma function will again converge when x > 1; for x < 1 it diverges.

As an exercise, we see that for z = x + iy and x > 0, we have

Complex Analysis: #27 The Gamma Function II equation pic 3

This is another standard formula for the gamma function, and in fact it is valid as long as Re(z) > 0. But we still haven’t shown that it is the same gamma function as that which was defined in the previous section.

For each n ∈ ℕ, let

Complex Analysis: #27 The Gamma Function II equation pic 4

and we see the connection with the definition in the previous section.

As a ﬁnal remark before proceeding with Stirling’s formula, I should mention that Legendre introduced the beta function, which is

Complex Analysis: #27 The Gamma Function II equation pic 5

This seems to be used these days in computer algebra programs. Also, when on that subject, one ﬁnds many interesting and obscure formulas related to the gamma function. (You can also look in books containing tables of mathematical formulas.) One such formula which caught my eye is the following. Let n ∈ ℕ. Then

Complex Analysis: #27 The Gamma Function II equation pic 6

Here n!! is the double factorial. That is

Complex Analysis: #27 The Gamma Function II equation pic 7

Friday, May 23, 2014

Complex Analysis: #26 Gamma Function I

Complex Analysis: #26 The Gamma Function I

In the analysis lecture, we defined the gamma function for real numbers x > 1 using the improper integral

Complex Analysis: #26 The Gamma Function I equation pic 1

We can do the same thing here in the realms of complex analysis. Just substitute the complex number z for the real number x in the formula. Using the same argument as in the real case, we ﬁnd that the integral converges if Re(z) > 1. But by doing this, we miss out on the fact that the gamma function can be defined everywhere in ℂ (with isolated poles). Of course it is again possible to look at the functional equation for the gamma function, then bring in analytic continuation. But this is a rather complicated manipulation! In reality, an inﬁnite product representation is simpler.

Let’s begin by looking at the following function

Complex Analysis: #26 The Gamma Function I equation pic 2

We will be using the second expression here, although the ﬁrst expression shows that we can apply theorem 48 and conclude that the product does converge to an entire function, whose zeros are simply the set of negative real integers. Taking a look at the product representation of the sine function which we obtained in the last section, we see that

Complex Analysis: #26 The Gamma Function I equation pic 3

The function zG(z) obviously has as it’s zeros the negative real integers, and also 0. But then, if we take the function G(z − 1), we see that it also has the negative real integers and 0 as it’s set of zeros. Since all of these zeros are simple zeros for both functions, we have

Complex Analysis: #26 The Gamma Function I equation pic 4

being an entire function which is never zero. It is now an exercise to show that there must exist an entire function γ : ℂ → ℂ such that

ze^γ(z)G(z) = G(z − 1)

for all z ∈ ℂ.

Let’s take the logarithmic derivative on both sides of the equation. We get

Complex Analysis: #26 The Gamma Function I equation pic 5

so the sum is certainly absolutely convergent and thus the third equation is valid. But this means that γ '(z) = 0, that is, γ(z) is a constant, which we simply denote by γ. In fact it is Euler’s constant.

To see this, consider the case z = 1. From the formula defining G, we certainly have G(0) = 1. Therefore G(0) = 1 = e^γG(1), or

Complex Analysis: #26 The Gamma Function I equation pic 6

So G is an entire function satisfying the functional equation G(z−1) = ze^γG(z). In order to improve the appearance of things, let us define H(z) = G(z)e^z·γ. Then we have the functional equation

H(z − 1) = zH(z).

This is beginning to look like the functional equation for the gamma function, but unfortunately (or fortunately?) it is going in the “wrong” direction. To ﬁx this up, we take

Complex Analysis: #26 The Gamma Function I equation pic 7

From the construction, we see that it has simple poles at the negative integers, and at 0, but it has no zeros. Also, looking at the equation for the sine function, we can express this in terms of the gamma function:

Complex Analysis: #26 The Gamma Function I equation pic 8

Thursday, May 22, 2014

Complex Analysis: #25 Some Infinite Products

Complex Analysis: #25 Some Infinite Products

Genus zero:
Given a sequence of non-zero points a_n with ∑ 1/|a_n| converging, we can simply say that

Complex Analysis: #25 Some Infinite Products equation pic 1

gives us an example of a genus zero function with zeros just where we want them. Of course if we also want a zero at 0, then we can multiply this product with the factor Cz^l say, where l is the order of this zero, and C is a non-zero constant which we can choose as we like.

The sine function:
Being more concrete, let us look at the function sin πz. According to the definition of the sine function, we have

Complex Analysis: #25 Some Infinite Products equation pic 2

But this can only be zero if e^iπz = e^−iπz. Writing z = x + iy, this means that e^−ye^iπx = e^ye^−iπx. In particular, y = 0 so that the zeros are just the familiar zeros which we know from real analysis, namely the integers, ℤ. As we know, the series ∑ 1/n² converges, thus the complex sine function must have genus one. Writing

Complex Analysis: #25 Some Infinite Products equation pic 3

the problem is then to determine the function g. For this, we take the logarithmic derivatives of both sides. We obtain

Complex Analysis: #25 Some Infinite Products equation pic 4

so that the sum is uniformly convergent in compact subsets of ℂ which do not contain points of ℤ, thus showing that the derivative of the sum is the sum of the derivatives.)

Therefore

Complex Analysis: #25 Some Infinite Products equation pic 5

where C is a constant. On the other hand, we have both cot(w) = −cot(−w) and χ(w) = −χ(−w). That is, they are both anti-symmetric. But this can only be true if the constant C = 0. But this implies that our function g satisfies the equation g '(z) = 0; that is, g is a constant. Since

Complex Analysis: #25 Some Infinite Products equation pic 6

Tuesday, May 20, 2014

Complex Analysis: #24 Infinite Products of Functions

Complex Analysis: #24 Infinite Products of Functions

For example, we have seen in exercise 6.3 that for z not a real integer, we have the series

Complex Analysis: #24 Infinite Products of Functions equation pic 1

converging. In fact it is absolutely convergent. Therefore, according to theorem 47, we must have

Complex Analysis: #24 Infinite Products of Functions equation pic 2

converging, and so defining a function, for all z ∉ ℤ. But is this function meromorphic? It is a small exercise to see that the sequence of partial products is uniformly convergent on compact subsets of ℂ \ ℤ, and thus according to Weierstrass’ convergence theorem, the function is analytic in ℂ \ ℤ. So this example shows one way to proceed in this special case.

On the other hand, polynomials seem to give us a more natural basis for generating analytic functions. Given a polynomial P(z), we can write it as

P(z) = a(a₁ − z) · · · (a_n − z),

where a₁, . . . , a_n are the zeros of the polynomial (perhaps some with multiplicity greater than one), and a ≠ 0 is some constant. If we generalize this to an inﬁnite product, and if we hope that things will converge, then we expect the terms to converge to 1. So it is natural to write something like this:

Complex Analysis: #24 Infinite Products of Functions equation pic 3

If the sequence of absolute values |a_n| grows sufficiently rapidly, then we might expect to have convergence. But even a simple sequence like a_n = n does not satisfy this property. Such considerations led to Weierstrass’ product theorem.

Theorem 48 (Weierstrass)
Let (a_n)_n∈ℕ be a sequence of complex numbers with a_n ≠ 0 for all n, and lim_n→∞ a_n = ∞. Then there exist integers m_n ≥ 0 such that the product

Complex Analysis: #24 Infinite Products of Functions equation pic 4

In fact, we can be more specific than this.

Corollary
Let f : ℂ → ℂ be a non-constant entire function. Then there exists an entire function g such that

Complex Analysis: #24 Infinite Products of Functions equation pic 5

we have that the function given by φ(z) = f(z)/Ψ(z) is an entire function (with removable singularities) with φ(z) ≠ 0, for all z ∈ ℂ. But then φ'(z)/φ(z) is also an entire function, with antiderivative g(z), say. Let’s look at the derivative of the function φ(z)e^−g(z). We get

So φ(z)e^−g(z) is a constant, which we can absorb into g in such a way that φ(z)e^−g(z) = 1

A simple consequence of this theorem is that every meromorphic function defined throughout ℂ is a rational function. That is, let f be meromorphic, with poles a₁, a₂, . . . . [ Of course if there are only ﬁnitely many poles, then this is trivial. On the other hand, since the poles are isolated, there can be at most countably many of them.] Regardless of the way the poles are numbered, since they are isolated we must have lim_n→∞ a_n = ∞. Therefore take the product Ψ · f, where Ψ is an entire function with the same set of zeros as f has poles (weighted according to their orders). Thus Ψ · f is — apart from a set of isolated, removable singularities — an entire function; call it g : ℂ → ℂ. Therefore we can write f = g/Ψ.

Looking at the the formula in Weierstrass’ theorem, we see that it can become a bit of a mess, particularly then the numbers m_n get larger and larger. Things are nicer if there exists a single number m, such that Weierstrass’ theorem works with all m_n ≤ m.

Deﬁnition 16
Given a countable set of non-zero complex numbers {a_n}, if there exists an integer m ≥ 0 such that the product

Complex Analysis: #24 Infinite Products of Functions equation pic 7

converges, then this is called the canonical product associated with the sequence (a_n)_n∈ℕ. The smallest such m is called the genus of the canonical product.

When does the canonical product exist? That is to say, for a ﬁxed non-negative integer m, and for each a_n, we have a “remainder term” of the form

Complex Analysis: #24 Infinite Products of Functions equation pic 8

Monday, May 26, 2014

Physics 9702 Notes | Worked Solutions for Past Papers

Physics 9702 (A-Level) Notes & Worked Solutions for Past Exam Papers

Physics 9702 (A-Level) Guide

Physics 9702 (A-Level) Notes

Physics 9702 (A-Level) Worked Solutions for Past Papers

Sunday, May 25, 2014

Complex Analysis: #28 Stirling`s Formula

Complex Analysis: #28 Stirling`s Formula

Saturday, May 24, 2014

Complex Analysis: #27 Gamma Function II

Complex Analysis: #27 The Gamma Function II

Friday, May 23, 2014

Complex Analysis: #26 Gamma Function I

Complex Analysis: #26 The Gamma Function I

Thursday, May 22, 2014

Complex Analysis: #25 Some Infinite Products

Complex Analysis: #25 Some Infinite Products

Tuesday, May 20, 2014

Complex Analysis: #24 Infinite Products of Functions

Complex Analysis: #24 Infinite Products of Functions