## Pages

News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

## Wednesday, 31 August 2011

### Math GRE - #30

What is $\int_{-3}^3{|x+1|\,dx}?$
1. 0
2. 5
3. 10
4. 15
5. 20

Solution :

This is a very basic question. The usual method of dealing with absolute values in integrals is to split it into a sum of two integrals and remove the absolute values. Since $|x+1|$ is negative on $[-3, 1)$ and positive on $[-1, 3]\,\,\,$ , we can split the integral into: $\int_{-3}^3{|x+1|\,dx}=-\int_{-3}^{-1}{(x+1)\,dx}+\int_{-1}^3{(x+1)\,dx}=10.$ Another way of doing this problem is to imagine the graph of $|x+1|$ and realize that the integral is the sum of two 45-45-90 triangles (one with base 2 and height 2,  the other with base 4 and height 4). Simply add the area of triangles up and we're done.

Bonus question: Given that $c$ is a constant, what is $\int_{-\infty}^{\infty}{e^{-|x+c|}\,dx}?$

### Physics GRE - #30

The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression:

1. $A\left(n+\frac{1}{2}\right)$
2. $A(1-n^2)$
3. $A\left(\frac{1}{n^2}-\frac{1}{4}\right)$
4. $An^2$
5. $-\frac{A}{n^2}$

Solution :

Recall that the energy levels of the hydrogen atom is given by $E_n = -\frac{13.6\,\text{eV}}{n^2}.$ It's just one of those things that you've got to remember, as it takes time you don't have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).

## Tuesday, 30 August 2011

### Math GRE - #29

If $z=e^{2\pi i/5},$ then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$
1. $0$
2. $4e^{3\pi i/5}$
3. $5e^{4\pi i/5}$
4. $-4e^{2\pi i/5}$
5. $-5e^{3\pi i/5}$

Solution :

Recall that the sum of all the $n-th$ roots of unity for $n>1$ ($n=5$ in this case) is 0. For this particular question, this means that $1+z+z^2+z^3+z^4 = 0.$ This motivates us to write the sum in the question as: $\begin{eqnarray*} (1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9 & = & 0+4z^4\cdot0+5z^9 \\ & = & 5z^9. \end{eqnarray*}$ Now we can easily evaluate the sum: $\begin{eqnarray*} 5z^9=5\left(e^{2\pi i/5}\right)^9 & = & 5\left(e^{2\pi i/5}\right)^4\cdot\left(e^{2\pi i/5}\right)^5 \\ & = & 5\left(e^{2\pi i/5}\right)^4 \\ & = &5e^{8\pi i/5}=-5e^{3\pi i/5}. \end{eqnarray*}$

### Physics GRE - #29

Solid argon is held together by which of the following bonding mechanisms?

1. Ionic bond only.
2. Covalent bond only.
3. Partly covalent and partly ionic.
4. Metallic bond.
5. Van der Waals bond

Solution :

Argon is not charged and stable as a monoatomic element (due to it's full orbitals). Thus, because it is stable alone, it does not ionically bond with itself since argon does not exist naturally as an ion. This eliminates choice 1.

Because covalent bonding occurs with electron sharing and the stability from filling unfilled orbitals, Argon does not covalently bond with itself since it already has full orbitals. This eliminates choice 2.

It cannot be partly ionic and partly covalent as we just stated above that it is to stable to do either. This eliminates choice 3.

Argon is not a metal and so it does not participate in metallic bonding (metallic bonding occurs in metals due to massive electron sharing among many metal atoms, but again, due to argon's completely filled orbitals, it is too stable to do this). This eliminates choice 4.

Van der Waals bonding occurs in all chemicals and compounds as they are mostly caused by instantaneously inducted dipoles due to movement of electrons in the atom. Argon is no exception to this rule. Hence choice 5 is the answer. Note that although Van der Waals bonding is always present, it is by no means the strongest bonding mechanism. If this question was applied to other elements, Van der Waals may not be the answer since it could be negligible compared to other bonding mechanisms.

## Monday, 29 August 2011

### Math GRE - #28

What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation: $y=\frac{1}{\sqrt{1+x^2}}?$
1. $\dfrac{\pi}{2}$
2. $\pi$
3. $\dfrac{\pi^2}{4}$
4. $\dfrac{\pi^2}{2}$
5. $\infty$

Solution :

Choice 4 is the correct answer.

Recall that the volume of revolution around the x-axis is given by the formula $V=\int{\pi y(x)^2}\,dx.$ In this case, we can substitute for $y$ and plug in the limits of integration to obtain: $V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.$
Almost everyone with some calculus experience has had the great displeasure of memorizing: $\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.$ Now we can put it to good use! We can see that the answer is (if you'll pardon my poor notation): $V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.$

Interesting tidbit: Here's a funny little thing you'll often see in first year calculus. It's called Gabriel's Horn. It's got finite volume but INFINITE surface area. Just something interesting I wanted to share.

### Physics GRE - #28

lalalala test element $^A_ZX$ decays by natural radioactivity in two stages to $^{A-4}_{Z-1}Y$. The two stages would most likely be which of the following?

 First stage Second stage 1. $\beta^-$ emission with an antineutrino $\alpha$ emission 2. $\beta^-$ emission $\alpha$ emission with a neutrino 3. $\beta^-$ emission $\gamma$ emission 4. Emission of a deuteron Emission of two neutrons 5. $\alpha$ emission $\gamma$ emission

Solution :

For choice 1, we have $^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e$ for the $\beta^-$ emission, and then $^A_{Z+1}X \rightarrow ^{A-4}_{Z-1}Y+^4_2\text{He}$ for the $\alpha$ emission. Everything works out, charge/lepton/baryon numbers are conserved as they should be.

For choice 2, lepton number is not conserved as the $\beta^-$ emission does not occur with an antineutrino.

For choice 3, the atomic number remains unchanged under both emissions.

For choice 4, the emissions specified are extremely rare. In other words, this type of emission is impossible for the 'natural radioactivity' specified in the question.

For choice 5, the atomic numbers do not match up (there are fewer protons than needed with this decay path).

## Sunday, 28 August 2011

### Physics GRE - #27

Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of $R/2$ where $R$ is the radius of the Earth. If $F(r)$ is the gravitational force of the Earth on a point mass at a distance $r$, what is: $\frac{F(R)}{F(2R)}?$

1. 32
2. 8
3. 4
4. 2
5. 1

Solution :

A very simple question today.

Recall that the gravitational force between two objects is given by $F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}$ where $k$ is a constant.
Hence we have: $\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.$

On the same exam, they also asked for $\frac{F(R)}{F\left(\frac{R}{2}\right)}.$ Try it if you want some extra practice.

### Math GRE - #27

Which of the following is the best approximation of $\sqrt{1.5}(266)^{3/2}$?

1. 1000
2. 2700
3. 3200
4. 4100
5. 5300

Solution :

With some simplification, we note that: $\begin{eqnarray*} \sqrt{1.5}(266)^{3/2}=\sqrt{\frac{3}{2}}\cdot(266)^{3/2} & = & 266\sqrt{\frac{3\cdot 266}{2}} \\ & = & 266\sqrt{399}\approx266\cdot20\approx 5300. \end{eqnarray*}$
Those of us familiar with Taylor series may try approximations with calculus, however this question teaches us that too much knowledge may be a little bit dangerous if misapplied. And that most of the time, there are simpler solutions than you'd think.

## Saturday, 27 August 2011

### Physics GRE - #26

An engine absorbs heat at a temperature of 727 degrees Celsius and exhausts heat at a temperature of 527 degrees Celcius. If the engine operates at maximum possible efficiency, for 2000 J of heat the amount of work the engine performs is most nearly:

1. 400 J
2. 1450 J
3. 1600 J
4. 2000 J
5. 2760 J

Solution :

Recall that the Carnot Efficiency of a heat engine is: $\eta=1-\frac{T_C}{T_H}$ where $T_C$ and $T_H$ are in Kelvins. For this problem, the efficiency is: $\eta\approx1-\frac{527+273}{727+273}=1-\frac{800}{1000}=0.2.$ Thus the maximum amount of work we can get is $W=\eta\cdot2000\,J=400\,J.$

### Math GRE - #26

A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?

1. $\dfrac{x^2}{9}$
2. $\dfrac{x^2}{8}$
3. $\dfrac{x^2}{4}$
4. $x^2$
5. $2x^2$

Solution :

Let's call one side of the fence $a$ and the other side $b$. Since we are only looking at 3 sides of a rectangular fence (e.g. we're fencing in our back yard and one side is the wall of the house), let's call the side length that we have to fence twice. So we have: $x = 2a+b.$ The area of the fence is $A = ab = a(x-2a)$ where the last equality comes from our equation for $x$. To maximize the area, we take the derivative of $A$ with respect to $a$ and set equal to 0 to obtain: $0 = x-4a\implies a=\frac{x}{4}.$ We know this represents the maximum since the second derivative of $A$ is always negative. Using our equation for the perimeter, we now know that: $x=2a + b\implies b=\frac{x}{2}.$ Hence the maximum possible area is: $A=ab=\frac{x}{4}\cdot\frac{x}{2}=\frac{x^2}{8}.$

## Friday, 26 August 2011

### Physics GRE - #25

A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally until the horizontal force is 10 N. The box is then in equilibrium as shown below.

The angle that the rope makes with the vertical is closest to:

1. $\arctan0.5$
2. $\arcsin0.5$
3. $\arctan2.0$
4. $\arcsin2.0$
5. $45^\circ$

Solution :

Since the block is in equilibrium, we must have $\vec{F}+\vec{F_g}=\vec{T}$ as shown below.

Hence $\theta=\arctan\frac{|\vec{F}|}{|\vec{F_g}|}=\arctan\frac{10}{2g}\approx\arctan\frac{10}{20}=\arctan0.5$.

### Math GRE - #25

Let $k$ be the number of roots to the equation $f(x)=e^x+x-2$ in the interval $[0, 1]$, and let $n$ be the number of roots that are not in $[0,1]$. Which of the following is true?

• $k=0$ and $n=1$
• $k=1$ and $n=0$
• $k=n=1$
• $k>1$
• $n>1$

Solution :

We can solve this problem by using the Intermediate Value Theorem.
By evaluating $f$ at 0 and 1, we see that: $f(0)=e^0+0-2<0$ $f(1)=e^1+1-2>0.$
By the Intermediate Value Theorem, there exists a $c\in[0,1]$ such that $f(c)=0.$ Thus we know for sure that there is at least one root in [0,1].

Taking the derivative of $f$, we note that: $f^\prime(x)=e^x+1>0$ for all $x$.
Thus $f$ is increasing everywhere and so it must only have one root. Since this root is in [0,1], $k=1.$ As we cannot have roots anywhere outside of [0,1], we also have $n=0$. This gives choice 2 as the answer.

Bonus question: The Intermediate Value Theorem may seem obvious, but it can also be used to prove many fun facts. One such fact is that if you draw a circle anywhere in the universe, there will be two points directly opposite to each other on the circle (i.e. 180 deg. apart) that have the exact same temperature. Can you prove this?

If you get stuck, here is some intuition as to why this interesting fact is true.

## Thursday, 25 August 2011

### Math GRE - #24

What is the greatest possible area of a triangular region with one vertex at the centre of a unit circle and the other two vertices on the circle?

1. $\frac{1}{2}$
2. $1$
3. $\sqrt{2}$
4. $\pi$
5. $\frac{1+\sqrt{2}}{4}$

Solution :

The area of any triangle with sides $a$, $b$, and angle $\theta$ between $a$ and $b$ is: $A = \frac{1}{2}ab\sin\theta$
Since the sides have the same length as the radius of the circle, $a=b=1.$
Therefore, $A=\frac{1}{2}\sin\theta$ which equals $\frac{1}{2}$ at the maximum of $\theta=\frac{\pi}{2}$.

### Physics GRE - #24

Assume $A$, $T$, $\lambda$ are positive constants. The equation $y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]$ represents a wave whose:

1. amplitude is $2A$
2. velocity is in the negative x-direction
3. period is $\frac{T}{\lambda}$
4. speed is $\frac{x}{t}$
5. speed is $\frac{\lambda}{T}$

Solution :

Choice 1 is incorrect because $\sin\theta\leq 1$ for any $\theta$ so $y\leq A$.

To see that choice 2 is incorrect, suppose that we are following the minimum wave as time passes (by changing $x$ with respect to time to a specific $x(t)$) in such a way that the wave seems to stand still (i.e. we are travelling at the same velocity as the wave). For the wave to 'stand still', we need $x(t)$ to satisfy $2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = const. = C$ Satisfying this relationship ensures that the wave always looks the same to us as: $y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right] = A\sin C = constant.$ Since we are following one of the minimums of the wave, we can choose $C = 0$ to get the minimum at $\sin 0$. Thus we get: $2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = 0 \implies x(t) = \frac{\lambda}{T}t.$ To get the sign of the velocity, we take the derivative of $x(t)$: $\frac{dx}{dt} = \frac{\lambda}{T}>0$ since $\lambda,T > 0$. This also tells us choice 5 is correct since the speed is  $\frac{\lambda}{T}.$

Choice 3 is wrong as the units do not work out.

Choice 4 is wrong because we just calculated the speed to be $\frac{\lambda}{T}.$

## Wednesday, 24 August 2011

### Physics GRE - #23

A particle is initially at rest at the top of a curved frictionless track.

The x and y coordinates of the track are related in dimensionless units by $y=\frac{x^2}{4}$, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration?
1. $0$
2. $g$
3. $\frac{gx}{2}$
4. $\frac{gx}{\sqrt{x^2+4}}$
5. $\frac{gx^2}{\sqrt{x^2+16}}$

Solution :

Choice 4 is the solution.

First we note that choice 2 cannot be the answer as there is a normal force from the track.

Now as $x\rightarrow\infty$, we should expect that the acceleration goes to $g$ as the tangent becomes nearly vertical. This makes choice 4 is the only possible answer.

We can also solve this question elegantly without resorting to limits and boundary conditions (however it will take a good half page). Can you see how it can be done?

### Math GRE - #23

Suppose B is a basis for a real vector space V of dimension greater than 1. Which of the following statements could be true?

1. The zero vector of V is an element of B.
2. B has a proper subset that spans V.
3. B is a proper subset of a linearly independent subset of V.
4. There is a basis for V that is disjoint from B.
5. One of the vectors in B is a linear combination of the other vectors in B.

Solution :

Let us first recall the definition of a basis:
"A basis for a vector space is a linearly independent set of vectors that spans the vector space".

B is linearly independent because it is a basis. This removes choice 5.

Because a basis is linearly independent, the zero vector cannot be part of it.
This removes choice 1.

For B to have a proper subset that spans V, there must exist vectors in B which are not linearly independent (since we can remove vectors and still span V). However, this is impossible since B is a basis (i.e. we cannot remove any vectors and still hope to span V). This removes choice 2.

Similarly, we cannot add any more linearly independent vectors to B because it already spans all of V. Thus B cannot be a proper subset of a linearly independent subset of V. This removes choice 3.

The only choice left is choice 4. Thus choice 4 is the answer (there are in fact an infinite number of bases that are disjoint from B).

## Tuesday, 23 August 2011

### Math GRE - #22

Consider the function $f$ defined by $f(x)=e^{-x}$ on the interval [0, 10]. Let $n>1$ and let $x_0,x_1,\ldots,x_n$ be numbers such that: $0= x_0 < x_1 < \cdots < x_n =10.$ Which of the following is greatest?

1. $\displaystyle\sum_{j=1}^n{f(x_j)(x_j-x_{j-1})}$
2. $\displaystyle\sum_{j=1}^n{f(x_{j-1})(x_j-x_{j-1})}$
3. $\displaystyle\sum_{j=1}^n{f\left(\frac{x_j+x_{j-1}}{2}\right)(x_j-x_{j-1})}$
4. $\displaystyle\int_0^{10}{f(x)dx}$
5. $0$

Solution :

The first three choices are Riemann sums with choice 1 the right sum, choice 2 the left sum and choice 3 the midpoint sum. Choice 4 is the actual area of $f$ from 0 to 10.

Since $e^{-x}$ is a decreasing function, the left Riemann sum is the greatest out of all the choices (see diagram below).

Thus, choice 2 (the left Riemann sum) is the answer.

### Physics GRE - #22

Two wedges, each of mass $m$, are placed next to each other on a flat floor. A cube of mass $M$ is balanced on the wedges as shown below.

Assume no friction between the cube and the wedges, but a coefficient of static friction $\mu < 1$ between the wedges and the floor. What is the largest $M$ that can be balanced as shown without motion of the wedges?

1. $\frac{m}{\sqrt{2}}$
2. $\mu\frac{m}{\sqrt{2}}$
3. $\mu\frac{m}{1-\mu}$
4. $2\mu\frac{m}{1-\mu}$
5. All $M$ will balance.

Solution :

We know immediately it's either choice 3 or 4 since any $M$ will balance if $\mu\rightarrow 1$, but it looks like we're going to have to do the work to figure out which one it is.

Due to the symmetry of the problem, we may assume that one wedge holds up a mass of $\frac{M}{2}$. By breaking down the force of the block on the wedge into components, we can see that the normal force on the wedge is $N = \left(m+\frac{M}{2}\right)g.$ So the friction force on the wedge is $F_{fr} = \mu\left(m+\frac{M}{2}\right)g.$ Since the friction force has to equal the mass of half the block (recall that we are only considering one wedge holding up a block of mass $M/2$), we know that $W_{1/2 - block} = \frac{M}{2}g = \mu\left(m+\frac{M}{2}\right)g.$ We can then solve this equation to obtain $M=2\mu\frac{m}{1-\mu}.$
What if the wedges weren't angled at 45 degrees? Can you find a general expression for $M$ at any angle?

## Monday, 22 August 2011

### Physics GRE - #21

The electric potential at point P (for the ring above), which is located on the axis of symmetry a distance $x$ from the centre of the ring, is given by:

• $\dfrac{Q}{4\pi\epsilon_0x}$
• $\dfrac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}$
• $\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)}$
• $\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)^{3/2}}$
• $\dfrac{QR}{4\pi\epsilon_0(R^2+x^2)}$

Solution :

Recall that $V=\int{\vec{E}\cdot d\vec{l}}=\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}.$
Since we are essentially summing up the charge over the entire ring (i.e. we are integrating around the ring with respect to $q$), we know that $int{dq}=Q$, where $Q$ is the charge of the entire ring. Therefore, $\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}=\frac{Q}{4\pi\epsilon_0|\vec{r}|}=\frac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}$ where $|\vec{r}|=\sqrt{R^2+x^2}$ as indicated in the diagram below (except they used $z$ for $x$).

### Math GRE - #21

Let $x$ and $y$ be uniformly distributed, independent random variables on [0, 1]. The probability that the distance between $x$ and $y$ is less than $\frac{1}{2}$ is:

• $\dfrac{1}{4}$

• $\dfrac{1}{3}$

• $\dfrac{1}{2}$

• $\dfrac{2}{3}$

• $\dfrac{3}{4}$

Solution :

The answer is $\frac{3}{4}$.

If $x$ and $y$ are uniformly distributed and independent random variables on [0, 1], this means that they are in the unit square.

We require $|x-y| < \frac{1}{2}$, i.e. we need $-\frac{1}{2} < x - y < \frac{1}{2}$.
On the unit square, we can draw two regions which represent the inequality above: $y < x+\frac{1}{2}$ $y > x - \frac{1}{2}.$
If we can find the area of this region (shown below), then we are done the question.

 The region we are looking at.

I'll leave it as an exercise to show that the filled out region above is $\frac{3}{4}$ of the unit square as we wanted.

## Sunday, 21 August 2011

### Physics GRE - #20

According to the Standard Model of elementary particles, which of the following is not a composite object?

• Muon
• Pi-meson
• Neutron
• Deuteron
• Alpha particle

Solution :

The answer is muon. A muon is an elementary particle (a lepton).

Pi - mesons are a quark with an antiquark.
A neutron is made of two down quarks and an up quark.
A deuteron is a proton and a neutron (a proton is two up quarks and a down quark).
An alpha particle is two protons and two neutrons.

### Math GRE - #20

A fair coin is tossed 8 times. What is the probability that more of the tosses will result in heads than tails?

• $\dfrac{1}{4}$
• $\dfrac{1}{3}$
• $\dfrac{87}{256}$
• $\dfrac{23}{64}$
• $\dfrac{93}{256}$

Solution :

The answer is $\dfrac{93}{256}$.

The chance of exactly 4 heads and 4 tails is: $\left(\frac{1}{2}\right)^8\cdot\binom{8}{4}=\frac{70}{256}.$

Thus, the chance of having more tails than heads or having more heads than tails is: $1-\frac{70}{256}=\frac{186}{256}.$

Because of symmetry, the probability of having more heads than tails is: $\frac{1}{2}\frac{186}{256}=\frac{93}{256}.$

## Saturday, 20 August 2011

### Math GRE - #19

Let $A$ be a $2\times 2$ matrix for which there is a constant $k$ such that the sum of the entries in each row and each column is $k$. Which of the following must be an eigenvector of $A$?

1. $\left[\begin{array}{c} 1\\ 0\end{array}\right]$
2. $\left[\begin{array}{c} 0\\ 1\end{array}\right]$
3. $\left[\begin{array}{c} 1\\ 1\end{array}\right]$
Choices:
• 1 only
• 2 only
• 3 only
• 1 and 2
• 1, 2, and 3

Solution :

Choice 3 is the only solution.

Since the rows add up to $k$, $\left[\begin{array}{c}1\\ 1\end{array}\right]$ is an eigenvector with eigenvalue $k$. To see that choice 1 and 2 do not work, take the matrix: $\left[\begin{array}{cc} 1 & 1\\ 1 & 1\end{array}\right]$ which satisfies the given conditions but do not have either choice 1 or choice 2 as eigenvectors.

Bonus problem:
Given a matrix with columns adding up to $k$, prove that $k$ is an eigenvalue (in fact, if the entries in the matrix are all greater than zero, then $k$ is the largest eigenvalue).

### Physics GRE - #19

As shown below, a ball of mass $m$, suspended on the end of a wire, is released from height $h$ and collides elastically, when it is at its lowest point, which a block of mass $2m$ at rest on a frictionless surface. After the collision, the ball rises to a final height equal to:

• $\dfrac{h}{9}$
• $\dfrac{h}{8}$
• $\dfrac{h}{3}$
• $\dfrac{h}{2}$
• $\dfrac{2h}{3}$

Solution :

Using conservation of energy, we know that: $2mgh = mv_0^2 = mv_1^2+2mv_2^2$ where $v_0$  is the initial velocity and $v_1$, $v_2$ are the resulting velocities after the collision.

Using conservation of momentum, we also know that $mv_0 = mv_1+2mv_2.$

Solving these equations give $v_1 = -\frac{v_0}{3}, v_2 = \frac{2v_0}{3}$ which make sense as the ball must bounce back after the collision. Now using conservation of energy again, we can determine the height that the ball will rise to: $mgh_{new} = \frac{1}{2}mv_1^2 = \frac{1}{2}m\frac{v_0^2}{9}= \frac{mgh}{9}\implies h_{new}=\frac{h}{9}.$

There is a faster way to find $v_1$ using the concept of a centre of mass. Can you see it?

## Thursday, 18 August 2011

### Math GRE - #18

If we have $F(x)=\int_e^x{\log{t}dt}$ for all positive $x$, then $F'(x)=$

• $x$
• $\frac{1}{x}$
• $\log{x}$
• $x\log{x}$
• $x\log{x}-1$

Solution :

I decided to use this question because of it's similarity to the question I posted yesterday. Since the derivative of an integral is itself (FTC - Part I), we have $F'(x)=\frac{d}{dx}\int_e^x{\log{t}dt}=\log{x}.$

### Physics GRE - #18

A string consists of two parts attached together.

The right part of the string has mass $\mu_r$ per unit length and the left part of the string has mass $\mu_l$ per unit length. If a wave of unit amplitude travels along the left part of the string, what is the amplitude of the wave that is transmitted to the right part of the string?

1. $1$
2. $\dfrac{2}{1+\sqrt{\mu_l / \mu_r}}$
3. $\dfrac{2\sqrt{\mu_l / \mu_r}}{1+\sqrt{\mu_l / \mu_r}}$
4. $\dfrac{\sqrt{\mu_l / \mu_r}-1}{\sqrt{\mu_l / \mu_r}+1}$
5. $0$

Solution :

The correct answer is choice 3.

To solve the problem we can look at some nice limiting cases.
First, suppose $\mu_l =\mu_r$, then there should be no difference in amplitude as the wave travels (i.e. amplitude is 1 under this condition). Choices 1, 2, and 3 satisfy this.

Second, suppose $\mu_r\rightarrow\infty$. Then the wave should be completely reflected at the part where the two strings join (imagine a string attached to a wall). Thus, we should have amplitude equal to 0 under this condition. Only choice 3 satisfy this.

## Wednesday, 17 August 2011

### Math GRE - #17

Let $h$ be the function defined by $h(x)=\int_0^{x^2}{e^{x+t}dt}.$ Then $h'(1)=$

• $e-1$
• $e^2$
• $e^2-e$
• $2e^2$
• $3e^2-e$

Solution :

The answer is $3e^2-e$.

This can be solved with applications of the product rule, the chain rule, and the fundamental theorem of calculus.

First let us recall that the derivative of an integral is itself, that is: $\frac{d}{dx}\int_a^x{f(x)}=f(x).$
When the upper limit is a function, we must apply the chain rule: $\frac{d}{dx}\int_a^{x^2}{f(x)}=2x\cdot f(x^2).$
For this particular question, we have: $h(x)=\int_0^{x^2}{e^{x+t}dt}=e^x\cdot\int_0^{x^2}{e^{t}dt}=e^x\cdot g(x)$ for $g(x)=\int_0^{x^2}{e^{t}dt}$. The product rule tells us that $h'(x)=e^x\cdot g(x)+e^x\cdot g'(x)=e^x\cdot\int_0^{x^2}{e^{t}dt}+e^x\cdot\left(2x\cdot e^{x^2}\right).$
Upon substituting $1$ for $x$, we obtain the answer: $h'(1)=e\cdot(e-1)+e\cdot(2\cdot e)=3e^2-e.$

### Physics GRE - #17

A block of mass $m$ sliding down an incline at constant speed is initially at height $h$ above the ground, as shown in the figure below.

The coefficient of kinetic friction between the mass and the incline is $\mu$. If the mass continues to slide down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reaches the bottom of the incline?

• $\dfrac{mgh}{\mu}$
• $mgh$
• $\dfrac{\mu mgh}{\sin\theta}$
• $mgh\sin\theta$
• $0$

Solution :

The answer is $mgh$.

Recall conservation of energy. Since the velocity stays the same at the top and bottom, friction dissipated all of the gravitational potential energy. Thus the value of energy dissipated is $mgh$.
In equation form: $E_i = E_f \implies mgh+\frac{1}{2}mv_i^2=E_{fr}+\frac{1}{2}mv_f^2.$ But since $v_i=v_f$, $mgh=E_{fr}.$

### Math GRE - #16

For how many positive integers $k$ does the ordinary decimal representation of the integer $k!$ end in exactly 99 zeros?

• None
• One
• Four
• Five
• Twenty-Four

Solution :

There are five positive integer $k$ for which this is true.

Note that the number of zeros depend directly on how many factors of 10 are present in $k!$. However, the number of 10s depend on the number of factors of 2s and 5s. Since there are an abundance of 2 in the factorization of $k!$, we just have to count the number of times a factor of 5 appears. Since one (or more) factor of 5 appear every five numbers, every five numbers have the same number of zeros at the end.

There is a simple flaw with the argument above, do you see it?
We haven't proven the existence of a single integer $k$ such that $k!$ ends in 99 zeros! It turns out that 400! has 99 zeros (and thus 401! to 404! do as well). I believe that we were supposed to assume the existence of such an integer $k$ and that the question is also a bit poorly worded. However, I could be wrong.

Can anybody see why there must be an integer $k$ such that $k!$ ends in 99 zeros?

### Physics GRE - #16

Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is $T$, the period of small oscillations of Y is:
• $T$
• $\dfrac{T}{2}$
• $\dfrac{T}{4}$
• $\dfrac{T}{8}$
• $\dfrac{T}{16}$

Solution :

The answer is $\dfrac{T}{2}$.

The key thing to note here is that the period of small oscillations of a circular hoop is the same as that of a normal pendulum with the mass where the centre of the hoop is.
Thus, the period is given by the formula (we'll disregard the constant of $2\pi$ at the front, since it will not matter to our calculations):
$T=\sqrt{\frac{r}{g}}.$ For hoops X, we have $T_x=\sqrt{\frac{r_x}{g}}.$ For hoop Y, we have $T_y=\sqrt{\frac{r_y}{g}}.$ Since $r_y = \frac{1}{4}r_x$, we have: $T_y=\sqrt{\frac{1}{4}\frac{r_x}{g}}=\frac{1}{2}\sqrt{\frac{r_x}{g}}=\frac{T}{2}.$

## Tuesday, 16 August 2011

### Physics GRE - #15

Two positive charges of $q$ and $2q$ coulombs are located on the x-axis at $x=0.5a$ and $1.5a$, respectively, as shown below.

There is an infinite, grounded conducting plane at $x=0$. What is the magnitude of the net force on charge q?

• $\dfrac{1}{4\pi\epsilon_0}\frac{q^2}{a^2}$
• $\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{2a^2}$
• $\dfrac{1}{4\pi\epsilon_0}\frac{2q^2}{a^2}$
• $\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{a^2}$
• $\dfrac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}$

Solution :

The last choice is correct.

Recall that Coulomb's Law states the force between two point charges to be: $F= \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$ where $q_1$ and $q_2$ are the size of the two charges, $r$ is the distance between the two charges, and all other variables are constants.

Since the conductor is grounded, image charges of $-q$ and $-2q$ are induced at $0.5a$ and $-1.5a$ respectively. These image charges pulls charge $q$ left, while the real charge $2q$ pushes charge $q$ left. Therefore we have three contributions (counted from left to right) to the force on charge $q$: $\begin{eqnarray*} \sum{F} & = & \frac{q}{4\pi\epsilon_0}\left(\frac{2q}{(-1.5a-0.5a)^2}+\frac{q}{(-0.5a-0.5a)^2}+\frac{2q}{(1.5a-0.5a)^2}\right) \\ & = & \frac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}. \end{eqnarray*}$

### Math GRE - #15

Suppose $A$ and $B$ are invertible $n\times n$ matrices. If $A$ and $B$ are similar, which of the following statements are true?

1. $A-2I$ and $B-2I$ are similar.
2. $A$ and $B$ have the same trace.
3. $A^{-1}$ and $B^{-1}$ are similar matrices.
Choices:
• 1 only.
• 2 only.
• 3 only.
• 1 and 2 only.
• 1, 2, and 3 only.

Solution :

1, 2, and 3 are all true.

Recall that $A$ and $B$ are similar if $B=P^{-1}AP$ for some invertible matrix $P$.

Since $\text{tr}(AB)=\text{tr}(BA)$, we know that 2 is true due to: $\text{tr}(B)=\text{tr}(P^{-1}AP)=\text{tr}(APP^{-1})=\text{tr}(A).$

Moreover, we have $B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P.$ But by the definition of being similar, this means that $A^{-1}$ and $B^{-1}$ are similar. So 3 is true. Since only the last choice offers 2 and 3 as true, we know that the answer is the last choice: 1, 2, and 3 are all true.

Can you prove why 1 must be true?

## Monday, 15 August 2011

### Math GRE - #14

If $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$ (i.e. the floor of x), then which of the choice below is the value of $\int_0^\infty{\lfloor x\rfloor e^{-x}dx}?$
• $\dfrac{e}{e^2-1}$

• $\dfrac{1}{e-1}$

• $\dfrac{e-1}{e}$

• $1$

• $\infty$

Solution :

Since the floor function is discrete, we can turn this integral into a discrete sum of integrals by noting that:
$\begin{eqnarray*} \int_0^\infty{\lfloor x\rfloor e^{-x}dx} & = & \int_0^1{0\cdot e^{-x}dx}+\int_1^2{1\cdot e^{-x}dx}+\int_2^3{2\cdot e^{-x}dx}+\cdots \\ & = & \sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}} \end{eqnarray*}$
The sum above turns out to be telescoping, as we can see that:
$\begin{eqnarray*} \sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}} & = & \sum_{n=0}^\infty{n\left(e^{-n}-e^{-(n+1)}\right)} \\ & = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{n\cdot e^{-(n+1)}} \\ & = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{(n+1)\cdot e^{-(n+1)}}+\sum_{n=0}^\infty {e^{-(n+1)}} \\ & = & \sum_{n=0}^\infty {e^{-(n+1)}} \end{eqnarray*}$
Where the last line is due to adjusting the summation indices on the previous line. We can also see that $\sum_{n=0}^\infty {e^{-(n+1)}}=\frac{1}{e-1}$ as the above is a geometric series.

Therefore, the answer is: $\int_0^\infty{\lfloor x\rfloor e^{-x}dx}=\sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}}=\frac{1}{e-1}$

### Physics GRE - #14

The figure below shows a small mass connected to a string, which is attached to a vertical post.

If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle $\theta$ is:

• $g\sin\theta$
• $2g\cos\theta$
• $2g\sin\theta$
• $g\sqrt{3\cos^2{\theta}+1}$
• $g\sqrt{3\sin^2{\theta}+1}$

Solution :

We know that at $\theta=0$, the acceleration of the mass is exactly $g$. The only solution that satisfies this boundary condition is the last choice.

## Sunday, 14 August 2011

### Physics GRE - #13

Two pendula are attached to a massless spring, as shown below. The arms of the pendula are both of length $l$, but the pendulum balls have unequal masses $m_1$ and $m_2$. The initial distance between the masses is the equilibrium length of the spring, which has spring constant $K$. What is the highest normal mode frequency of this system?

1. $\sqrt{\frac{g}{l}}$
2. $\sqrt{\frac{K}{m_1+m_2}}$
3. $\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$
4. $\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$
5. $\sqrt{\frac{2g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$

Solution :

The correct answer is choice 4.

The easy way of solving this problem is to look at the limiting behaviour of the system in several ways. First we see that if $m_1\rightarrow\infty$ and $l\rightarrow\infty$, the resulting behaviour should be the same as a single mass, $m_2$, attached to a wall by a spring, thus we should have $\sqrt{\frac{K}{m_2}}$. This just leaves options 3 and 4 as possibilities. Then we note that if $K\rightarrow 0$, it would be as if there was no spring between the two pendula, and so we should get $\sqrt{\frac{g}{l}}$. This leaves choice 4 as the right answer. This solution sure beats deriving the normal frequencies from scratch (which I'm not even sure I remember how to do).

A note on looking at limiting behaviours
Looking at limiting behaviour is a useful technique for doing a quick check of your answer or in multiple choice questions such as these. Sometimes looking at a system this way can also give you nice approximate answers. Common methods of looking at limiting behaviour usually involves making all masses/lengths/constants equal in a problem, or taking them to 0/∞.

### Math GRE - #13

Determine the set of real numbers x for which the series below is convergent: $\sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^{n}\left(1+x^{2n}\right)}$
• $\{0\}$
• $\{x: -1 \leq x \leq 1\}$
• $\{x: -1 < x < 1\}$
• $\{x: -\sqrt{e} \leq x \leq \sqrt{e}\}$
• $\mathbb{R}$

Solution :

The answer is $\mathbb{R}$.

Note that for all numbers $x\in\mathbb{R}$, $x^{2n}\geq 0\implies \frac{x^{2n}}{1+x^{2n}}\leq \frac{x^{2n}}{x^{2n}}=1.$
This means that $\sum_{n = 1}^\infty \frac{n! x^{2n}}{n^n(1 + x^{2n})} < \sum_{n = 1}^\infty \frac{n!}{n^n}.$ Thus, if we can prove that $\sum_{n = 1}^\infty \frac{n!}{n^n}$ is convergent, then by the comparison test our original series is convergent for all $x$.

By the ratio test,  $\sum_{n = 1}^\infty \frac{n!}{n^n}$ converges as:
$\begin{eqnarray*} \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}} & = & \lim_{n\rightarrow\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!} \\ & = & \lim_{n\rightarrow\infty}\frac{n+1}{n}\cdot\left(\frac{n}{n+1}\right)^{n+1} \\ & = & \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} \\ & = & \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n} \\ & = & \frac{1}{e} \end{eqnarray*}$
Where the last line is essentially a result of the limit definition of e.

Note that if we hadn't guessed that the answer was $\mathbb{R}$, it would have been safer to do the ratio test on the original sequence directly and obtain the same answer (the method above just saves a bit of work).

## Saturday, 13 August 2011

### Math GRE - #12

A simple one today:
What is the value of the integral: $\int_0^1{\frac{x}{1+x^2}}dx?$
• $1$
• $\frac{\pi}{4}$
• $\arctan\frac{\sqrt{2}}{2}$
• $\log 2$
• $\log\sqrt{2}$

Solution :

We can use the substitution $u=1+x^2\implies du = 2x\,dx$. With this substitution, the original integral becomes: $\int_0^1{\frac{x}{1+x^2}}dx=\frac{1}{2}\int_1^2{\frac{1}{u}}du$
Upon computing this integral, we obtain: $\frac{1}{2}\int_1^2{\frac{1}{u}}du=\frac{1}{2}\left. \log{u}\right|_1^2=\frac{1}{2}\log{2}=\log{\sqrt{2}}.$

### Physics GRE - #12

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of about:

• 0.94 h
• 0.80 h
• 0.75 h
• 0.64 h
• 0.50 h

Solution :

Conservation of energy tells us that $mgh = \frac{1}{2}mv^2$. Therefore, we know that $h\propto v^2$. Since $v_{new}=0.8v_{old}$, $h_{new} = (0.8)^2h_{old} = 0.64h$.

It's a simple problem, but it may just mess you up if you aren't careful.

## Friday, 12 August 2011

### Physics GRE - #11

A pendulum of length $l$ is attached to the roof of an elevator near the surface of the Earth.

The elevator moves upward with acceleration $\frac{1}{2}g$. The angular frequency is:

• $\sqrt{\frac{3g}{2l}}$
• $\sqrt{\frac{2g}{3l}}$
• $\sqrt{\frac{g}{l}}$
• $\sqrt{\frac{g}{2l}}$
• $\sqrt{\frac{2g}{l}}$

Solution :

Quick solution:
The angular frequency of a pendulum is $\omega=\sqrt{\frac{g}{l}}$ when the gravitational field has value $g$. In this problem, the effective gravity is $g_{eff} = g + \frac{1}{2}g = \frac{3}{2}g$. $\therefore \omega=\sqrt{\frac{g_{eff}}{l}}=\sqrt{\frac{3g}{2l}}.$

Full solution:
By Newton's 2nd law for rotational motion, $\sum{\tau}=I\alpha \implies -mg_{eff}l\sin\theta=I\ddot{\theta}.$
Substituting $I = ml^2$ and $g_{eff} = \frac{3}{2}g$, we have
$-\frac{3}{2}mgl\sin\theta=ml^2\ddot{\theta}\implies \ddot{\theta}=-\frac{3g}{2l}\sin\theta\approx-\frac{3g}{2l}\theta$ $\therefore \omega^2 = \frac{3g}{2l} \implies \omega=\sqrt{\frac{3g}{2l}}.$

### Math GRE - #11

For all real numbers $x$ and $y$, which of the answers below is the expression $\frac{x+y+|x-y|}{2}$ equal to?

• The maximum of $x$ and $y$
• The minimum of $x$ and $y$
• $|x+y|$
• The average of $|x|$ and $|y|$
• The average of $|x+y|$ and $x-y$

Solution :

The expression is equal to the maximum of $x$ and $y$.
Suppose $x>y$. Then $|x-y|=x-y$ and so $\frac{x+y+|x-y|}{2}=\frac{x+y+x-y}{2}=x.$
Now suppose $x<y$. Then  $|x-y|=-(x-y)$ and so $\frac{x+y+|x-y|}{2}=\frac{x+y-(x-y)}{2}=y.$
$\therefore\quad\frac{x+y+|x-y|}{2}=\max\{x,y\}.$

Can you find a representation for $\min\{x,y\}$?

## Thursday, 11 August 2011

### Math GRE - #10

What is the coefficient of $x^3$ in the expansion of $(1+x)^3(2+x^2)^{10}$?

• $2^{14}$
• $31$
• $\dbinom{3}{1}+\dbinom{10}{1}$
• $\dbinom{3}{1}+2\dbinom{10}{1}$
• $\dbinom{3}{3}\dbinom{10}{1}2^9$

Solution :

### Physics GRE - #10

If $\frac{\partial L}{\partial q_n}=0$, where $L$ is the Lagrangian for a conservative system without constraints and $q_n$ is a generalized coordinate, then the generalized momentum $p_n$ is:

• an ignorable coordinate
• constant
• undefined
• equal to $\frac{d}{dt}\left(\frac{\partial L}{\partial q_n}\right)$
• equal to the Hamiltonian for the system

Solution :

The generalized momentum is constant.
Recall that the generalized momentum $p_n$ is $p_n=\frac{\partial L}{\partial \dot{q_n}}.$
By the Euler-Lagrange equation, $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_n}}\right) = \frac{\partial L}{\partial q_n}.$
Therefore,  $\frac{\partial L}{\partial q_n}=0\implies\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q_n}}\right) = 0 \implies \frac{\partial L}{\partial\dot{q_n}} = p_n = const.$

## Wednesday, 10 August 2011

### Physics GRE - #9

A rock is thrown vertically upward with initial speed $v_0$. Assume a friction force proportional to $-v$, where $v$ is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

• The acceleration of the rock is always equal to $g$.
• The acceleration of the rock is only equal to $g$ at the top of the flight.
• The acceleration of the rock is always less than $g$.
• The speed of the rock upon return is $v_0$.
• The rock can attain a terminal speed greater than $v_0$ before it returns to its starting point.

Solution :

The second choice is the answer.
• Choice 1 is not true as it does not take into account the frictional force (which makes the acceleration going up greater than $g$ and the acceleration going down less than $g$).
• Choice 2 is true as $v=0$ at the top of the flight, so the frictional force is zero and the only other force is gravity.
• Choice 3 is not true as the acceleration going up is greater than $g$.
• Choice 4 is not true as it does not take into account the frictional force. Friction would cause the rock to land with less kinetic energy than it started with, thus leading to a speed lower than $v_0$ upon return.
• Choice 5 cannot be true as energy would not be conserved if speed was greater than $v_0$ (speed can only be greater than $v_0$ if friction gave the rock more kinetic energy, which is impossible since friction slows the rock by definition).

### Math GRE - #9

A circular region is divided by 5 radii into sectors as shown below. Twenty-one points are chosen in the circular region, none of which are on any of the 5 radii. Which of the following statements must be true?

1. Some sector contains at least 4 of the points.
2. Some sector contains at most 3 of the points.
3. Some pair of adjacent sectors contain a total of at least 9 of the points.
• 1 only.
• 3 only.
• 1 and 2.
• 1 and 3.
• 1, 2, and 3.

Solution :

Statement 1 is true. If all sectors contained less than 5 points (i.e. max 4 points in each sector), there would only be a maximum of 5*4 = 20 points in the circle.
Statement 2 is false. {4, 4, 4, 4, 5} is a counter example.
Statement 3 is true. We can label the number of points in our sectors as {a, b, c, d, e}. Suppose no pair of adjacent sectors contain more than 9 points. This means that:

a + b <= 8
b + c <= 8
c + d <= 8
d + e <= 8
e + a <= 8

Adding all of the above inequalities together, we obtain:

2(a + b + c + d + e) <= 40
a + b + c + d + e <= 20

which is a contradiction as a + b + c + d +e = 21.

## Tuesday, 9 August 2011

### Physics GRE - #8

A particle of mass $m$ moves in the potential shown above. The period of the motion when the particle has energy $E$ is:
• $\sqrt{\frac{k}{m}}$
• $2\pi\sqrt{\frac{m}{k}}$
• $2\sqrt{\frac{2E}{mg^2}}$
• $\pi\sqrt{\frac{m}{k}}+2\sqrt{\frac{2E}{mg^2}}$
• $2\pi\sqrt{\frac{m}{k}}+4\sqrt{\frac{2E}{mg^2}}$

Solution :

The potential function $V(x)$ is half of a simple harmonic oscillator potential and a full gravitational potential. To find the period, we can simply add half the period of a simple harmonic oscillator to twice the period of a falling object (since we know there are 2 parts, we know that the answer is one of the last two choices at this point).

Recall that the period of a harmonic oscillator is $T=2\pi\sqrt{\frac{m}{k}}$. Half of this period is $T_1=\pi\sqrt{\frac{m}{k}}$. From this we know that the second last choice must be the answer as the last choice has a factor of 2 in front of it. To check our answer, we can find the period of a falling object from the kinematics equation: $x=\frac{1}{2}gt^2$, giving us $T=\sqrt{\frac{2x}{g}}$. At the endpoints of the oscillation, we have $v=0$ so $mgx=E$. Substituting for $x$ in $T$  as well as taking into account of the fact that the particle has to go to the endpoint and back (this introduces a factor of 2), the period contribution from the gravitational potential is $T_2=2\sqrt{\frac{2E}{mg^2}}$. Thus the period is: $T_{total}=T_1+T_2=\pi\sqrt{\frac{m}{k}}+2\sqrt{\frac{2E}{mg^2}}$ as we suspected.