Question 1
Find the eigenvalues and corresponding eigenvectors of the following matrix:
$$\left(\matrix{1 & 3 \\ 4 & 2}\right)$$
Question 2
The gravitational field inside a hollow spherical shell of mass $$m$$ is zero, and outside the shell the field is as if all the mass were at the centre of the sphere.
(a) Show that the gravitational field inside the Earth at radius $$r$$ is given by $$g=-Kr$$, where $$K=\frac{4\pi G \rho}{3}$$, $$G$$ is the universal gravitational constant and $$\rho$$ is the density of the Earth (taken to be uniform).
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Imagine that a straight tunnel is drilled through the centre of the Earth from surface to opposite surface. A small mass $$m$$ is positioned inside the hole at distance $$r$$ from the centre of the Earth and released from rest.
(b) Write an expression in terms of the parameters defined above for the force on the mass when it is in the given position.
(c) Explain why the mass will oscillate about the centre of the Earth and specify the type of motion that it will undergo.
(d) Show that if the mass is released into the hole at the Earth's surface, it will reach the opposite side of the Earth after a time $$t=\pi\sqrt{1/K}$$.
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Question 3
A sinusoidal wave is travelling along a string with a wave speed of 40 cm/s. The displacement of an element of the string located at $$x=10~cm$$ is found to vary with time according to the following equation:
$$y(t) = (5.0~cm)\sin [1.0 - (4.0~s^{-1})t]$$
If the linear density of the string is $$4.0~g/cm$$ find:
(a) The amplitude of the wave.
(b) The frequency of the wave.
(c) The wavelength of the wave.
(d) The general equation giving the transverse displacement of the string as a function of position and time.
(e) An expression for the transverse velocity of the string element at $$x=10~cm$$ as a function of time.
(f) The tension in the string.
Question 1
Well, I know that the eigenvalues of a 2x2 matrix can be obatined from the quadratic formula. But there are simpler ways.
The eigenvalue is often denoted as $$\lambda$$.
So we have,
$$\left(\matrix{1 & 3 \\ 4 & 2}\right)$$
The eigenvalues, could be given by values of r which we know make the det of a matrx equal to zero.
$$\left(\matrix{1-r & 3 \\ 4 & 2-r}\right)$$
$$(1-r)(2-r)-3-4 = 0$$
So the Eigenvalues equal $$\lambda_1 = -2$$ and $$\lambda_2 = 5$$.
Question 2 (a)
Right, well straight away I can see Gauss' Law equation divided by 3, probably 3 comes from the dimensions working on a three dimensional sphere (or atleast an approximation).
I haven't admittedly ever seen it expressed like this, equalled to K. What is K?
The gravitational field of the earth is $$\frac{\mu}{r} = \frac{GM}{r}$$ where $$\mu$$ is the gravitational parameter. So in your equation, it describes the volume times the density.
So is, $$K$$ simply another way to write
$$\frac{\mu}{r^2}= \frac{4\pi G \rho}{3}$$
? This is just the gravitational field of an homogeneous sphere. Now, $$g$$ is traditionally given as
$$g= \frac{GM}{r^2}$$
So I assume that
$$g= \frac{4 \pi G \rho}{3}$$
So if we have
$$K= \frac{4 \pi G \rho}{3}$$
and
$$g = -Kr$$
So, up till now, if you haven't guessed, I've been guessing
.... there is an equation which comes to mind which is similar to the appearance of the negative $$K$$ in the context I gave it, with the radius $$r$$ and that is
$$-\frac{GM(r)}{r^2} = -\frac{4 \pi G \rho}{3}$$
Hows that?
Question 2 (b)
I am to write an expression in terms of the parameters for the force on a mass when at any single position?
Well, force is generally given by the equation $$F=\frac{GM^2}{r^2}$$.
So the expression would be
$$\frac{GM^2}{r^2}$$
? Not sure if this is what you wanted?
Question 2 (c)
''Explain why the mass will oscillate about the centre of the Earth and specify the type of motion that it will undergo.''
As an object (take it to be ourselves) falls past the surface of the earth, the portion of the Earth's mass which is above you is further from the center than you are. The force of gravity which attracts you is contributed from the center of the Earth. Since the surface is no longer contributing to pull you back, you will continue to fall towards the center of the Earth.
As you fall however, mass tends to zero, as the radius tends to zero. As you continue to fly towards the center, there will be kinetic energy left over from your decent, so you do not settle in the center of the spherical mass. Instead, you will oscillate back and forth around the center, for each time you turn back and move past the center, the same rules apply: the surface is no longer pulling at you, but rather the center of mass, the origin of the Earth's radius is attracting your body.
Question 2 (d)
The distance traversed by an object falling for some time is if I am correct, usually given by the equation
$$d = \frac{1}{2} gt^2$$
Solving for the time, we have an equation which looks (only) similar to your equation, but are essentially the same
$$t = \sqrt{\frac{2d}{g}}$$
Where I have surmised before $$K = \frac{GM}{r^2}$$ and that $$g = \frac{GM}{r^2}$$ So I've replaced this with the acceleration due to gravity and I am going to replace your equation numerator with the radius.
$$t=\pi \sqrt{\frac{r}{g}}$$
If the acceleration of the gravity is $$9.8 m/s$$ and (I've had to look this up) the radius $$6,353$$ km.
But I feel I have unecesserily made this complicated. This is essentially a problem concerning Hooks Law.
Taken that the spring constant would be $$\frac{k}{M}$$ would be $$\frac{9.8 m/s}{6300 km}$$ works out then that to fall to the other side of the earth would take about 42 mins, considering the period of oscillation is
$$t = 2\pi \sqrt{\frac{Mr}{Mg}} = 2\pi \sqrt{\frac{r}{g}}$$