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Solution

Black body radiation, spectral density: ε (ν) dν = hνρ (ν) n (ν)

The photon energy, E = hν where h is Planck’s constant and ν is the photon frequency.

The density of states, ρ (ν) = Aνⁿ⁻¹ where A is a constant independent of the frequency and the frequency term is the scaling of surface area of an n-dimensional sphere.

The Bose–Einstein distribution,

n(v)$=\frac{1}{e\frac{hv}{kT}-1}$

where k is the Boltzmann constant and T is the temperature.

We let

$x=\frac{hv}{kT}$

and get

ε(x)$\propto =\frac{x^n}{e^x-1}$

We do not need the constant of proportionality (which is not simple to calculate in 4D) to find the maximum of ε (x). Working out the constant just tells us how tall the peak is, but we are interested in where the peak is, not the total radiation.

$\frac{d\epsilon }{dx}\propto \frac{n{x}^{n-1}}{e^x-1}-\frac{x^n{e}^x}{{\left(e^x-1\right)}^2}$

We set this equal to zero for the maximum of the distribution,

$\frac{x^{n-1}{e}^x}{{\left(e^x-1\right)}^2}\left(n\left(1-e^{-x}\right)-x\right)=0$

This yields x = n (1 − e^−x) where

$x=\frac{h{v}_{max}}{kT}$

and we can relate

${\lambda }_{max}=\frac{c}{v_{max}}$

and c being the speed of light.

This equation has the solution x = n +W (−ne⁻ⁿ) where W is the Lambert W function z = W (y) that solves ze^z = y (although there is a subtlety about which branch of the function). This is kind of useless to do anything with, though. One can numerically solve this equation using bisection/Newton–Raphson/iteration. Alternatively, one could notice that as the number of dimensions increases, e^−x is small, so to leading approximation x ≈ n. One can do a little better iterating this, x ≈ n − ne⁻ⁿ which is what we will use. Note the second iteration yields

$x\approx n-n{e}^{\left(n-n{e}^{-n}\right)}$

Number of dimensions, n	Numerical solution	Approximation
2	1.594	1.729
3	2.821	2.851
4 (the one we want)	3.921	3.927
5	4.965	4.966
6	5.985	5.985

Using the result above,

${\lambda }_{max}=\frac{hc}{kT{x}_{max}}=\frac{6.63 \times {10}^{-34}·3\times {10}^8}{1.38\times {10}^{-23}·6\times {10}^3·3.9}=616 nm$

616 nm is middle of the spectrum, so it will look white with a green-blue tint. Note, we have used T = 6000 K for the temperature here, as given in the question.

It would also be valid to look at ε (λ) dλ instead of ε (ν) dν.

Solution

First we will set up the equations of motion for our system. We will set one mass to be at position −x and the other to be at x, so the masses are at a distance of 2x from each other. Starting from Newton’s law of gravity:

$F=-\frac{G{m}^2}{{\left(2x\right)}^2}$

we can then use Newton’s second law to rewrite the LHS,

$m\ddot{x}=-\frac{G{m}^2}{4x^2}$

which we can simplify to

$\ddot{x}=-\frac{Gm}{4x^2}$

It is important that you get the right factor here depending on your choice for the particle coordinates at the start. Note there are other methods of getting this point, e.g. reduced mass.

We can now solve the second order ODE above. We will not show the whole process here but present the starting point and key results. We can write the acceleration in terms of the velocity. The initial velocity is zero and the initial position

$x_i=\frac{d}{2}$

So,

$v\frac{dv}{dx}=-\frac{Gm}{4x^2}\to \underset{0}{\overset{v}{\int }}vdv=-\frac{Gm}{4}\underset{x_i}{\overset{x}{\int }}\frac{dx}{x^2}$

and once the integrals are solved we can rearrange for the velocity,

$v=\frac{dx}{dt}=-\sqrt{\frac{Gm}{2}}\sqrt{\frac{1}{x}-\frac{1}{x_i}}$

Now we can form an expression for the total time taken for the masses to meet in the middle,

$T=\sqrt{\frac{2}{Gm}}\underset{0}{\overset{x_i}{\int }}\frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{x_i}}}$

There are quite a few steps involved in solving this integral, for these solutions, we shall make use of the following (but do attempt to solve it for yourselves in full).

$\underset{0}{\overset{1}{\int }}\sqrt{\frac{y}{1-y}}dy=si{n}^{-1}\left(1\right)=\frac{\pi }{2}$

Hence,

$T=\frac{\pi }{2}\sqrt{\frac{2x_i^3}{Gm}}=\frac{\pi }{2}\sqrt{\frac{d^3}{4Gm}}$

We can now rearrange for G and substitute in the values given in the question, don’t forget to convert the time into seconds.

$G=\frac{d^3}{4m}{\left(\frac{\pi }{2T}\right)}^2=6.67\times {10}^{-11} m^{3}k{g}^{-1}{s}^{-2}$

This is the generally accepted value for the gravitational constant of our universe as well.

Solution

We will write the period of oscillation of the pendulum at the surface of the Earth to be

$T_0=2\pi \sqrt{\frac{l}{g_0}}$.

At a height h above the surface of the Earth the period of oscillation will be

$T_h=2\pi \sqrt{\frac{l}{g_h}}$,

where g₀ and g_h are the acceleration due to gravity at the surface of the Earth and a height h above it respectively.

We can define τ₀ to be the total duration of the day which is 8.64 × 10⁴ seconds and equal to N complete oscillations of the pendulum at the surface. The lag is then τ_h which will equal N times the difference in one period of the two clocks, τ_h = NΔT, where ΔT = (T_h − T₀). We can now take a ratio of the lag over the day and the total duration of the day:

$\frac{{\tau }_h}{{\tau }_0}=\frac{N\left(T_h-T_0\right)}{N{T}_0}\Rightarrow {\tau }_h={\tau }_0\frac{T_h-T_0}{T_0}={\tau }_h={\tau }_0\left(\frac{T_h}{T_0}-1\right)$

Then by substituting in the expressions we have for the period of a pendulum at the surface and height h we can write this in terms of the gravitational constant,

${\tau }_h={\tau }_0\left(\sqrt{\frac{g_0}{g_h}}-1\right)$

[Award 1 mark for finding the ratio of the lag over the day and the total period of the day.]

The acceleration due to gravity at the Earth’s surface is

$g_0=\frac{GM}{R^2}$

where G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the Earth. At an altitude h, it will be

$g_h=\frac{GM}{{\left(R+h\right)}^2}$

[Award 1 mark for finding the expression for the acceleration due to gravity at height h.]

Substituting into our expression for the lag, we get:

${\tau }_h={\tau }_0\left(\sqrt{\frac{{\left(R+h\right)}^2}{R^2}}-1\right)={\tau }_0\left(\sqrt{1+\frac{2h}{R}+\frac{h^2}{R^2}}-1\right)={\tau }_0\left(\frac{\sqrt{R^2+2hR+h^2}}{R}-1\right)={\tau }_0\left(\frac{R+h}{R}-1\right)$

This simplifies to an expression for the lag over a day. We can then substitute in the given values to find,

${\tau }_h=\frac{{\tau }_{0}h}{R}=\frac{8.64\times {10}^4 s·300 m}{8.3781\times {10}^6 m} =4.064 s\approx 4 s$

[Award 2 marks for completing the simplification of the ratio and finding the lag to be ≈ 4 s.]

Solution

We can imagine this as an inverted pendulum, with gravity acting from the centre of mass $\left(\frac{l}{2}\right)$ and at an angle θ from the unstable equilibrium point.

[Award 1 mark for a suitable diagram of the system.]

We must now find the equations of motion of the system. For this we can use Newton’s second law $\overrightarrow{F}=m\overrightarrow{a}$ in its rotational form τ = Iα (torque = moment of inertia × angular acceleration). We have another equation for torque we can use as well

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta \widehat{n}$

where r is the distance from the pivot to the centre of mass $\left(\frac{l}{2}\right)$ and F is the force, which in this case is gravity mg. We can then equate these giving

$\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta =I\alpha$

Substituting in the given moment of inertia of the stick and that the angular acceleration

$\alpha =\frac{{\delta }^2\theta }{\delta t^2}=\ddot{\theta }$

We can cancel a few things and rearrange to get a differential equation of the form:

$\ddot{\theta }-\frac{3g}{2l}\sin \theta =0$

we then can take the small angle approximation sin θ ≈ θ, resulting in

$\ddot{\theta }-\frac{3g}{2l}\theta =0$

[Award 2 marks for finding the equation of motion for the system and using the small angle approximation.]

Solve with ansatz of θ = Ae^ωt + Be^−ωt, where we have chosen

${\omega }^2=\frac{3g}{2l}$

We can clearly see that this will satisfy the differential equation

$\dot{\theta }=\omega A{e}^{\omega t}-\omega B{e}^{-\omega t} and \ddot{\theta }={\omega }^{2}A{e}^{\omega t}+{\omega }^{2}B{e}^{-\omega t}$

Now we can apply initial conditions to find A and B, by looking at the two cases from the uncertainty principle

$\Delta x\Delta p=\Delta xm\Delta v\frac{\hslash }{2}$

Case 1: The stick is at an angle but not moving

At t = 0, θ = Δθ

θ = Δθ = A + B

At t = 0, $\dot{\theta }=0$

$\dot{\theta }=0=\omega A{e}^{{\omega }_0}-\omega B{e}^{-{\omega }_0}$, A=B

This implies Δθ = 2A and we can then find

$A=\frac{\Delta \theta }{2}=\frac{2\Delta x}{2l}=\frac{\Delta x}{l}$

So we can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right)$ or $\theta =\frac{2\Delta v}{\omega l}\sin h \omega t$

Case 2: The stick is at upright but moving

At t = 0, θ = 0

This condition gives us A = −B.

At t = 0, $\ddot{\theta }=\frac{2v}{l}$

This initial condition has come from the relationship between the tangential velocity, Δv which equals the distance to the centre of mass from the pivot point, $\frac{l}{2}$ and the angular velocity $\dot{\theta }$. Using the above initial condition gives us $\dot{\theta }=2\omega A$ where $A=\frac{\Delta v}{\omega l}$

We can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right) or \theta =\frac{2\Delta v}{\omega l}\sinh \omega t$

[Award 4 marks for finding the two expressions for θ by using the two cases of the uncertainty principle.]

Now there are a few ways we can finish off this problem, we shall look at three different ways. In each case when the stick has fallen on the ground $\theta \left(t_f\right)=\frac{\pi }{2}$.

Method 1

Take $\theta =\frac{2\Delta x}{l}\cosh \omega t$ and $\theta =\frac{2\Delta v}{\omega l}\sinh \omega t$, use $\theta \left(t_f\right)=\frac{\pi }{2}$ then rearrange for t_f in both cases. We have

$t_f=\frac{1}{\omega }\cos h{}^{-1}\frac{\pi l}{4\Delta x} and t_f=\frac{1}{\omega }\sinh {}^{-1}\frac{\pi \omega l}{4\Delta v}$

Look at the expression for cosh⁻¹ x and sinh⁻¹ x given in the question. They are almost identical, we can then approximate the two arguments to each other and we find,

$\Delta x=\frac{\Delta v}{\omega }$

we can then substitute in the uncertainty principle $\Delta x\Delta p=\frac{\hslash }{2}$ as $\Delta v=\frac{\hslash }{2m\delta x}$ and then write an expression of $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$, which we can put back into our arccosh expression (or do it for Δv and put into arcsinh).

$t_f=\frac{1}{\omega }\cosh {}^{-1}\frac{\pi l}{4\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 2

In this next method, when you get to the inverse hyperbolic functions, you can take an expansion of their natural log forms in the tending to infinity limit. To first order both functions give ln 2x, we can then equate the arguments and find Δx or Δv in terms of the other and use the uncertainty principle. This would give the time taken as,

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 3

Rather than using hyperbolic functions, you could do something like above and do an expansion of the exponentials in the two expressions for t_f or we could make life even easier and do the following.

Disregard the e^−ωt terms as they will be much smaller than the e^ωt terms. Equate the two expressions for $\theta \left(t_f\right)=\frac{\pi }{2}$ and then take the natural logs, once again arriving at an expression of

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

This method efficiently sets B = 0 when applying the initial conditions.

[Award 2 marks for reaching an expression for t using one of the methods above or a suitable alternative that gives the correct units for time.]

Then, by using one of the expressions above for time, substitute in the values and find that t = 10.58 seconds.

[Award 1 mark for finding the correct time value of t = 10.58 seconds.]

• If you’re a student who wants to sign up for the 2025 edition of PLANCKS UK and Ireland, entries are now open at plancks.uk

The post PLANCKS physics quiz – the solutions appeared first on Physics World.

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