A problem proposal for IPhO 2019 participants

All the best to all the 5 participants will be selected to represent India at IPhO 2019, to be held at Tel Aviv, Israel.

Here is a doable problem for you guys :P

A Dielectric slab of thickness tt, relative permittivity ϵr\epsilon_r is between two fixed metallic parallel plates. Faces of the slab and the plates are parallel. Distance between the plates is dd. Find the minimum voltage applied between the plates sufficient to rupture the slab, if the breaking stress of slab is ω\omega. Consider a uniform string of length ll, tension TT, mass per unit length ρ\rho that is stretched between two immovable walls. Show that the total energy of the string, which is the sum of its kinetic and potential energies E=12∫01[ρ(∂y∂t)+T(∂y∂x)2]dxE=\frac {1}{2}\int_0^1 [\rho (\frac {\partial y}{\partial t})+T (\frac {\partial y}{\partial x})^2]dx. Where y(x,t)y (x,t) is the string's transverse displacement relatively small. The general motion of the string can be represented as a linear superposition of the normal modes: that is y(x,t)=∑n=1,∞(Sin(nπxl))Cos(nπvtl−ϕn)y (x,t)=\sum_{n=1,\infty}(Sin(n\pi\frac {x}{l}))Cos (n\pi\frac {vt}{l}-\phi_n). Here v=Tρv={\sqrt {\frac{T}{\rho}}}. Demonstrate that E=∑n=1,∞EnE=\sum_{n=1,\infty}E_n, where En=14mωn2An2E_n=\frac {1}{4}m\omega_n^2A_n^2 is the energy of nthn^{th} normal mode. Here, m=ρlm=\rho l is the mass of the string, and ωn=nπvl\omega_n=n\pi\frac {v}{l} is the angular frequency of the nthn^{th} normal place.

Note by Pawan Goyal
2 years, 1 month ago

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Comments

@Archit Boobna @Rajdeep Dhingra best of luck for the TSTs of IPhO 2019.

Pawan Goyal - 2 years, 1 month ago

Ya, best of luck. @Pawan Goyal, do you know them? How?

Mr. India - 2 years, 1 month ago

first part is too well known and trivial lol

Solution: Kinetic energy: dK=12dm(∂y∂t)2=12ρdx(∂y∂t)2  ⟹  dKdx=12ρ(∂y∂t)2 dK = \frac{1}{2} dm {(\frac{\partial y}{\partial t})}^2 = \frac{1}{2} \rho dx {(\frac{\partial y}{\partial t})}^2 \implies \frac{dK}{dx} = \frac{1}{2} \rho {(\frac{\partial y}{\partial t})}^2 For potential energy, dU=T(dl−dx)=12Tdx(∂y∂x)2  ⟹  dUdx=12ρv2(∂y∂x)2 dU = T(dl-dx) = \frac{1}{2} T dx {(\frac{\partial y}{\partial x})}^{2} \implies \frac{dU}{dx} = \frac{1}{2} \rho v^2 {(\frac{\partial y}{\partial x})}^{2} Adding the two, we get dEdx=12(ρ(∂y∂t)2+T(∂y∂x)2) \frac{dE}{dx} = \frac{1}{2} (\rho(\frac {\partial y}{\partial t})^2+T(\frac{\partial y}{\partial x})^2) and so E=∫x=0L12(ρ(∂y∂t)2+T(∂y∂x)2)dx\boxed{E = \int_{x=0}^{L}{\frac{1}{2} (\rho(\frac {\partial y}{\partial t})^2+T(\frac{\partial y}{\partial x})^2)}dx}This is in general true for any one-dimensional linear wave, transverse or longitudinal, propagating or non-propagating.

And for the second part

Let y(x,t)=∑n=1,∞yn(x,t) y(x,t)=\sum_{n=1,\infty}y_n(x, t). Plug ∂y∂x\frac{\partial y}{\partial x} and ∂y∂t\frac{\partial y}{\partial t} in the energy equation. You should be able to show that ∫x=0L∂ya∂x⋅∂yb∂x dx=0\int_{x=0}^{L}{\frac{\partial y_a}{\partial x} \cdot \frac{\partial y_b}{\partial x}}\ dx = 0 if a≠ba \neq b and likewise for tt. Then, the only terms that remain are the squared terms, and the answer is immediate.

Overall this question is quite easy and almost everybody would get a full score in this one.

Kushal Thaman - 1 year, 4 months ago
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