BE Civil Engineering (IOE, TU) Engineering Physics (IOE, SH 402) Question Paper 2079 Nepal

Differential equation

A particle of mass $m$ acted on by a linear restoring force $-kx$ and a velocity-proportional damping force $-b\dfrac{dx}{dt}$ obeys Newton's second law:

$$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$$ $$\frac{d^2x}{dt^2} + \frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$$

Writing $2\beta = b/m$ (damping coefficient) and $\omega_0^2 = k/m$ (natural frequency):

$$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0$$

General solution

Trying $x = e^{\alpha t}$ gives the auxiliary equation $\alpha^2 + 2\beta\alpha + \omega_0^2 = 0$, so

$$\alpha = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$$

For the under-damped case $\beta < \omega_0$, let $\omega = \sqrt{\omega_0^2 - \beta^2}$:

$$x(t) = A_0\, e^{-\beta t}\cos(\omega t + \phi)$$

The motion is oscillatory with an exponentially decaying amplitude $A_0 e^{-\beta t}$.

Relaxation time $\tau = 1/\beta$: the time for the amplitude to fall to $1/e$ of its value.

Quality factor $Q = \dfrac{\omega_0}{2\beta} = \dfrac{\omega_0 m}{b}$: a measure of how lightly damped the oscillator is.

Numerical part

Given $m=0.20\ \text{kg}$, $k=80\ \text{N/m}$, $b=0.40\ \text{kg/s}$.

$$\omega_0 = \sqrt{k/m} = \sqrt{80/0.20} = \sqrt{400} = 20\ \text{rad/s}$$ $$\beta = \frac{b}{2m} = \frac{0.40}{2(0.20)} = 1.0\ \text{s}^{-1}$$

(i) Damped angular frequency:

$$\omega = \sqrt{\omega_0^2 - \beta^2} = \sqrt{400 - 1} = \sqrt{399} = 19.975\ \text{rad/s} \approx \mathbf{19.97\ rad/s}$$

(ii) Amplitude falls to $1/e$ when $e^{-\beta t}=e^{-1}$, i.e. $t = \tau = 1/\beta$:

$$\tau = \frac{1}{1.0} = \mathbf{1.0\ s}$$

(iii) Quality factor:

$$Q = \frac{\omega_0}{2\beta} = \frac{20}{2(1.0)} = \mathbf{10}$$

Formation

A plano-convex lens of large radius of curvature $R$ rests on a flat glass plate, enclosing a thin air film whose thickness $t$ increases radially. Monochromatic light reflected from the top and bottom of the air film interferes. Because reflection at the lower (denser) surface introduces an extra phase of $\pi$, the condition for dark rings is

$$2t = n\lambda \quad (n = 0,1,2,\dots)$$

For a sphere of radius $R$, the film thickness at radius $r$ satisfies $r^2 = 2Rt$ (since $t \ll R$), so $2t = r^2/R$. Hence for the $n$-th dark ring:

$$\frac{r_n^2}{R} = n\lambda \;\Rightarrow\; r_n = \sqrt{nR\lambda} \propto \sqrt{n}$$

Centre dark: at the point of contact $t=0$, the only path difference is the $\lambda/2$ from reflection, giving destructive interference — so the centre is dark.

Numerical part

For dark rings the diameter satisfies $D_n^2 = 4nR\lambda$. Using two rings eliminates uncertainty in the contact point:

$$D_{n+p}^2 - D_n^2 = 4pR\lambda \;\Rightarrow\; R = \frac{D_{20}^2 - D_{10}^2}{4(10)\lambda}$$

Data: $D_{10} = 4.00\ \text{mm}$, $D_{20} = 5.66\ \text{mm}$, $\lambda = 589\times10^{-9}\ \text{m}$.

$$D_{20}^2 = (5.66\times10^{-3})^2 = 3.2036\times10^{-5}\ \text{m}^2$$ $$D_{10}^2 = (4.00\times10^{-3})^2 = 1.6000\times10^{-5}\ \text{m}^2$$ $$D_{20}^2 - D_{10}^2 = 1.6036\times10^{-5}\ \text{m}^2$$ $$R = \frac{1.6036\times10^{-5}}{40 \times 589\times10^{-9}} = \frac{1.6036\times10^{-5}}{2.356\times10^{-5}} = 0.6806\ \text{m}$$

$R \approx 0.68\ \text{m}$.

Maxwell's equations (integral form)

1. Gauss's law (electricity): $\displaystyle \oint_S \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$ — electric flux through a closed surface equals the enclosed charge over $\varepsilon_0$; charges are sources of $\vec{E}$.

2. Gauss's law (magnetism): $\displaystyle \oint_S \vec{B}\cdot d\vec{A} = 0$ — net magnetic flux through any closed surface is zero; no magnetic monopoles.

3. Faraday's law: $\displaystyle \oint_C \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$ — a changing magnetic flux induces an electric field (emf).

4. Ampère–Maxwell law: $\displaystyle \oint_C \vec{B}\cdot d\vec{l} = \mu_0\left(I_c + \varepsilon_0\frac{d\Phi_E}{dt}\right)$ — both conduction current and a changing electric flux (displacement current) produce a magnetic field.

Displacement current

Maxwell noted that Ampère's law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c$ fails for a charging capacitor: choosing a surface that passes between the plates encloses no conduction current, yet $\vec B \ne 0$. He added the displacement current

$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$

Equality for a parallel-plate capacitor

Let plate area $A$, charge $Q$. The field between plates is $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, so the electric flux is

$$\Phi_E = EA = \frac{Q}{\varepsilon_0}$$

Then the displacement current is

$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{1}{\varepsilon_0}\frac{dQ}{dt} = \frac{dQ}{dt} = I_c$$

Thus the displacement current between the plates exactly equals the conduction current $I_c = dQ/dt$ in the wire, restoring continuity of current and making Ampère's law consistent.

Set-up

Inside the well ($0 < x < L$) the potential $V=0$; outside it is infinite, forcing $\psi=0$ there. The time-independent Schrödinger equation inside is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \;\Rightarrow\; \frac{d^2\psi}{dx^2} + k^2\psi = 0,\quad k^2 = \frac{2mE}{\hbar^2}$$

General solution $\psi(x) = A\sin kx + B\cos kx$.

Boundary conditions

$\psi(0)=0 \Rightarrow B=0$. $\psi(L)=0 \Rightarrow \sin kL = 0 \Rightarrow kL = n\pi$, $n=1,2,3,\dots$

So $k_n = n\pi/L$.

Energy eigenvalues

$$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}$$

Normalized wavefunctions

Normalization $\int_0^L A^2\sin^2(k_n x)\,dx = 1$ gives $A^2 (L/2) = 1$, so $A=\sqrt{2/L}$:

$$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right)$$

Numerical part (ground state, $n=1$)

$L = 0.20\times10^{-9}\ \text{m} = 2.0\times10^{-10}\ \text{m}$.

$$E_1 = \frac{h^2}{8mL^2} = \frac{(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(2.0\times10^{-10})^2}$$

Numerator: $(6.626\times10^{-34})^2 = 4.390\times10^{-67}\ \text{J}^2\text{s}^2$.

Denominator: $8 \times 9.11\times10^{-31} \times 4.0\times10^{-20} = 2.915\times10^{-49}$.

$$E_1 = \frac{4.390\times10^{-67}}{2.915\times10^{-49}} = 1.506\times10^{-18}\ \text{J}$$

Convert: $E_1 = \dfrac{1.506\times10^{-18}}{1.602\times10^{-19}} = \mathbf{9.40\ eV}$.

Laser action

An atom in an excited state can drop to a lower state by spontaneous emission (random, incoherent) or, when struck by a photon of matching energy, by stimulated emission, which produces a second photon identical in phase, direction, frequency and polarization. Light amplification requires more atoms in the upper than the lower level — a population inversion — which is impossible in a two-level system at equilibrium. A metastable state (long lifetime) lets atoms accumulate in the upper level so that stimulated emission dominates. An optical resonator (mirrors) feeds photons back, building a coherent, monochromatic, highly directional beam.

Spontaneous vs stimulated emission

Numerical part

Numerical aperture:

$$\text{NA} = \sqrt{n_1^2 - n_2^2} = \sqrt{1.50^2 - 1.46^2} = \sqrt{2.25 - 2.1316} = \sqrt{0.1184} = 0.3441$$

NA $\approx 0.344$.

Acceptance angle (in air, $n_0=1$): $\sin\theta_a = \text{NA}$, so

$$\theta_a = \sin^{-1}(0.3441) = \mathbf{20.1^{\circ}}$$

Sabine's formula

$$T = \frac{0.161\,V}{A} = \frac{0.161\,V}{\sum a_i S_i}$$

where $T$ = reverberation time (s) — the time for sound intensity to fall to $10^{-6}$ of its initial value (a 60 dB drop); $V$ = volume of the hall (m³); $A = \sum a_i S_i$ = total absorption (m² sabin), with $a_i$ the absorption coefficient and $S_i$ the area of each surface; the constant $0.161\ \text{s/m}$ follows from the speed of sound.

Calculation

$V = 2500\ \text{m}^3$, $A = 400\ \text{m}^2$ sabin.

$$T = \frac{0.161 \times 2500}{400} = \frac{402.5}{400} = 1.006\ \text{s}$$

$T \approx 1.01\ \text{s}$.

With extra absorption

New total absorption $A' = 400 + 90 = 490\ \text{m}^2$ sabin.

$$T' = \frac{0.161 \times 2500}{490} = \frac{402.5}{490} = 0.8214\ \text{s}$$

$T' \approx 0.82\ \text{s}$ — the added absorption shortens the reverberation time, improving clarity.

Diffraction grating

A plane diffraction grating is an optical element with a large number of equally spaced parallel slits (or lines) of period $d$. When light passes through, the diffracted beams interfere, producing sharp principal maxima where

$$d\sin\theta = n\lambda \quad (n = 0, 1, 2, \dots)$$

Here $d$ is the grating element (slit separation), $\theta$ the diffraction angle and $n$ the order.

Calculation

Grating constant:

$$d = \frac{1\ \text{cm}}{5000} = \frac{1\times10^{-2}\ \text{m}}{5000} = 2.0\times10^{-6}\ \text{m}$$

First order ($n=1$), $\lambda = 600\times10^{-9}\ \text{m}$:

$$\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 600\times10^{-9}}{2.0\times10^{-6}} = 0.300$$ $$\theta = \sin^{-1}(0.300) = \mathbf{17.46^{\circ}}$$

Derivation

By symmetry the field $\vec E$ is radial and uniform on a coaxial Gaussian cylinder of radius $r$ and length $\ell$. Flux through the curved surface only (ends contribute zero):

$$\oint \vec E\cdot d\vec A = E\,(2\pi r\,\ell)$$

Enclosed charge $Q_{enc} = \lambda\ell$. Gauss's law $\oint\vec E\cdot d\vec A = Q_{enc}/\varepsilon_0$ gives

$$E\,(2\pi r\ell) = \frac{\lambda\ell}{\varepsilon_0} \;\Rightarrow\; \boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$

Calculation

$\lambda = 4.0\times10^{-8}\ \text{C/m}$, $r = 0.20\ \text{m}$, $\dfrac{1}{2\pi\varepsilon_0} = 2k = 1.798\times10^{10}\ \text{N·m}^2/\text{C}^2$ (using $k = 8.99\times10^9$).

$$E = \frac{2k\lambda}{r} = \frac{2(8.99\times10^9)(4.0\times10^{-8})}{0.20} = \frac{719.2}{0.20} = 3596\ \text{N/C}$$

$E \approx 3.6\times10^{3}\ \text{N/C}$, directed radially outward.

Poynting vector

$$\vec S = \frac{1}{\mu_0}\,\vec E\times\vec B$$

It gives the instantaneous rate of energy flow per unit area (W/m²) carried by the electromagnetic field, directed along the propagation of the wave. Its time average is the wave intensity.

Calculation

(i) In vacuum $E_0 = cB_0$, so

$$B_0 = \frac{E_0}{c} = \frac{50}{3.0\times10^8} = 1.667\times10^{-7}\ \text{T} \approx \mathbf{1.67\times10^{-7}\ T}$$

(ii) Average intensity:

$$I = \langle S\rangle = \frac{1}{2}\,c\,\varepsilon_0 E_0^2$$ $$I = \frac{1}{2}(3.0\times10^8)(8.85\times10^{-12})(50)^2$$ $$= \frac{1}{2}(3.0\times10^8)(8.85\times10^{-12})(2500)$$

Step: $3.0\times10^8 \times 8.85\times10^{-12} = 2.655\times10^{-3}$. Then $\times 2500 = 6.6375$. Half: $3.319$.

$I \approx 3.32\ \text{W/m}^2$.

Hall effect

When a current-carrying conductor is placed in a magnetic field perpendicular to the current, the moving charge carriers experience a Lorentz force $q\vec v\times\vec B$ that pushes them to one side. Charge accumulates until the transverse electric field balances the magnetic force, producing a measurable Hall voltage $V_H$ across the width. Its sign reveals the carrier type (n- or p-).

The Hall coefficient is

$$R_H = \frac{1}{nq} = \frac{V_H\,t}{I\,B}$$

where $t$ is the thickness along $\vec B$, $I$ the current and $B$ the field.

Calculation

$V_H = 6.0\times10^{-3}\ \text{V}$, $t = 1.0\times10^{-3}\ \text{m}$, $I = 3.0\ \text{A}$, $B = 0.50\ \text{T}$.

(i)

$$R_H = \frac{V_H\,t}{I\,B} = \frac{(6.0\times10^{-3})(1.0\times10^{-3})}{(3.0)(0.50)} = \frac{6.0\times10^{-6}}{1.5} = 4.0\times10^{-6}\ \text{m}^3/\text{C}$$

$R_H = 4.0\times10^{-6}\ \text{m}^3/\text{C}$.

(ii) From $R_H = 1/(nq)$ with $q = 1.602\times10^{-19}\ \text{C}$:

$$n = \frac{1}{R_H\,q} = \frac{1}{(4.0\times10^{-6})(1.602\times10^{-19})} = \frac{1}{6.408\times10^{-25}} = 1.56\times10^{24}\ \text{m}^{-3}$$

$n \approx 1.56\times10^{24}\ \text{m}^{-3}$.

Superconductivity and the Meissner effect

Superconductivity is the complete loss of electrical resistance in certain materials below a critical temperature $T_c$. The Meissner effect is the expulsion of magnetic flux from the interior of a superconductor as it is cooled below $T_c$: the material becomes a perfect diamagnet ($B = 0$ inside), which is a stronger condition than mere zero resistance.

Type-I vs Type-II

Calculation

The critical field varies with temperature as

$$H_c(T) = H_c(0)\left[1 - \left(\frac{T}{T_c}\right)^2\right]$$

With $H_c(0) = 0.030\ \text{T}$, $T = 3.0\ \text{K}$, $T_c = 7.0\ \text{K}$:

$$\left(\frac{T}{T_c}\right)^2 = \left(\frac{3.0}{7.0}\right)^2 = (0.4286)^2 = 0.1837$$ $$H_c(3.0) = 0.030\,(1 - 0.1837) = 0.030 \times 0.8163 = 0.02449\ \text{T}$$

(i) $H_c(3.0\ \text{K}) \approx 0.0245\ \text{T}$.

(ii) Setting $H_c(T) = \tfrac{1}{2}H_c(0)$:

$$1 - \left(\frac{T}{T_c}\right)^2 = 0.5 \;\Rightarrow\; \frac{T}{T_c} = \sqrt{0.5} = 0.7071$$ $$T = 0.7071 \times 7.0 = 4.95\ \text{K}$$

$T \approx 4.95\ \text{K}$.