Problem set 70

1. Consider the differential equation $\frac{dy}{dx}=x^2-y$ with initial condition $y=2$ at $x=0$. Let $y_{(1)}$ and $y_{(1/2)}$ be the solutions at $x=1$ obtained using Euler's forward algorithm with step size 1 and $\frac{1}{2}$ respectively. The value of $\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}$ is
1. $-1/2$
2. $-1$
3. $1/2$
4. 1

Euler's forward formula is $$y_{n+1}=y_n+hf(x_n,y_n)$$ where, $f(x,y)=x^2-y$. \begin{align*} y_{1}&=y_0+f(x_0,y_0)\\ &=2+f(0,2)\\ &=0 \end{align*} \begin{align*} y_{1/2}&=y_0+\frac{1}{2}f(x_0,y_0)\\ &=2+\frac{1}{2}f(0,2)\\ &=1 \end{align*} $$\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}=-1$$

Hence, answer is (B)

2. Let $f(x,t)$ be a solution of the wave equation $\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2}$ in 1-dimension. If at $t=0$, $f(x,0)=e^{-x^2}$ and $\frac{\partial f}{\partial t}(x,0)=0$ for all $x$, then $f(x,t)$ for all future times $t > 0$ is described by
1. $e^{-(x^2-v^2t^2)}$
2. $e^{-(x-vt)^2}$
3. $\frac{1}{4}e^{-(x-vt)^2}+\frac{3}{4}e^{-(x+vt)^2}$
4. $\frac{1}{2}\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]$

According to d’Alambert’s formula solution of wave equation is given by \begin{align*} f(x,t)&=\frac{f(x-vt,0)+f(x+vt,0)}{2}\\ &+\frac{1}{2v}\int\limits_{x-vt}^{x+vt}\frac{\partial f(x,0)}{\partial t}dx\\ &=\frac{\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]}{2}\\ &+\frac{1}{2v}\int\limits_{x-vt}^{x+vt}0 dx\\ &=\frac{\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]}{2} \end{align*}

Hence, answer is (D)

3. Let $q$ and $p$ be canonical coordinate and momentum of a dynamical system. Which of the following transformation is canonical?

   A: $Q_1=\frac{1}{\sqrt{2}}q^2$ and $P_1=\frac{1}{\sqrt{2}}p^2$

   B: $Q_2=\frac{1}{\sqrt{2}}(p+q)$ and $P_2=\frac{1}{\sqrt{2}}(p-q)$

1. neither A nor B
2. both A and B
3. only A
4. only B

A transformation is said be canonical if it satisfies Poisson Bracket $\left\{Q,P\right\}_{q,p}=1$ i.e. $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$ Clearly, $$\frac{\partial Q_2}{\partial q}\frac{\partial P_2}{\partial p}-\frac{\partial Q_2}{\partial p}\frac{\partial P_2}{\partial q}=1$$

Hence, answer is (D)

4. The differential cross-section for scattering by a target is given by $\frac{d\sigma}{d\Omega}(\theta,\phi)=a^2+b^2\cos^2\theta$. If $N$ is the flux of the incoming particles, the number of particles scattered per unit time is
1. $\frac{4\pi}{3}N(a^2+b^2)$
2. $4\pi N(a^2+\frac{1}{6}b^2)$
3. $4\pi N(\frac{1}{2}a^2+\frac{1}{3}b^2)$
4. $4\pi N(a^2+\frac{1}{3}b^2)$

$$\frac{d\sigma}{d\Omega}(\theta,\phi)=a^2+b^2\cos^2\theta$$ \begin{align*} \sigma&=\int\frac{d\sigma}{d\Omega}d\Omega\\ &=\int\limits_0^{2\pi}\int\limits_0^{\pi}\frac{d\sigma}{d\Omega}\sin\theta d\theta d\phi\\ &=\int\limits_0^{2\pi}d\phi\int\limits_0^{\pi}\left(a^2+b^2\cos^2\theta\right)\sin\theta d\theta \\ &=\!2\pi\!\!\left[\!a^2\!\!\!\int\limits_0^{\pi}\!\!\sin\theta d\theta+b^2\!\!\!\int\limits_0^{\pi}\!\!\cos^2\!\theta\sin\theta d\theta\right] \\ &=2\pi\left[2a^2+b^2\int\limits_0^{\pi}\cos^2\theta\sin\theta d\theta\right] \\ &=2\pi\left[2a^2+b^2\int\limits_{-1}^{1}x^2 dx\right] \\ &=2\pi\left[2a^2+\frac{2}{3}b^2\right] \\ \sigma&=4\pi\left[a^2+\frac{1}{3}b^2\right] \\ \end{align*} Hence, number of particles scattered $=4\pi N\left[a^2+\frac{1}{3}b^2\right]$

Hence, answer is (D)

5. Consider a rectangular wave guide with transverse dimensions 2 m $\times$ 1 m driven with an angular frequency $\omega=10^9\: rad/s$. Which transverse electric (TE) modes will propagate in this wave guide?
1. $TE_{10}$, $TE_{01}$, and $TE_{20}$
2. $TE_{10}$, $TE_{11}$, and $TE_{20}$
3. $TE_{01}$, $TE_{10}$, and $TE_{11}$
4. $TE_{01}$, $TE_{10}$, and $TE_{22}$

Cut off frequency for rectangular waveguide is given by $$f_{c_{mn}}=\frac{v}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ where, $v=\frac{1}{\sqrt{\epsilon_0\mu_0}}\frac{1}{\sqrt{\epsilon_r}}=\frac{c}{\sqrt{\epsilon_r}}$ is velocity propagation in dielectric.

The cutoff frequency for $TE_{mn}$ mode is the operating frequency below which attenuation occurs and above which propagation takes place.

For air filled wave guide $v=c$ $$f_{c_{mn}}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ Now, $\omega=10^9\: rad/s$, hence, $f=\frac{\omega}{2\pi}=1.6\times10^8\:1/sec$ \begin{align*} f_{c_{22}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{2}{1}\right)^2}\\ &=3.35\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{20}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{11}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.68\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{10}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=0.75\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{01}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{0}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} As, $f < f_{c_{22}}$ and $f < f_{c_{11}}$, these modes will attenuate and as $f < f_{c_{22}}$, $f > f_{c_{10}}$, $f > f_{c_{01}}$, these modes will propagate.

Hence, answer is (A)