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Sunday 12 February 2017
Problem set 70
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
$-1/2$
$-1$
$1/2$
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)
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
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)
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 $P_2=\frac{1}{\sqrt{2}}(p-q)$
neither A nor B
both A and B
only A
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)
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
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?
$TE_{10}$, $TE_{01}$, and $TE_{20}$
$TE_{10}$, $TE_{11}$, and $TE_{20}$
$TE_{01}$, $TE_{10}$, and $TE_{11}$
$TE_{01}$, $TE_{10}$, 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.
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