Problem set 47

1. The Lagrangian of a particle moving in a plane is given in Cartesian coordinates as $$L=\dot x\dot y-x^2-y^2$$ In polar coordinates the expression for the canonical momentum (conjugate to the radial coordinate ) is
1. $\dot r\sin\theta+r\dot\theta\cos\theta$
2. $\dot r\cos\theta+r\dot\theta\sin\theta$
3. $2\dot r\cos{2\theta}-r\dot\theta\sin{2\theta}$
4. $\dot r\sin{2\theta}+r\dot\theta\cos{2\theta}$

Substituting $x=r\cos\theta$ and $y=r\sin\theta$ in $L$ $$\begin{align*} L&={\scriptstyle\left(\dot r\cos\theta-r\dot\theta\sin\theta\right)\left(\dot r\sin\theta+r\dot\theta\cos\theta\right)}\\ &-r^2\cos^2\theta-r^2\sin^2\theta\\ &=\dot r^2\cos\theta\sin\theta+r\dot r\dot\theta\cos^2\theta\\ &-r\dot r\dot\theta\sin^2\theta- r^2\dot\theta^2\cos\theta\sin\theta-r^2\\ &=\frac{1}{2}\dot r^2\sin{2\theta}+r\dot r\dot\theta\cos{2\theta}\\ &-\frac{1}{2} r^2\dot\theta^2\sin{2\theta}-r^2 \end{align*}$$ $$\begin{align*} p_r=\frac{\partial L}{\partial\dot r}=\dot r\sin{2\theta}+r\dot\theta\cos{2\theta} \end{align*}$$

Hence, answer is (D)

2. The Hermite polynomial $H_n(x)$ satisfies the differential equation $$\frac{d^2H_n}{dx^2}-2x\frac{dH_n}{dx}+2nH_n(x)=0$$ . The corresponding generating function $G(x,t)=\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n(x)t^n$ satisfies the
1. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}=0$
2. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}-2t^2\frac{\partial G}{\partial t}=0$
3. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial G}{\partial t}=0$
4. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial^2 G}{\partial x\partial t}=0$
Let us consider option (A)
$\frac{\partial^2G}{\partial x^2}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{d^2H_n}{d x^2}t^n$, $\frac{\partial G}{\partial x}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{dH_n}{d x}t^n$, $2t\frac{\partial G}{\partial t}=2n\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n( x)t^n$ $$\begin{align*} &\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}\\ &{\scriptstyle =\sum\limits_{n=0}^{\infty}\frac{1}{n!}\left(\frac{d^2H_n}{d x^2}-2x\frac{1}{n!}\frac{dH_n}{d x}+2n\frac{1}{n!}H_n( x)\right)t^n}\\ &=0 \end{align*}$$ Hence, answer is (A)

3. A sinusoidal signal of peak to peak amplitude 1V and unknown time period is input to the following circuit for 5 seconds duration. If the counter measures a value (3E8)_H in hexadecimal then the time period of the input signal is
   
1. 2.5 ms
2. 4 ms
3. 10 ms
4. 5 ms

In general n-bit counter counts $2^n$ numbers in a sequence ranging from 0 to $2^n-1$ and the counter cycles through the same sequence of numbers continuously so long as there is an incoming clock pulse. For one complete cycle we require $2^n-1$ pulses.

If $T$ is time period of a pulse then time required for one complete cycle of counts = $T\times(2^n-1)$.

Now, to count a decimal number $N$ we require $N$ pulses. This can be seen from a table below for a 2-bit counter.

Input pulse	Count	decimal equivalent
0	00	0
1	01	1
2	10	2
3	11	3

The decimal equivalent of $(3E8)_H={\scriptstyle 3\times16^2+14\times16^1+8\times16^0}=1000$ Hence, 1000 pulses of time period $T$ will be required to count the number $(3E8)_H$ in 5 seconds. $$5 \:sec=1000\times T$$ $$T=\frac{5}{1000}=5\:msec$$

Hence, answer is (D)

4. For a dynamical system governed by the equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, with $|x|\leq1$,
1. $x=-1$ and $x=1$ are both unstable fixed points
2. $x=-1$ and $x=1$ are both stable fixed points
3. $x=-1$ is an unstable fixed point and $x=1$ is a stable fixed point
4. $x=-1$ is a stable fixed point and $x=1$ is an unstable fixed point
Let $\frac{dx}{dt}=2\sqrt{1-x^2}=f(x)$

As $f(-1)=0$, x=-1 is a fixed point.

As $f(1)=0$, x=1 is a fixed point. $$f'(x)=\frac{-2x}{\sqrt{1-x^2}}$$

As $f'(-1)=\frac{2}{0}=\infty > 0$, $x=-1$ is a unstable point.

As $f'(1)=\frac{-2}{0}=-\infty < 0$, $x=1$ is a stable point.

Hence, answer is (C)

5. The value of the integral $\int_0^8\frac{1}{x^2+5}dx$, evaluated using Simpson’s $\frac{1}{3}$ rule with $h=2$, is
1. 0.565
2. 0.620
3. 0.698
4. 0.736
According to Simpson's 1/3 rule $$\begin{align*} I&=\frac{h}{3}\left[y_0+y_n\right.\\ &+4\left(y_1+y_3+\dots+y_{n-1}\right)\\ &\left.+2\left(y_2+y_4\dots+y_{n-2}\right)\right] \end{align*}$$ $$n=\frac{x_n-x_0}{h}=\frac{8-0}{2}=4$$ $$I=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2\left(y_2\right)\right]$$ $$y_0=f(x_0)=f(0)=\frac{1}{5}$$ $$y_1=f(x_1)=f(2)=\frac{1}{9}$$ $$y_2=f(x_2)=f(4)=\frac{1}{21}$$ $$y_3=f(x_3)=f(6)=\frac{1}{41}$$ $$y_4=f(x_4)=f(8)=\frac{1}{69}$$ $$\begin{align*} I&={\scriptstyle \frac{2}{3}\left[\frac{1}{5}+\frac{1}{69}+4\left(\frac{1}{9}+\frac{1}{41}\right)+2\left(\frac{1}{21}\right)\right]}\\ &=0.568 \end{align*}$$

Hence, answer is (A)