Problem set 14

1. A particle of unit mass moves in a potential $V(x)=ax^2+\frac{b}{x^2}$, where $a$ and $b$ are positive constants. The angular frequency of small oscillations about the minimum of the potential is
1. $\sqrt{8b}$
2. $\sqrt{8a}$
3. $\sqrt{8a/b}$
4. $\sqrt{8b/a}$
The equation of motion is $$m\ddot{x}=F=-kx$$ $$\ddot{x}=-\frac{k}{m}x=-\omega^2 x$$ \begin{align*} \omega&=\sqrt{-\frac{\ddot{x}}{x}}=\sqrt{-\frac{m\ddot{x}}{mx}}\\ &=\sqrt{-\frac{F}{mx}}=\sqrt{\frac{\frac{dV}{dx}}{mx}}\\ &=\sqrt{\frac{\frac{d^2V}{dx^2}}{m}} \end{align*} Minimum of potential occurs at mean position $x_0$. Hence frequency of oscillation about the minimum of the potential is $$\omega=\sqrt{\frac{\left.\frac{d^2V}{dx^2}\right|_{x=x_0}}{m}}$$ $$\frac{dV}{dx}=2ax-\frac{2b}{x^3}$$ At minimum potential energy, we have $$\left.\frac{dV}{dx}\right|_{x=x_0}=0$$ $$\therefore 2ax_0-\frac{2b}{x_0^3}=0$$ $$\therefore x_0^4=\frac{b}{a}$$ $$\frac{d^2V}{dx^2}=2a+\frac{6b}{x^4}$$ $$\left.\frac{d^2V}{dx^2}\right|_{x=x_0}=2a+\frac{6b}{x_0^4}=8a$$ $$\omega=\sqrt{\frac{\left.\frac{d^2V}{dx^2}\right|_{x=x_0}}{m}}=\sqrt{\frac{8a}{m}}=\sqrt{8a}$$
2. A signal of frequency 10 kHz is being digitized by an A/D converter. A possible sampling time which can be used is
1. $100\quad \mu s$
2. $40\quad \mu s$
3. $60\quad \mu s$
4. $200\quad \mu s$
The conversion time is the time required to complete a conversion of the input signal. It establishes the upper signal frequency limit that can be sampled without aliasing $$f_{max}=\frac{1}{2\times conversion~time}$$ \begin{align*} conversion~time &=\frac{1}{2\times f_{max}}\\ &=\frac{1}{2\times 10\times10^3}\\ &=5\times10^{-5}=50\mu S \end{align*} Hence answer is (B).
Supplementary information: AD converter specifications:
• The conversion time is the time required to complete a conversion of the input singnal. It establishes the upper signal frequency limit that can be sampled without aliasing $$f_{max}=\frac{1}{2\times conversion~time}$$
• Resolution: The number of bits in the converter gives the resolution $${\scriptstyle Resolution=\frac{full-scale~signal}{2^n}}$$
• Accuracy: Accuracy describes how close the measurement is to the actual value. $$Accuracy=\frac{V_{resolution}}{V_{signal}}\times 100\%$$
3. The electrostatic potential $V ( x, y)$ in free space in a region where the charge density $\rho$ is zero is given by $V ( x, y)=4e^{2x}+f(x)-3y$. Given that the x-component of the electric field, $E_x$, and $V$ are zero at the origin, $f (x)$ is
1. $3x^2-4e^{2x}+8x$
2. $3x^2-4e^{2x}+16x$
3. $4e^{2x}-8$
4. $3x^2-4e^{2x}$
$V ( x, y)$ is zero at origin is zero gives $f(0)=-4$. This requirement is satisfied by all options. So let us calculate electric field. $$\vec F=q_0\vec E$$ If test charge $q_0=1$ then \begin{align*} \vec E&=\vec F=-\vec \nabla V\\ &={\scriptstyle -\left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z}\right)\left(4e^{2x}+f(x)-3y\right)} \end{align*} $$\vec E=-\hat i \left(8e^{2x}+\frac{\partial f(x)}{\partial x}\right)+\hat j(3)$$ Hence, x-component of electric field is $$E_x(x)=-\left(8e^{2x}+\frac{\partial f(x)}{\partial x}\right)$$ However, $E_x=0$ at origin. $$E_x(0)=-\left(8e^{0}+\frac{\partial f(0)}{\partial x}\right)=0$$ Hence, $$\frac{\partial f(0)}{\partial x}=-8$$ For $f(x)=3x^2-4e^{2x}+8x$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}+8$$ $$\frac{\partial f(0)}{\partial x}=0-8+8=0$$
For $f(x)=3x^2-4e^{2x}+16x$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}+16$$ $$\frac{\partial f(0)}{\partial x}=0-8+16=8$$
For $f(x)=4e^{2x}-8$, $$\frac{\partial f}{\partial x}=8e^{2x}$$ $$\frac{\partial f(0)}{\partial x}=8$$
For $f(x)=3x^2-4e^{2x}$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}$$ $$\frac{\partial f(0)}{\partial x}=0-8=-8$$ Hence, Answer is (D).
4. Consider the transition of liquid water to steam as water boils at a temperature of $100^oC$ under a pressure of 1 atmosphere. Which one of the following quantities does not change discontinuously at the transition?
1. The Gibbs free energy
2. The internal energy
3. The entropy
4. The specific volume
(A) The Gibbs free energy.
5. The value of the integral $\int\limits_Cdz\:z^2\:e^z$, where C is an open contour in the complex z-plane as shown in the figure below, is:
   
   1. $\frac{5}{e}+e$
   2. $e-\frac{5}{e}$
   3. $\frac{5}{e}-e$
   4. $-\frac{5}{e}-e$
   \begin{align*} &\int_Cdz\:z^2\:e^z\\ &=\int_Cd(x+iy)\:(x+iy)^2\:e^{(x+iy)} \end{align*} Along path (1) $x=1$, $dx=0$ and $y$ changes 0 to 1 \begin{align*} I_1&=i\int_0^1 (1+iy)^2e^{(1+iy)}dy\\ &=i\int_0^1 (1-y^2+2iy)e^{(1+iy)}dy \end{align*} Along path (2) $y=1$, $dy=0$ and $x$ changes 1 to $-1$ \begin{align*} I_2&=\int_{-1}^1 (x+i)^2e^{(x+i)}dx\\ &=\int_{-1}^1 (x^2-1+2ix)e^{(x+i)}dx \end{align*} Along path (3) $x=-1$, $dx=0$ and $y$ changes 1 to 0 \begin{align*} I_3&=i\int_{-1}^0 (-1+iy)^2e^{(-1+iy)}dy\\ &=i\int_{-1}^0 (1-y^2-2iy)e^{(-1+iy)}dy \end{align*} $$I=I_1+I_2+I_3=\frac{5}{e}+e$$