Problem set 27

1. The Fourier transform of $f(x)$ is $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$. If $f(x)=\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)$, where $\delta(x)$ is the Dirac delta-function (and prime denotes derivative), what is $\tilde{f}(k)$?
1. $\alpha+i\beta k+i\gamma k^2$
2. $\alpha+\beta k-\gamma k^2$
3. $\alpha-i\beta k-\gamma k^2$
4. $i\alpha+\beta k-i\gamma k^2$
For Dirac delta function, we have, $$\int\limits f(x)\delta(x-y)\:dx=f(y)$$ $$\int\limits f(x)\delta'(x-y)\:dx=-f'(y)$$ $$\int\limits f(x)\delta''(x-y)\:dx=(-1)^2f''(y)$$ $$\int\limits f(x)\delta^n(x-y)\:dx=(-1)^nf^n(y)$$ $$\begin{align*} \tilde{f}(k)&={\scriptstyle\int\limits_{-\infty}^{\infty}dx\:e^{ikx}\left(\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)\right)}\\ &=\alpha f(0)+\beta(-f'(0))+\gamma f''(0)\\ &=\alpha-i\beta k-\gamma k^2 \end{align*}$$ Hence, answer is (C)
2. A particle moves in three-dimensional space in a central potential $V(r)=kr^4$, where $k$ is a constant. The angular frequency $\omega$ for a circular orbit depends on its radius $R$ as
1. $\omega\propto R$
2. $\omega\propto R^{-1}$
3. $\omega\propto R^{1/4}$
4. $\omega\propto R^{-2/3}$
Force acting on particle is given by $$\begin{eqnarray} \begin{array}{c}\vec F(r)=-\nabla V(r)\\ {\scriptstyle=-\left(\hat r\frac{\partial}{\partial r}+\hat\theta\frac{1}{r}\frac{\partial}{\partial\theta}+\hat\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\right)\left(kr^4\right)}\\ =-4kr^3\hat r\end{array} \end{eqnarray}$$ For circular motion in radius $R$, we must have $$\frac{mv^2}{R}=F(R)=-4kR^3$$ Using $V=R\omega$ $$\frac{mR^2\omega^2}{R}=-4kR^3$$ $$\omega^2=\frac{-4kR^2}{m}$$ $$\omega\propto R$$ Hence, answer is (A)
3. The Lagrangian of a system is given by ${\scriptstyle L=\frac{1}{2}m\dot q_1^2+2m\dot q_2^2-k\left(\frac{5}{4}q_1^2+2q_2^2-2q_1q_2\right)}$ where $m$ and $k$ are positive constants. The frequencies of its normal modes are
1. $\sqrt{\frac{k}{2m}}\sqrt{\frac{3k}{m}}$
2. $\sqrt{\frac{k}{2m}}(13\pm\sqrt{73})$
3. $\sqrt{\frac{5k}{2m}}\sqrt{\frac{k}{m}}$
4. $\sqrt{\frac{k}{2m}}\sqrt{\frac{6k}{m}}$
Lagrange equation is $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_j}\right)-\frac{\partial L}{\partial q_j}$$ Substituting value of $L$ for generalized coordinates $q_1$ and $q_2$ we get $$m\ddot q_1+k\left(\frac{5}{2}q_1-2q_2\right)=0\quad (1)$$ $$4m\ddot q_2+k\left(4q_2-2q_1\right)=0\quad (2)$$ Form normal modes, both particles moves simple harmonically with same frequency $\omega$, hence, we have $$\ddot q_1=-\omega^2q_1$$ $$\ddot q_2=-\omega^2q_2$$ Substituting values of $\ddot q_1$ and $\ddot q_2$ in equations (1) and (2), we get $$\left(\frac{5}{2}k-m\omega^2\right)q_1=2kq_2\quad (3)$$ $$\left(4k-4m\omega^2\right)q_2=2kq_1\quad (4)$$ From equation (3) and (4) we get $$\frac{q_2}{q_1}=\frac{\left(\frac{5}{2}k-m\omega^2\right)}{2k}$$ $$\frac{q_2}{q_1}=\frac{2k}{\left(4k-4m\omega^2\right)}$$ Hence, we have $$\frac{\left(\frac{5}{2}k-m\omega^2\right)}{2k}=\frac{2k}{\left(4k-4m\omega^2\right)}$$ Solving this equation, we get $$2m^2\omega^4-7km\omega^2+3k^2=0$$ This is a quadratic equation in $\omega^2$. Solving this equation we get $$\omega^2=\frac{k}{2m}\quad\text{or }\frac{3k}{m}$$ or we get $$\omega=\sqrt{\frac{k}{2m}}\quad\text{or }\sqrt{\frac{3k}{m}}$$ Hence, answer is (A)
4. Consider a particle of mass $m$ moving with a speed $v$. If $T_R$ denotes the relativistic kinetic energy and $T_N$ its non-relativistic approximation, then the value of $(T_R-T_N)/T_R$ for $v=0.01\:c$, is
1. $1.25\times10^{-5}$
2. $5.0\times10^{-5}$
3. $7.5\times10^{-5}$
4. $1.0\times10^{-4}$
$$T_R=(\gamma-1)mc^2$$ where, $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1.00005000375031$ $$T_R=5.000375031\times10^{-5}mc^2$$ $$T_N=0.5\times mv^2=5\times10^{-5}mc^2$$ $$\frac{(T_R-T_N)}{T_R}=7.5\times10^{-5}$$ Hence, answer is (C).
5. Two masses, $m$ each, are placed at the points $(x,y)=(a,a)$ and $(-a,-a)$. Two masses, $2m$ each, are placed at the points $(a,-a)$ and $(-a,a)$. The principal moments of inertia of the system are
1. $2ma^2$, $4ma^2$
2. $4ma^2$, $8ma^2$
3. $4ma^2$, $4ma^2$
4. $8ma^2$, $8ma^2$
Inertia tensor is given by $${\scriptscriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2+x_{\alpha,3}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,3}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,3}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,3}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,3}x_{\alpha,1}&-\sum\limits_\alpha m_\alpha x_{\alpha,3}x_{\alpha,2}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,2}^2\right) \end{pmatrix} }$$ where, $\alpha$ runs over number of particles. Here, $x_{\alpha,1}$ means x-coordinate of particle number $\alpha$, $x_{\alpha,2}$ means y-coordinate of particle number $\alpha$ and $x_{\alpha,3}$ means z-coordinate of particle number $\alpha$

For a two-dimensional case, we have $${\scriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right) \end{pmatrix} }$$ $$\begin{align*} I_{11}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)\\ &=ma^2+ma^2+2ma^2+2ma^2\\ &=6ma^2 \end{align*}$$ $$\begin{align*} I_{12}&= -\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ &=-(ma^2+ma^2-2ma^2-2ma^2)\\ &=2ma^2 \end{align*}$$ $$\begin{align*} I_{21}&= -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}\\ &=-(ma^2+ma^2-2ma^2-2ma^2)\\ &=2ma^2 \end{align*}$$ $$\begin{align*} I_{22}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)\\ &=ma^2+ma^2+2ma^2+2ma^2\\ &=6ma^2 \end{align*}$$ $${I}=\begin{pmatrix} 6ma^2&2ma^2\\ 2ma^2&6ma^2 \end{pmatrix}$$ The principal moments of inertia are obtained by solving $${I}=\begin{vmatrix} 6ma^2-\lambda&2ma^2\\ 2ma^2&6ma^2-\lambda \end{vmatrix}=0$$ Solving this determinant we get $$\lambda^2-12ma^2\lambda+32m^2a^4=0$$ Solving this quadratic equation in $\lambda$ we get, $\lambda_1=8ma^2$ and $\lambda_2=4ma^2$
Hence, answer is (B)