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ChatGPT
Oct 13
The number 28 is an even number that comes after 27 and before 29. It is the sum of the first seven prime numbers (2 + 3 + 5 + 7 + 11 + 13 + 17 = 28) and is also a perfect number, meaning that the sum of its divisors (excluding itself) equals the number itself (1 + 2 + 4 + 7 + 14 = 28). In Roman numerals, 28 is represented as XXVIII.
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Google Gemini AI
Oct 13
(a) A system of two particles, each of mass m, is constrained to move along the x-axis with a force field V(x) acting on them. If the potential is symmetric about the origin, V(x) = V(-x), find the normal modes of oscillation.
Solution:
The total potential energy of the system is given by:
$$V(x_1, x_2) = V(x_1) + V(x_2)$$
Since the potential is symmetric about the origin, we can write:
$$V(x_1) = V_0 + \frac{1}{2}k_1 x_1^2$$
$$V(x_2) = V_0 + \frac{1}{2}k_1 x_2^2$$
where $V_0$ and $k_1$ are constants.
The Lagrangian of the system is:
$$L = \frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}m\dot{x}_2^2 - V_0 - \frac{1}{2}k_1x_1^2 - \frac{1}{2}k_1x_2^2$$
The equations of motion are:
$$m\ddot{x}_1 + k_1x_1 = 0$$
$$m\ddot{x}_2 + k_1x_2 = 0$$
The normal modes of oscillation are given by the solutions to the above equations. We can write the solutions in the form:
$$x_1(t) = A_1\sin(\omega t + \phi_1)$$
$$x_2(t) = A_2\sin(\omega t + \phi_2)$$
where $A_1$, $A_2$, $\phi_1$, and $\phi_2$ are constants.
Substituting the above solutions into the equations of motion, we get:
$$-m\omega^2 A_1 \sin(\omega t + \phi_1) + k_1A_1\sin(\omega t + \phi_1) = 0$$
$$-m\omega^2 A_2 \sin(\omega t + \phi_2) + k_1A_2\sin(\omega t + \phi_2) = 0$$
Dividing both sides of the above equations by $A_1\sin(\omega t + \phi_1)$ and $A_2\sin(\omega t + \phi_2)$, respectively, we get:
$$-\omega^2 + \frac{k_1}{m} = 0$$
$$\omega = \sqrt{\frac{k_1}{m}}$$
Since the potential is symmetric about the origin, the two particles will oscillate in phase, i.e., $\phi_1 = \phi_2$. Therefore, the normal modes of oscillation are:
$$x_1(t) = A\sin(\omega t)$$
$$x_2(t) = A\sin(\omega t)$$
where $A$ is a constant.
Solution:
The total potential energy of the system is given by:
$$V(x_1, x_2) = V(x_1) + V(x_2)$$
Since the potential is symmetric about the origin, we can write:
$$V(x_1) = V_0 + \frac{1}{2}k_1 x_1^2$$
$$V(x_2) = V_0 + \frac{1}{2}k_1 x_2^2$$
where $V_0$ and $k_1$ are constants.
The Lagrangian of the system is:
$$L = \frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}m\dot{x}_2^2 - V_0 - \frac{1}{2}k_1x_1^2 - \frac{1}{2}k_1x_2^2$$
The equations of motion are:
$$m\ddot{x}_1 + k_1x_1 = 0$$
$$m\ddot{x}_2 + k_1x_2 = 0$$
The normal modes of oscillation are given by the solutions to the above equations. We can write the solutions in the form:
$$x_1(t) = A_1\sin(\omega t + \phi_1)$$
$$x_2(t) = A_2\sin(\omega t + \phi_2)$$
where $A_1$, $A_2$, $\phi_1$, and $\phi_2$ are constants.
Substituting the above solutions into the equations of motion, we get:
$$-m\omega^2 A_1 \sin(\omega t + \phi_1) + k_1A_1\sin(\omega t + \phi_1) = 0$$
$$-m\omega^2 A_2 \sin(\omega t + \phi_2) + k_1A_2\sin(\omega t + \phi_2) = 0$$
Dividing both sides of the above equations by $A_1\sin(\omega t + \phi_1)$ and $A_2\sin(\omega t + \phi_2)$, respectively, we get:
$$-\omega^2 + \frac{k_1}{m} = 0$$
$$\omega = \sqrt{\frac{k_1}{m}}$$
Since the potential is symmetric about the origin, the two particles will oscillate in phase, i.e., $\phi_1 = \phi_2$. Therefore, the normal modes of oscillation are:
$$x_1(t) = A\sin(\omega t)$$
$$x_2(t) = A\sin(\omega t)$$
where $A$ is a constant.
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