Differential Equation for Spring Oscillation, General Solution, Period, Velocity, and Acceleration

1. Setting Up the Differential Equation

Physical Setup

Consider a spring of negligible mass with one end fixed. A particle of mass m is attached to the free end. When the mass is displaced by a distance x (or l) from its equilibrium (natural) position, a restoring force acts on it.

Applying Hooke's Law

From Hooke's Law, the restoring force developed in the spring is proportional to the displacement:

$$F = -kx$$

where:

• $k$ = spring constant (force per unit extension, in N/m)
• $x$ = displacement from equilibrium
• The negative sign indicates the restoring force acts opposite to the direction of displacement

Applying Newton's Second Law

From Newton's Second Law of Motion:

$$F = ma = m\frac{d^2x}{dt^2}$$

Combining Both Laws

Equating the two expressions:

$$m\frac{d^2x}{dt^2} = -kx$$

$$\boxed{\frac{d^2x}{dt^2} + \frac{k}{m}x = 0}$$

This is the differential equation of Simple Harmonic Motion (SHM) for a spring-mass system.

Let $\omega^2 = \dfrac{k}{m}$, then:

$$\frac{d^2x}{dt^2} + \omega^2 x = 0$$

where $\omega$ is the angular frequency (rad/s).

2. General Solution of the Differential Equation

The equation:

$$\frac{d^2x}{dt^2} + \omega^2 x = 0$$

is a second-order linear homogeneous ODE with constant coefficients.

Auxiliary (Characteristic) Equation

Assume a solution of the form $x = e^{\lambda t}$. Substituting:

$$\lambda^2 + \omega^2 = 0$$

$$\lambda^2 = -\omega^2 \implies \lambda = \pm i\omega$$

General Solution (complex roots)

Since the roots are purely imaginary ($\pm i\omega$), the general solution is:

$$\boxed{x(t) = A\sin(\omega t) + B\cos(\omega t)}$$

or equivalently written as:

$$x(t) = r\sin(\omega t + \phi)$$

where:

• $A$, $B$ (or $r$, $\phi$) are constants determined by initial conditions
• $r = \sqrt{A^2 + B^2}$ is the amplitude
• $\phi$ is the initial phase angle

Note: Using circular motion analogy with a particle moving on a circle of radius $r$, the displacement is written as $y = r\sin(\omega t)$, which is the special case when $\phi = 0$ (particle starts from equilibrium).

3. Expression for Period

The angular frequency is:

$$\omega = \sqrt{\frac{k}{m}}$$

The time period $T$ is the time for one complete oscillation. Since $\omega = \dfrac{2\pi}{T}$:

$$\frac{2\pi}{T} = \sqrt{\frac{k}{m}}$$

$$\boxed{T = 2\pi\sqrt{\frac{m}{k}}}$$

The frequency of oscillation is:

$$f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

4. Expression for Velocity

Starting from the displacement equation:

$$x = r\sin(\omega t)$$

Taking the time derivative:

$$v = \frac{dx}{dt} = r\omega\cos(\omega t)$$

Using the identity $\cos(\omega t) = \sqrt{1 - \sin^2(\omega t)}$ and $\sin(\omega t) = \dfrac{x}{r}$:

$$v = r\omega\sqrt{1 - \frac{x^2}{r^2}}$$

$$\boxed{v = \omega\sqrt{r^2 - x^2}}$$

Key observations:

5. Expression for Acceleration

Taking the time derivative of velocity:

$$a = \frac{dv}{dt} = \frac{d}{dt}(r\omega\cos(\omega t)) = -r\omega^2\sin(\omega t)$$

Since $x = r\sin(\omega t)$:

$$\boxed{a = -\omega^2 x}$$

Key observations:

The negative sign confirms that acceleration is always directed opposite to displacement, which is the defining characteristic of SHM.

Summary Table

Magnetic Dipole Moment: Effect on Atoms and Molecules

1. Magnetic Dipole Moment

A magnetic dipole moment is a measure of the strength of a magnetic dipole (a tiny current loop or a spinning charged particle). It is defined as the product of the current flowing in a loop and the area enclosed by that loop.

$$\vec{\mu} = I \cdot \vec{A}$$

where:

• $I$ = current in the loop
• $A$ = area of the loop
• Direction: perpendicular to the plane of the loop (by right-hand rule)

SI Unit: A·m² (Ampere-square metre)

2. Magnetic Dipole Moment of an Electron in an Atom

Consider an electron moving in a circular orbit of radius $r$ with linear velocity $v$ and angular velocity $\omega$ around the nucleus.

• Charge of electron = $e$
• Mass of electron = $m$
• The revolving electron constitutes a current loop.

Current due to orbital motion:

The electron completes $f$ revolutions per second, so the equivalent current is:

$$I = \frac{e}{T} = ef = \frac{e\omega}{2\pi}$$

Magnetic dipole moment:

$$\mu = I \cdot A = \frac{e\omega}{2\pi} \cdot \pi r^2$$

$$\boxed{\mu = \frac{e\omega r^2}{2}}$$

Since angular momentum $L = m\omega r^2$, we can write:

$$\mu = \frac{e}{2m} \cdot L$$

This is the orbital magnetic moment of the electron.

3. Effect on Atoms

(a) Orbital Magnetic Moment

Every electron orbiting the nucleus acts as a tiny current loop and possesses an orbital magnetic dipole moment as derived above.

(b) Spin Magnetic Moment

In addition to orbital motion, every electron has an intrinsic spin, which also contributes a spin magnetic moment. The total magnetic moment of an atom is the vector sum of all orbital and spin magnetic moments of its electrons.

(c) Torque in External Field

When an atom having magnetic dipole moment $\vec{\mu}$ is placed in an external magnetic field $\vec{B}$, it experiences a torque:

$$\vec{\tau} = \vec{\mu} \times \vec{B}$$

$$\tau = \mu B \sin\theta$$

This torque tends to align the magnetic dipole moment along the direction of the applied field.

(d) Potential Energy

The potential energy of the dipole in the field is:

$$U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta$$

• Minimum energy (stable equilibrium): $\theta = 0°$ (parallel to field)
• Maximum energy (unstable equilibrium): $\theta = 180°$ (antiparallel to field)

(e) Translatory Force (Stern-Gerlach Effect)

In a non-uniform magnetic field (gradient $dB/dz$), atoms experience a translatory force:

$$F = \mu \frac{dB}{dz}$$

Atoms with dipole moments aligned at different angles are deflected by different amounts. This is the basis of the Stern-Gerlach experiment, where a beam of atoms (e.g., Ag or H atoms) splits into discrete components, demonstrating quantization of magnetic moment.

4. Effect on Molecules

The magnetic behavior of molecules depends on the net magnetic dipole moment of all electrons:

• In paramagnetic molecules, the permanent dipole moments tend to align with the applied field, but thermal agitation opposes complete alignment.
• In diamagnetic molecules, the applied field induces a small opposing magnetic moment (Lenz's law effect), causing weak repulsion.

Summary

The magnetic dipole moment arises from the orbital and spin motion of electrons. In atoms, it leads to torque, potential energy, and translatory force in magnetic fields. In molecules, it determines whether the substance is diamagnetic, paramagnetic, or ferromagnetic.

Digital Circuit: $y = (CB + \bar{C}A)(\bar{B}A)$

Step 1: EXTRACT - Given Data

Boolean expression: $$y = (CB + \bar{C}A)(\bar{B}A)$$

Inputs: $A, B, C$ (three inputs → $2^3 = 8$ combinations).

Gates allowed: AND, OR, NOT only.

Step 2: SOLVE

Gates required

Circuit Diagram

C ──────────────┬────────────────[AND G3]───(CB)───────────┐
                │                                            │
B ──────────────┼──────────┬─────────────────────           ├──[OR G5]──┐
                │          [NOT G2]──(B̄)──┐         (CB+C̄A) │           │
C ──[NOT G1]──(C̄)──[AND G4]──(C̄A)────────┼───────────────────┘          │
                                          │                              ├──[AND G7]── y
A ───────┬────────────────────[AND G4 in]─┘                              │
         │                                                               │
         └──────────────────[AND G6]───(B̄A)──────────────────────────────┘

Cleaner signal flow:

• $\bar{C}$ from G1 (NOT on C)
• $\bar{B}$ from G2 (NOT on B)
• $CB$ from G3 (AND of C, B)
• $\bar{C}A$ from G4 (AND of $\bar{C}$, A)
• $CB + \bar{C}A$ from G5 (OR of G3, G4)
• $\bar{B}A$ from G6 (AND of $\bar{B}$, A)
• $y$ from G7 (AND of G5, G6)

Truth Table

Check This Row Carefully

The row A=1, B=0, C=1 is the one most often got wrong:

• $CB = C \cdot B = 1 \cdot 0 = 0$ (not $1$)
• $\bar{C}A = 0 \cdot 1 = 0$
• So $CB + \bar{C}A = 0$, not $1$
• $\bar{B}A = 1 \cdot 1 = 1$
• $y = 0 \cdot 1 = 0$, not $1$

So $y = 1$ occurs only for A=1, B=0, C=0.

Simplified expression

$$y = (CB + \bar{C}A)(\bar{B}A)$$

Distribute: $$y = CB\bar{B}A + \bar{C}A\bar{B}A = 0 + \bar{C}\bar{B}A = A\bar{B}\bar{C}$$

(since $B\bar{B}=0$ and $A\cdot A = A$)

Result

$$\boxed{y = A\bar{B}\bar{C}}$$

Output $y = 1$ only when $A=1, B=0, C=0$, so $y = A\bar{B}$ (independent of $C$) is not correct.

Probability of Finding a Particle in an Infinite Square Well

STEP 1 - Given Data

• Width of well: $a$
• Position: $x = \dfrac{a}{4}$
• Normalized wavefunction: $\psi(x) = \left(\dfrac{2}{a}\right)^{1/2}\sin\left(\dfrac{n\pi x}{a}\right)e^{-iEt/\hbar}$
• Quantum numbers: $n = 1, 2, 3$

Note: The wording asks for "the probability of finding a particle at a position $a/4$." Strictly, the probability of finding a particle at an exact single point is zero. What is physically meaningful is the probability density $|\psi|^2$ at that point (probability per unit length). This is the standard interpretation for this classic problem.

STEP 2 - Solve

The probability density:

$$P(x) = |\psi(x)|^2 = \frac{2}{a}\sin^2\left(\frac{n\pi x}{a}\right)$$

since $|e^{-iEt/\hbar}|^2 = 1$.

At $x = a/4$:

$$P\left(\frac{a}{4}\right) = \frac{2}{a}\sin^2\left(\frac{n\pi}{4}\right)$$

Case $n = 1$

$$P = \frac{2}{a}\sin^2\left(\frac{\pi}{4}\right) = \frac{2}{a}\left(\frac{1}{\sqrt 2}\right)^2 = \frac{2}{a}\cdot\frac{1}{2} = \boxed{\frac{1}{a}}$$

Case $n = 2$

$$P = \frac{2}{a}\sin^2\left(\frac{\pi}{2}\right) = \frac{2}{a}\cdot 1 = \boxed{\frac{2}{a}}$$

Case $n = 3$

$$P = \frac{2}{a}\sin^2\left(\frac{3\pi}{4}\right) = \frac{2}{a}\left(\frac{1}{\sqrt 2}\right)^2 = \frac{2}{a}\cdot\frac{1}{2} = \boxed{\frac{1}{a}}$$

Summary

Interpretation: For $n = 2$, $x = a/4$ coincides with an antinode of the wavefunction, giving maximum density $2/a$. For $n = 1$ and $n = 3$ the density is $1/a$, half that value.