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Goal: To prove that for a wave on a string, $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$.

1. System Setup: Consider a small element of string of length $dx$ and mass $dm = \mu dx$. The string has tension $T$.

2. Force Analysis (Vertical):

Net force in y-direction $F_y = T \sin \theta_2 - T \sin \theta_1$.

For small angles, $\sin \theta \approx \tan \theta = \frac{\partial y}{\partial x}$ (Slope).

$$ F_y \approx T \left[ \left(\frac{\partial y}{\partial x}\right)_{x+dx} - \left(\frac{\partial y}{\partial x}\right)_x \right] $$

By definition of derivative, term in bracket is $\frac{\partial^2 y}{\partial x^2} dx$.

3. Newton's Second Law:

Mass of element $dm = \mu dx$. Acceleration $a_y = \frac{\partial^2 y}{\partial t^2}$.

$$ F_y = (dm) a_y \Rightarrow T \frac{\partial^2 y}{\partial x^2} dx = (\mu dx) \frac{\partial^2 y}{\partial t^2} $$

4. The Result:

Canceling $dx$ and rearranging:

Comparing with General Wave Equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$, we get:

$$ \frac{1}{v^2} = \frac{\mu}{T} \Rightarrow v = \sqrt{\frac{T}{\mu}} $$

Derivation of the Equation of a Plane Progressive Harmonic Wave:

Goal: To mathematically describe a wave where every particle performs Simple Harmonic Motion (SHM).

Step 1: Motion at the Source ($x=0$)

Assume the particle at the origin ($x=0$) oscillates in SHM starting from mean position:

$$ y(0, t) = A \sin(\omega t) $$

Step 2: Propagation & Time Lag

The disturbance travels with wave velocity $v$. To reach a particle at position $x$, the wave takes time $t_{lag} = \frac{x}{v}$.

Step 3: Motion at Position $x$

The particle at $x$ does exactly what the source particle did earlier at time $(t - t_{lag})$.

$$ y(x, t) = y(0, t - t_{lag}) $$

Substituting the SHM equation:

$$ y(x, t) = A \sin[\omega (t - \frac{x}{v})] $$

$$ y(x, t) = A \sin(\omega t - \frac{\omega}{v}x) $$

Step 4: Defining Wave Constants

We define Propagation Constant (or Angular Wave Number) as $k = \frac{\omega}{v} = \frac{2\pi \nu}{\nu \lambda} = \frac{2\pi}{\lambda}$.

Substituting $k$:

$$ y(x, t) = A \sin(\omega t - kx) $$

Note: Since $\sin(-\theta) = -\sin(\theta)$, and usually we start with $kx$, we often write it as:

This represents a wave traveling in the Positive X direction.

Why $(kx - \omega t)$?

• As $t$ increases, the term $\omega t$ increases.
• To keep the phase $(kx - \omega t)$ constant (tracking a specific crest), $x$ must also INCREASE.
• Hence, wave moves in $+x$ direction.
• For $(kx + \omega t)$, $x$ must DECREASE to keep phase constant $\rightarrow$ Wave moves in $-x$ direction.

Goal: To derive $v = \sqrt{T/\mu}$ using the concept of Centripetal Force.

Logic: Imagine moving with the pulse at speed $v$. In this frame, the pulse appears stationary, and the string moves backward with speed $v$.

Step 1: Dynamics of Element

Consider a small element of length $dl = R(2\theta)$ (where $\theta$ is small half-angle). Mass $dm = \mu (2R\theta)$.

The element moves on a curved path of radius $R$ with speed $v$. Required Centripetal Force:

$$ F_c = \frac{(dm)v^2}{R} = \frac{(2\mu R \theta)v^2}{R} = 2\mu v^2 \theta $$

Step 2: Restoring Force

The tension $T$ at both ends provides the downward radial force. Vertical component is $2T \sin\theta$.

For small $\theta$, $\sin\theta \approx \theta$. So, $F_{res} \approx 2T\theta$.

Step 3: Equating Forces

$$ 2T\theta = 2\mu v^2 \theta $$

$$ T = \mu v^2 \Rightarrow v = \sqrt{\frac{T}{\mu}} $$

General Formula: $v = \sqrt{\frac{B}{\rho}}$ where $B$ is Bulk Modulus.

Definition of Bulk Modulus: $B = -V \frac{dP}{dV}$.

1. Newton's Formula (Isothermal):

Assumption: Sound travels slowly, temperature remains constant ($PV = \text{constant}$).

Differentiating: $P dV + V dP = 0 \Rightarrow P dV = -V dP \Rightarrow P = -V \frac{dP}{dV}$.

This implies $B_{iso} = P$. So, $v = \sqrt{\frac{P}{\rho}}$.

(Result: ~280 m/s for air. Incorrect.)

2. Laplace's Correction (Adiabatic):

Assumption: Compressions are rapid, heat cannot escape ($PV^\gamma = \text{constant}$).

Differentiating: $\gamma P V^{\gamma-1} dV + V^\gamma dP = 0$.

Divide by $V^{\gamma-1}$: $\gamma P dV + V dP = 0 \Rightarrow \gamma P = -V \frac{dP}{dV}$.

This implies $B_{adia} = \gamma P$. So, $v = \sqrt{\frac{\gamma P}{\rho}}$.

(Result: ~331 m/s. Correct!)

Goal: To find the energy carried by a wave.

1. Kinetic Energy ($dK$):

Consider string element of length $dx$ and mass $dm = \mu dx$.

Transverse velocity $v_p = \frac{\partial y}{\partial t} = -\omega A \cos(kx - \omega t)$.

$$ dK = \frac{1}{2} (dm) v_p^2 = \frac{1}{2} (\mu dx) [-\omega A \cos(kx - \omega t)]^2 $$

2. Potential Energy ($dU$):

The string element stretches from length $dx$ to $dl$. Work done against Tension $T$ is stored as PE.

$$ dU = T(dl - dx) $$

Geometry: $dl = \sqrt{dx^2 + dy^2} = dx \sqrt{1 + (\frac{\partial y}{\partial x})^2}$.

Using Binomial approximation $(1+x)^n \approx 1+nx$: $dl \approx dx [1 + \frac{1}{2}(\frac{\partial y}{\partial x})^2]$.

So, extension $dl - dx = \frac{1}{2} (\frac{\partial y}{\partial x})^2 dx$.

Slope $\frac{\partial y}{\partial x} = k A \cos(kx - \omega t)$. And $T = \mu v^2 = \mu (\omega/k)^2$.

$$ dU = \frac{1}{2} T [kA \cos(kx-\omega t)]^2 dx = \frac{1}{2} (\mu \frac{\omega^2}{k^2}) k^2 A^2 \cos^2(\dots) dx $$

Conclusion: $dK = dU$. At any instant, Kinetic and Potential energies are EQUAL in a traveling wave.

3. Energy Density ($u$) and Power ($P$):

Total Energy $dE = dK + dU = \mu \omega^2 A^2 \cos^2(kx - \omega t) dx$.

Average value of $\cos^2\theta$ over one cycle is $1/2$.

Average Energy per unit length (Energy Density):

(Volume energy density if using $\rho$ volume density, or Linear if $\mu$).

Power (Rate of Energy Transfer):

Energy flowing past a point in time $dt$ is energy contained in length $v dt$.

$$ P = \text{Energy Density} \times \text{Speed} = \langle u \rangle v $$

Goal: To show that superposition of two identical counter-propagating waves creates a stationary wave pattern.

1. Wave Equations:

Incident Wave ($+x$): $y_1 = A \sin(kx - \omega t)$

Reflected Wave ($-x$): $y_2 = A \sin(kx + \omega t)$ (Assuming reflection from free end/ open pipe for in-phase reflection).

2. Superposition Principle:

$$ y = y_1 + y_2 = A [\sin(kx - \omega t) + \sin(kx + \omega t)] $$

Using Identity: $\sin(C) + \sin(D) = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$

• $C+D = 2kx \Rightarrow \text{Avg} = kx$
• $C-D = -2\omega t \Rightarrow \text{Avg} = -\omega t$ (and $\cos(-\theta) = \cos\theta$)

3. Interpretation:

• Amplitude Term ($A_x$): $2A \sin(kx)$. The amplitude is NOT constant; it depends on position $x$.
• Oscillation Term: $\cos(\omega t)$. Every particle oscillates with same frequency $\omega$.
• No Travelling Term: There is no $(kx \pm \omega t)$ term. Energy is not transported along $x$. It is "Standing".

4. Nodes and Antinodes:

Nodes (Zero Displacement):

$\sin(kx) = 0 \Rightarrow kx = n\pi$

$\frac{2\pi}{\lambda} x = n\pi \Rightarrow x = \frac{n\lambda}{2}$

Positions: $x = 0, \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2} \dots$

Antinodes (Max Displacement $\pm 2A$):

$\sin(kx) = \pm 1 \Rightarrow kx = (2n+1)\frac{\pi}{2}$

Positions: $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4} \dots$

Goal: Deriving the Beat Frequency ($\nu_b = \nu_1 - \nu_2$) from superposition.

Consider two sound waves at $x=0$ with slightly different frequencies $\omega_1$ and $\omega_2$:

$$ y_1 = A \cos(\omega_1 t) \quad , \quad y_2 = A \cos(\omega_2 t) $$

Superposition: $y = y_1 + y_2 = A [\cos(\omega_1 t) + \cos(\omega_2 t)]$

Using $\cos C + \cos D = 2 \cos(\frac{C-D}{2}) \cos(\frac{C+D}{2})$:

Let $\omega_{avg} = \frac{\omega_1 + \omega_2}{2}$ and $\Delta \omega = \frac{\omega_1 - \omega_2}{2}$.

Interpretation: This is a wave with frequency $\nu_{avg}$ but its amplitude varies with time.

Amplitude Modulation: $A(t) = 2A \cos(2\pi \frac{\nu_1-\nu_2}{2} t)$.

Maxima (Waxing): Amplitude is max when $\cos(\dots) = \pm 1$.

Arguments: $0, \pi, 2\pi \dots$

Time interval between two maxima: $\Delta t = \frac{1}{|\nu_1 - \nu_2|}$.

Therefore, Frequency of Beats:

Derivation using Wave Crest Counting Logic:

Let Source ($S$) emit frequency $\nu_0$ and move with velocity $v_s$. Observer ($O$) moves with $v_o$. Speed of sound is $v$.

1. Effect of Source Motion (Change in Wavelength):

In time $t=1$ sec, Source emits $\nu_0$ waves. These waves occupy distance $(v - v_s)$ if Source moves towards observer.

New Wavelength $\lambda' = \frac{\text{Distance}}{\text{Number}} = \frac{v - v_s}{\nu_0}$.

2. Effect of Observer Motion (Change in Relative Velocity):

Observer moves towards incoming waves. Relative velocity of wave w.r.t Observer is $v_{rel} = v + v_o$.

3. Apparent Frequency:

$$ \nu' = \frac{\text{Relative Speed}}{\text{Effective Wavelength}} = \frac{v + v_o}{\lambda'} $$

Substituting $\lambda'$:

$$ \nu' = (v + v_o) \cdot \frac{\nu_0}{v - v_s} $$

Sign Convention (Top Signs): Use Top signs ($+v_o, -v_s$) when approaching each other. Use Bottom signs when receding.