Waves 1

1. \( y \) must be continous.(otherwise string will break)
2. \( y \) must be differentiable.(sudden slope change would cause infinite acceleration)
3. \( y \) must be bounded.(Any material particle cannot have infinite displacement)

If for a certain location(\( x_0 \)), the disturbance is given by \( y = f(t) \), then at a general location \( x \) it will be given by \[ y = f\left( t - \frac{x-x_0}{v} \right) \] Similarly, if at a general time(\( t_0 \) ), the equation of wave is given by \( y = f(x) \) , then at a general time it will be given by \[ y = f (x - v(t-t_0)) \] These two shifting properties are used in transformation of waves such as \( \pi \) shifts.

In general: \[ P = -T\left( \frac{\partial y}{\partial x} \right) \frac{\partial y}{\partial t} \] \[ = -T v \left( \frac{\partial y }{\partial x} \right)^2 \]

For progressive waves: \[ = \frac{1}{2} \mu\ v \ G \]

Average power generated: \[ \left< P \right> = \frac{1}{2} \mu v A^2 \omega^2 \]

\[ I = \frac{P}{S} = - \frac{T}{S} \left( \frac{\partial y}{\partial x} \right) \frac{\partial y}{\partial t} \] For progressive wave: \[ = \frac{1}{2} \frac{\mu v G}{S} \] \[ = \frac{1}{2} \rho vG \] Average Intensity: Same as average power, \[ \left< I \right> = \frac{1}{2} \rho v^2 A^2 \omega^2 \]

Suppose we have reflecting boundary at \( x = a \) and our incident wave is \( y_i = f(x-v_1t) \), then we follow these steps:

1. Evaluate incident disturbance at \( x = a \) , \( y_i = f (a-v_1t) \)
2. Calculate transmitted disturbance as \(\displaystyle y_t = \frac{2v_2}{v_1 + v_2} f(a-v_1t) \)
3. Calculate transmitted wave by applying a future shift of \( \displaystyle \left( \frac{x-a}{v_2} \right) \) to \( y_t \) at the boundary i.e. \[ y_t = \frac{2v_2}{v_1 + v_2} \times f \left( a - v_1 \left[ t- \frac{(x-a)}{v_2} \right] \right) \]
4. Calculate reflected disturbance at the boundary as \( \displaystyle y_r = \frac{v_2 - v_1}{v_1 + v_2} \times f (a-v_1t) \)
5. Calculate reflected wave by future shifting by \( \displaystyle \left( \frac{a-x}{v_1} \right) \) i.e. \[ y_r = \left( \frac{v_2 -v_1}{v_1 + v_2} \right) \times f \left( a - v_1 \left[ t- \frac{(a-x)}{v_1} \right] \right) \]

At fixed boundary, there can be no displacement, thus \[ y_r = - y_i \] \[ \implies y_r = \left( \frac{v_2 - v_1}{v_1 + v_2} \right) y_i = - y_i \] i.e. the wave just inverts. It can be modelled as an superposition with an inverted imaginary wave coming from boundary.

A free boundary is modelled as end of a string attached to a massless ring free to slide in a frictionless rod. Since the ring is massless therefore it can’t have any vertical component of tension acting on it. Therefore the slope of the string at this boundary must always be zero. \[ \dot{y}_i = \dot{y}_r \] \[ \implies y_i = y_r \] \[ y_r = \left( \frac{v_2 - v_1}{v_1 + v_2} \right) y_i \] Since, \( v_2 \to \infty \) , \[ y_r = y_i \] It can be modelled as an superposition with an imaginary wave coming from boundary.

Incident power: \( \mu_1 v_1 \dot{y}^2_i \) Reflected power: \( \mu_1 v_1 \dot{y}^2_r \) Thus, reflected fraction: \[ \eta = \left( \frac{\dot{y}_r}{\dot{y}_i} \right)^2 \] \[ \implies \eta = \left( \frac{v_2 - v_1}{v_1 + v_2} \right)^2 \] Transmitted fraction: \[ \eta_t = 1 - \left( \frac{v_2 - v_1}{v_1 + v_2} \right)^2 \] \[ = \frac{4v_1 v_2 }{(v_1 + v_2)^2} \]

The reflected and transmitted fraction are same for \( \displaystyle \frac{v_2}{v_1} = \lambda \) or \( \displaystyle \frac{v_1}{v_2} = \lambda \) : \[ \eta_r = \left( \frac{v_2 - v_1}{v_1 + v_2} \right)^2 = \left( \frac{\frac{v_2}{v_1} - 1}{\frac{v_2}{v_1} + 1} \right)^2 = \left( \frac{\frac{v_1}{v_2} - 1}{\frac{v_1}{v_2} + 1}\right)^2 \]

Since, \( \displaystyle P_i = \frac{1}{2} \mu v A_i^2 \omega^2 \) and \( \displaystyle P_r = \frac{1}{2}\mu v A_r^2 \omega^2 \) , we get: \[ A_r = \left( \frac{v_2 - v_1}{v_1 + v_2} \right)A_i \qquad \text{and} \qquad A_t = \left( \frac{2v_2}{v_1 + v_2} \right) A_i \]

Whenever two waves of same amplitude travelling in opposite direction superpose, we get what are known as stationary waves.

Let there be two waves: \[ y_1 = A\sin(kx - \omega t) \] \[ y_2 = A \sin (kx + \omega t) \] \[ y = y_1 + y_2 = 2A \sin (kx) \cos (\omega t) \]