Progressive waves
By The general equation of a wave is given by \begin{align} \frac{\partial^2 y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}, \end{align} where $y$ is the disturbance function, $t$ is the time and $v$ is the wave speed. The disturbance function can be particle's displacement, pressure, electric field, etc.
The amplitude of a wave is denoted by $A$, frequency by $\nu$, wavelength by $\lambda$, period by $T$, angular frequency by $\omega$, and wave number by $k$. These are related by following relations \begin{align} T&=\frac{1}{\nu}=\frac{2\pi}{\omega},\\ v&=\nu \lambda,\\ k&=\frac{2\pi}{\lambda}. \end{align}
The equation of progressive wave travelling with speed $v$ is given by \begin{align} y & = f(t-x/v), \rightsquigarrow+x;\\ y & = f(t+x/v),\rightsquigarrow-x \end{align} The progressive sine wave is given by \begin{align} y &=A\sin(kx-\omega t) \\ &=A\sin(2\pi\left({x}/{\lambda}-{t}/{T}\right)) \nonumber \end{align}
Interesting Questions
Question 1 The disturbance $y$ on a very long string kept along the $x$-axis is given by \begin{align} y=\frac{x^2+at^2}{b}. \end{align} Is this a travelling wave? If yes, what is the wave speed? If not, why?
Question 2: When do we use the progressive wave equation as \begin{align} y=A\sin(\omega t - kx) \end{align} and when do we use \begin{align} y=A\sin(kx - \omega t)? \end{align}
Answer: These represents two progressive waves travelling along $+x$ direction and having same physical properties. The phase difference between these waves is $\pi$. On the other hand, if displacement of the particle at $(kx-\omega t)=\pi/2$ is $+A$ then $y=A\sin(kx-\omega t)$ is the correct expression. If displacement of the particle at $(kx-\omega t)=\pi/2$ is $-A$ then $y=A\sin(\omega t-kx)$ is the correct expression. The choice largely depend on selection of origin $(x=0)$ and time $(t=0)$. The readers are encouraged to draw the displacement of two waves at $t=0$.
Problems from IIT JEE
Problem (IIT JEE 1997): A plane progressive wave of frequency 25 Hz, amplitude ${2.5\times{10}^{-5}}\;\mathrm{m}$ and initial phase zero, propagates along the negative $x$ direction with a velocity of 300 m/s. At any instant, the phase difference between the oscillations at two points 6 m apart along the line of propagation is ________ and the corresponding amplitude difference is ________ m.
Solution: Equation of the given progressive wave is, \begin{align} y=A\sin\left(2\pi\nu t+{2\pi x}/{\lambda}\right),\nonumber \end{align} where $A={2.5\times{10}^{-5}}\;\mathrm{m}$, $\nu={25}\;\mathrm{Hz}$, and $\lambda={v}/{\nu}={300}/{25}={12}\;\mathrm{m}$. Let $x_1$ and $x_2$ be the two points separated by 6 m i.e., $x_2=x_1+6$. The phase at $x_1$ is, \begin{align} \phi_1=2\pi\nu t+{2\pi x_1}/{\lambda}, \end{align} and the phase at $x_2$ is, \begin{align} \phi_2 &=2\pi\nu t+{2\pi x_2}/{\lambda} \\ &=2\pi\nu t+{2\pi (x_1+6)}/{\lambda}. \end{align} Subtract first equation from second equation to get the phase difference between $x_1$ and $x_2$ as $\phi_2-\phi_1=\pi$. The amplitude does not change in a plane progressive wave.
