IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Progressive and Stationary Waves

## Interesting Questions

Question: When do we use the progressive wave equation as $y=A\sin(\omega t - kx)$ and when do we use $y=A\sin(kx - \omega t)$? (Source FB)

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 $y=Asin(kx-\omega t)$ is the correct expression. If displacement of the particle at $(kx-\omega t)=\pi/2$ is $-A$ then $y=Asin(\omega t-kx)$ 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} \label{lza:eqn:1} \phi_1=2\pi\nu t+{2\pi x_1}/{\lambda}, \end{align} and the phase at $x_2$ is, \begin{align} \label{lza:eqn:2} \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.

Problem (IIT JEE 1988): A wave represented by the equation $y=a\cos(kx-\omega t)$ is superimposed with another wave to form a stationary wave such that point $x=0$ is a node. The equation for the other wave is,

1. $a\sin(kx+\omega t)$
2. $-a\cos(kx-\omega t)$
3. $-a\cos(kx+\omega t)$
4. $-a\sin(kx-\omega t)$

Solution: A stationary wave is formed by the superposition of two identical waves travelling in the opposite directions. The point $x=0$ is a node if displacement of the resultant wave is zero at this point at all times. Thus, the wave superimposed on $y=a\cos(kx-\omega t)$ is $y^\prime=-a\cos(kx+\omega t)$.