Q.Let $y=1+\dfrac{1}{1+x^2}$ at t = 0 s be the amplitude of the wave propagating in the positive x-direction. At t = 2 s, the amplitude of the wave propagating becomes $y=1+\dfrac{1}{1+(x-2)^2}$. Assume that the shape of the wave does not change during propagation. The velocity of the wave is

A progressive (or travelling) wave is a continuous disturbance that keeps advancing through a medium indefinitely, as opposed to a single, isolated pulse. If a pulse's shape is $y = f(x)$ at time $t = 0$ and it moves rightward at constant speed $v$ without changing shape, then at a later time $t$ the same shape has simply shifted, giving $y(x,t) = f(x - vt)$; a pulse moving leftward instead satisfies $y(x,t) = f(x + vt)$. Both forms satisfy the one-dimensional wave equation $\dfrac{\partial^2 y}{\partial x^2} = \dfrac{1}{v^2}\dfrac{\partial^2 y}{\partial t^2}$, which every genuine wave function must obey (though not every solution of this equation is itself a physically realisable wave, since a physical wave must also stay finite for all $x$ and $t$). For a sinusoidal progressive wave the standard form is $y(x,t) = A\sin(kx - \omega t)$, where $A$ is the amplitude, $k = 2\pi/\lambda$ the angular wave number and $\omega = 2\pi f$ the angular frequency; a $\sin(kx+\omega t)$ form instead describes a wave travelling in the $-x$ direction. Such a wave can be visualised with two complementary graphs. The space (spatial) variation graph, $y = A\sin(kx)$ at one fixed instant, plots displacement against position and directly shows the wavelength $\lambda$ as the horizontal repeat distance. The time (temporal) variation graph, $y = A\sin(\omega t)$ at one fixed position, plots displacement against time and directly shows the time period $T$ as the repeat interval. Two distinct velocities must be carefully told apart. The particle velocity $v_P = \partial y/\partial t = -\omega A\cos(kx-\omega t)$ is the instantaneous rate of change of displacement of one specific medium particle at its fixed location -- it oscillates sinu …