Travelling Waves - Practice Questions & MCQ

The function f(x,t) represents the displacements y of the particle at t=0 and x=x'

$y=f(x=x^{\prime },t=0)=A\sin \left(k{x}^{\prime }\right)$

k- propagation constant or angular wave number
A- Amplitude
For a given time, between position $\mathrm{x}=0$ to $\mathrm{x}={\mathrm{x}}^{\prime }$ the phase changes from 0 to ${\mathrm{kx}}^{\prime }$ similarly, $\mathrm{x}=0$ to $\mathrm{x}=\lambda$ the phase changes from 0 to $2\pi$

$\begin{array}{rl}& x^{\prime }\to k{x}^{\prime }\\ & \lambda \to 2\pi \end{array}$

$k=\frac{2\pi }{\lambda }$

kx ' represents phase of wave at $\mathrm{x}=\mathrm{x}$ '
The disturbance travels on the strings towards along positive $x$-axis with a constant speed ' $v$ '. Thus, the displacement produced at the left end at time ' $t=0$ ', reaches the point ' $x$ ' at time ' $t=(x-x^{\prime })/v^{\prime }$.

As wave shape remains same for progressive wave, particle's displacement at $\mathrm{x}={\mathrm{x}}^{\prime },\mathrm{t}=0$ and $\mathrm{x}={\mathrm{x}}^{\prime }+\mathrm{vt},\mathrm{t}=\mathrm{t}$ are same
i.e., $y=f(x=x^{\prime },t=0)$ is same as $y=f(x,t)$

Let's now write the equation in terms of the stationary coordinate x , where ${\mathrm{x}}^{\prime }=\mathrm{x}-\mathrm{vt}$

$\begin{array}{rl}& y=f(x^{\prime },0)=f(x,t)\\ & \therefore y(x,t)=A\sin \left(k{x}^{\prime }\right)=A\sin (k(x-vt))\\ & y(x,t)=A\sin (k(x-vt))=A\sin (\frac{2\pi x}{\lambda }-\frac{2\pi vt}{\lambda })=A\sin (kx-\omega t)\\ & v=f\lambda =\frac{\lambda }{T}\\ & T=\frac{2\pi }{\omega }\end{array}$
 

$y(x,t)=A\sin (k(x-vt))=A\sin (\frac{2\pi x}{\lambda }-\frac{2\pi t}{T})=A\sin (kx-\omega t)$

For wave travelling along negative $x$-axis,

$y(x,t)=A\sin (k(x+vt))=A\sin (\frac{2\pi x}{\lambda }+\frac{2\pi t}{T})=A\sin (kx+\omega t)$

GENERAL EQUATION OF TRAVELLING WAVE

$y(x,t)=A\sin (k(x\pm vt)+\varphi )=A\sin (\frac{2\pi x}{\lambda }\pm \frac{2\pi t}{T}+\varphi )=A\sin (kx\pm \omega t+\varphi )$