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\left(x=x^{\prime}, t=0\right)=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{aligned}
& x^{\prime} \rightarrow k x^{\prime} \\
& \lambda \rightarrow 2 \pi
\end{aligned}
$

$
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 ' $t=\left(x-x^{\prime}\right) / v '$.

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., $y=f\left(x=x^{\prime}, t=0\right)$ 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{aligned}
& \quad y=f\left(x^{\prime}, 0\right)=f(x, t) \\
& \therefore y(x, t)=A \sin \left(k x^{\prime}\right)=A \sin (k(x-v t)) \\
& y(x, t)=A \sin (k(x-v t))=A \sin \left(\frac{2 \pi x}{\lambda}-\frac{2 \pi v t}{\lambda}\right)=A \sin (k x-\omega t) \\
& \quad v=f \lambda=\frac{\lambda}{T} \\
& T=\frac{2 \pi}{\omega}
\end{aligned}
$
 

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

For wave travelling along negative $x$-axis,

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

GENERAL EQUATION OF TRAVELLING WAVE

$
y(x, t)=A \sin (k(x \pm v t)+\phi)=A \sin \left(\frac{2 \pi x}{\lambda} \pm \frac{2 \pi t}{T}+\phi\right)=A \sin (k x \pm \omega t+\phi)
$