Question

Which of the following options represent a travelling wave? Here, $y$ is the displacement of the particle, $x$ is the position of the particle from origin and $t$ is time.

• A. $y=2\cos \left(3x\right)\sin \left(10t\right)$
• B. $y=2\sqrt{x-5t}$
• C. $y=2\sin (5x-2t)+2\cos (5x-2t)$
• D. $y=5\sin (100\pi t-5x)$

Answer 1

The correct option is D $y=5\sin (100\pi t-5x)$

For option (a),
The equation is of form
$y=2A\cos \left(kx\right)\sin \left(\omega t\right)$
This represents a stationary wave.

For option (b),
The term is of form $(ax\pm bt)$, but non-linear. Hence, it does not represent any kind of wave.

For option (c),
It can be written as,
$y=a\sin \theta +a\cos \theta$
$\Rightarrow y=\sqrt{2}a\sin (\theta +\varphi )$
Here,
$\tan \varphi =\frac{a}{a}\Rightarrow \varphi =45{}^{\circ }$

Hence, it is not of the form $(ax\pm bt)$. Therefore, it can never represent a wave function.

In option $\left(d\right)$, the format of $f(ax\pm bt)$ is fulfilled, apart from that it is representing a sinusoidal travelling wave moving along $+vex$ axis.
Thus $\left(d\right)$ is correct answer.