Question

If the equation of transverse wave is $y=2\sin (kx–2t)$, then the maximum particle velocity is

• A. 4 unit
• B. 2 unit
• C. zero
• D. 6 unit

Answer 1

The correct option is A

4 unit

Step 1: Given Data

$y=2\sin (kx–2t)$

$$\begin{gathered} \omega =2 \\ A=2 \end{gathered}$$

Step 2: Formula Used
$\text{wave velocity,}v={\displaystyle \frac{\text{coefficient of t}}{\text{coefficient of x}}}$

Maximum particle velocity, $v_{max}=\omega A$
Standard equation, $y=A\sin (kx–\omega t)$

Step 3: Calculation of Maximum particle velocity

The given equation of transverse wave is:
$y=2\sin (kx–2t)$

Compared to the standard equation, $y=A\sin (kx–\omega t)$

We get the velocity of a wave is mathematically equal to the product of its wavelength and frequency i.e., the number of vibrations per second, and is independent of the intensity of the wave itself.
$\text{wave velocity,}v={\displaystyle \frac{\text{coefficient of t}}{\text{coefficient of x}}}$
Mathematically, the maximum value of particle velocity is given by:

$$\begin{gathered} \omega =2 \\ A=2 \end{gathered}$$
$v_{max}=\omega A$

$v_{max}=2\times 2$
$v_{max}=4unit$

Therefore the maximum particle velocity is $4units$.

Hence option A is the correct answer.