Question

A wave is represented by the equation $y=$ $(0.02m)\sin (5\pi x-20t)$. The minimum distance between the two particles always having the same speed is. (Assume $x$ and $t$ are in SI units)

• A. $0.02m$
• B. $0.4m$
• C. $0.8m$
• D. $0.2m$

Answer 1

Answer: (D)

Given, wave equation, $y=0.02\sin (5\pi x-20t)m$
As we know that,
Minimum distance between two particles
$=\frac{\text{ Wavelength }(\lambda )}{2}$
and according to wave equation,
Wave number, $k=5\pi m^{-1}$
Angular frequency, boldsymbol $\omega =20$ rads ${}^{-1}$
Since, $k=\frac{2\pi }{\lambda }$
$\therefore \lambda =\frac{2\pi }{k}=\frac{2\pi }{5\pi }$
$=\frac{2}{5}=0.4m$
$\therefore$ Minimum separation between two particles
$=\frac{0.4}{2}=0.2m$