Question

The equation of a simple harmonic wave is given by $y=3 \sin \frac{\pi}{2}(50t-x), $ where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is

• A. $2\pi$
• B. $\frac{3}{2}\pi$
• C. $3\pi$
• D. $\frac{2}{3}\pi$

Answer 1

Answer: (B)

The given wave equation is
$y=3 \sin \frac{\pi}{2}(50 t-x)$
$y=3 \sin \left(25 \pi t-\frac{\pi}{2} x\right) \ldots(i)$
The standard wave equation is
$y=A \sin (\omega t-k x) \ldots(i i)$
Comparing (i) and (ii), we get
$\omega=25 \pi, k=\frac{\pi}{2}$
Wave velocity,
$v=\frac{\omega}{k}=\frac{25 \pi}{(\pi / 2)}=50\, ms ^{-1}$
Particle velocity, $v_{p}=\frac{d y}{d t}\left(3 \sin \left(\left(25 \pi t-\frac{\pi}{2}\right)\right)\right.$
$=75 \pi \cos \left(25 \pi t-\frac{\pi}{2}\right)$
Maximum particle velocity, $\left(v_{p}\right)_{\max }=75 \pi \,ms ^{-1}$
$\therefore \frac{\left(v_{p}\right)_{\max }}{v}=\frac{75 \pi}{50}=\frac{3}{2} \pi$