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
$y=3\sin \frac{\pi }{2}(50t-x)$
$y=3\sin \left(25\pi t-\frac{\pi }{2}x\right)\dots (i)$
The standard wave equation is
$y=A\sin (\omega t-kx)\dots (ii)$
Comparing (i) and (ii), we get
$\omega =25\pi ,k=\frac{\pi }{2}$
Wave velocity,
$v=\frac{\omega }{k}=\frac{25\pi }{(\pi /2)}=50m{s}^{-1}$
Particle velocity, $v_p=\frac{dy}{dt}(3\sin \left(\left(25\pi t-\frac{\pi }{2}\right)\right)$
$=75\pi \cos \left(25\pi t-\frac{\pi }{2}\right)$
Maximum particle velocity, ${\left(v_p\right)}_{\max }=75\pi m{s}^{-1}$
$\therefore \frac{{\left(v_p\right)}_{\max }}{v}=\frac{75\pi }{50}=\frac{3}{2}\pi$