When a wave travels in a medium, displacement of a particle is given by $y = a \sin 2\pi (bt - cx)$, where…
When a wave travels in a medium, displacement of a particle is given by $y = a \sin 2\pi (bt - cx)$, where $a$, $b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity, if
$b = ac$
$b = \frac{1}{ac}$
$c = \pi a$
$c = \frac{1}{\pi a}$
Solution
Given, $y = a\sin 2\pi (bt - cx)$
On comparing this equation with general equation
$y = r\sin \left(\frac{2\pi t}{T} - \frac{2\pi}{\lambda} x\right)$, we get
$\frac{2\pi}{T} = \omega = 2\pi b$, $r = a$
$\lambda = \frac{1}{c}$ and $T = \frac{1}{b}$
Maximum particle velocity, $\omega r = 2\pi ba$
Wave velocity, $v = \frac{\lambda}{T} = \frac{b}{c}$
Given, maximum particle velocity $= 2 \times \text{wave velocity}$
$2\pi ba = 2 \times \frac{b}{c} \Rightarrow c = \frac{1}{\pi a}$