The equation of a transverse wave is given by $y = 0.05 \sin \pi (2t - 0.02 x)$, where $x, y$ are in metre…
The equation of a transverse wave is given by $y = 0.05 \sin \pi (2t - 0.02 x)$, where $x, y$ are in metre and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively [KCET 2013]
$50\text{ m}, 50\text{ ms}^{-1}$
$100\text{ m}, 100\text{ ms}^{-1}$
$50\text{ m}, 100\text{ ms}^{-1}$
$100\text{ m}, 50\text{ ms}^{-1}$
Solution
Given, $y = 0.05 \sin \pi (2t - 0.02x)$
or $y = 0.05 \sin (2\pi t - 0.02\pi x)$
On comparing this equation with standard equation,
$y = A \sin (\omega t - kx)$
where, $\omega = \frac{2\pi}{T}$ and $k = \frac{2\pi}{\lambda}$, we have
$\frac{2\pi}{\lambda} = 0.02\pi \Rightarrow \lambda = 100\text{ m}$
Also, $\omega = \frac{2\pi}{T} = 2\pi$ or $\frac{1}{T} = 1 \Rightarrow \nu = 1\text{ Hz}$
So, $v = \nu\lambda = 1 \times 100 = 100\text{ ms}^{-1}$