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

$A$ wave travelling along a string is described by,
$y(x, t) = 0.005 \sin (80.0 x - 3.0 t)$
In which the numerical constants are in $SI$ units ($0.005 \, m, 80.0 \, rad \, m^{-1},$ and $3.0 \, rad \, s^{-1}$). Calculate
$(a)$ the amplitude,
$(b)$ the wavelength,and
$(c)$ the period and frequency of the wave.
Also,calculate the displacement $y$ of the wave at a distance $x = 30.0 \, cm$ and time $t = 20 \, s$.

$A$ wave represented by the given equation $Y = A\sin(10\pi x + 15\pi t + \frac{\pi}{3})$,where $x$ is in meters and $t$ is in seconds. The expression represents:

$A$ simple wave motion is represented by $y=5(\sin 4 \pi t+\sqrt{3} \cos 4 \pi t)$. Its amplitude is

The equation of a transverse wave is given by $y=0.05 \sin \pi(2 t-0.02 x)$,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

If the equation of a transverse wave is $y = 5\sin 2\pi \left[ \frac{t}{0.04} - \frac{x}{40} \right]$,where distance is in $cm$ and time is in seconds,then the wavelength of the wave is .... $cm$.