The phase difference between two points separated by $0.8 \ m$ in a wave of frequency $120 \ Hz$ is $\frac{\pi}{2}$. The velocity of the wave is ..... $m/s$.

Consider the snapshot of a wave traveling in the positive $x$-direction.

$A$ transverse wave is described by the equation $y = y_0 \sin 2 \pi \left( \nu t - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if $\lambda =$

$A$ transverse wave is given by $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to $4$ times the wave velocity when:

$A$ transverse wave propagates in a medium with a velocity of $1450 \, m/s$. The distance between the nearest points at which the oscillations of the particles are in the opposite phase (phase difference of $\pi$) is $0.1 \, m$. What is the frequency of the wave in $Hz$?

$A$ transverse sinusoidal wave moves along a string in the positive $x$-direction at a speed of $10 \text{ cm/s}$. The wavelength of the wave is $0.5 \text{ m}$ and its amplitude is $10 \text{ cm}$. At a particular time $t$,the snapshot of the wave is shown in the figure. The velocity of point $P$ when its displacement is $5 \text{ cm}$ is: