When a sound wave of wavelength $\lambda$ is propagating in a medium, the maximum velocity of the particle is equal to the wave velocity. The amplitude of the wave is:

$A$ transverse wave travelling along a stretched string has a speed of $30 \ m/s$ and a frequency of $250 \ Hz$. The phase difference between two points on the string $10 \ cm$ apart at the same instant is

$A$ 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 $\lambda =$

In a transverse wave,the distance between a crest and the neighbouring trough at the same instant is $4.0 \, cm$,and the distance between a crest and trough at the same place is $1.0 \, cm$. The next crest appears at the same place after a time interval of $0.4 \, s$. The maximum speed of the vibrating particles in the medium is

For the travelling harmonic wave $y(x, t) = 2.0 \cos 2 \pi(10 t - 0.0080 x + 0.35)$,where $x$ and $y$ are in $cm$ and $t$ is in $s$. Calculate the phase difference between oscillatory motion of two points separated by a distance of:
$(a)$ $4 \, m$
$(b)$ $0.5 \, m$
$(c)$ $\lambda / 2$
$(d)$ $3 \lambda / 4$

$A$ 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: