For a travelling harmonic wave $y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$, where $x$ and $y$ are in $\text{cm}$ and $t$ in $\text{s}$. The phase difference between oscillatory motion of two points separated by a distance of $0.5 \text{ m}$ is: (in $\pi \text{ rad}$)

The equation of a plane progressive wave is given by $y = 0.025 \sin (100t + 0.25x)$. The frequency of this wave would be

$A$ transverse sinusoidal wave of amplitude '$a$',wavelength $\lambda$,and frequency $f$ is travelling on a stretched string. The maximum speed at any point on the string is $v/10$,where $v$ is the speed of propagation of the wave. If $a = 10^{-3} \ m$ and $v = 10 \ ms^{-1}$,then $\lambda$ and $f$ are given by:

Obtain the relation between wave velocity,angular frequency,and angular wave number.

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

$A$ transverse wave propagating along the $x-$ axis is represented by $y(x,t) = 8.0 \sin \left( 0.5\pi x - 4\pi t - \frac{\pi}{4} \right)$,where $x$ is in metres and $t$ is in seconds. The speed of the wave is ..... $m/s$.