Two waves are represented by the equations $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$. The first wave

$A$ transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \ s$. If $x$ and $y$ are in $cm$ and $t$ in $s$,then the length of the string is: (in $m$)

$A$ sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then:

The equation of a simple harmonic wave is given by $y = 6 \sin 2 \pi (2t - 0.1x)$,where $y$ and $x$ are in $mm$ and $t$ is in seconds. The phase difference between two particles $2 \ mm$ apart at any instant is (in $^{\circ}$)

$A$ 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 $Hz$?

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$. This represents: