In a wave,the path difference corresponding to a phase difference of $\phi$ is

Two waves are represented by: $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$. Then the phase difference between them is

The phase difference between two points separated by $0.8 \ m$ in a wave of frequency $120 \ Hz$ is $90^o$. Then the velocity of the wave will be ............ $m/s$.

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

The equation of a wave is given by $y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$. If the displacement is $5 \ cm$ at $t = 0 \ s$,then the total phase angle at $t = 7.5 \ s$ will be:

$A$ wave is given by $Y = 3 \sin 2 \pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$ where $Y$ is in $cm$. The frequency of the wave and the maximum acceleration will be $(\pi^2 = 10)$.