The equation of $S.H.M.$ is $y = a \sin(2\pi nt + \alpha)$,then its phase at time $t$ is:

The diagram shows two oscillations. What is the phase difference between the oscillations?

$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

$A$ particle performs simple harmonic motion between $x = -A$ and $x = +A$. How much time does it take to travel from $x = A$ to $x = A/2$?

The displacements of two particles executing $S.H.M.$ on the same line are given as $y_1 = a \sin \left(\frac{\pi}{2} t + \phi\right)$ and $y_2 = b \sin \left(\frac{2 \pi}{3} t + \phi\right)$. The phase difference between them at $t = 1 \, s$ is .........