$A$ particle executes simple harmonic motion between $x=-A$ and $x=+A$. If it takes a time $T_1$ to go from $x=0$ to $x=A/2$ and $T_2$ to go from $x=A/2$ to $x=A$,then:

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be: