The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

Distance travelled by a particle in $\text{SHM}$ when its phase changes from $\frac{\pi}{6}$ to $\frac{5 \pi}{6}$ is:

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

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

$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

Two particles are in $SHM$ on the same straight line with amplitudes $A$ and $2A$ and with the same angular frequency $\omega$. It is observed that when the first particle is at a distance $A/\sqrt{2}$ from the origin and moving toward the mean position,the other particle is at the extreme position on the other side of the mean position. Find the phase difference between the two particles. (in $^o$)