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 .........

$A$ particle executes $SHM$ of amplitude $25\, cm$ and time period $3\, s$. What is the minimum time required for the particle to move between two points $12.5\, cm$ on either side of the mean position?

$A$ particle oscillates in a straight line simple harmonically with a period of $8 \ s$ and an amplitude of $4 \sqrt{2} \ m$. The particle starts from the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ second of its motion to that in the $2^{\text{nd}}$ second is: $(\sin 45^{\circ} = 1 / \sqrt{2}, \sin \frac{\pi}{2} = 1)$

Explain with plots the position of a particle executing simple harmonic motion at different times.

$A$ 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.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.