$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $x = \frac{A}{2}$ is $2 \text{ s}$, then time taken by particle in going from $x = \frac{A}{2}$ to $x = A$ is (in $\text{ s}$)

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 is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

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 $^o$)

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.