The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

Which of the following statements is correct for $S.H.M.$?