The time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement is

Which of the following curves represents correctly the oscillation given by $y = y_0 \sin(\omega t - \phi)$,where $0 < \phi < 90^\circ$?

$A$ 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:

$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is: