$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

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

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 performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is: