At $t=0$,a particle executing $SHM$ with a time period $3 \ s$ is in phase with another particle executing $SHM$. The time period of the second particle is $T$ (less than $3 \ s$). If they are again in the same phase for the third time after $45 \ s$,then the value of $T$ is .... . (in $s$)

The displacement-time graph of a particle executing $S.H.M.$ is given in the figure: (sketch is schematic and not to scale). Which of the following statements is/are true for this motion?
$(A)$ The force is zero at $t = \frac{3T}{4}$
$(B)$ The acceleration is maximum at $t = T$
$(C)$ The speed is maximum at $t = \frac{T}{4}$
$(D)$ The $P.E.$ is equal to $K.E.$ of the oscillation at $t = \frac{T}{2}$

For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then:

$A$ horizontal platform with an object placed on it is executing $S.H.M.$ in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \, m$. What must be the least period of these oscillations,so that the object is not detached from the platform (in $, s$)?

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

$A$ small block is connected to one end of a massless spring of un-stretched length $4.9 \ m$. The other end of the spring is fixed at $O$. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \ m$ and released from rest at $t = 0$. It then executes simple harmonic motion with angular frequency $\omega = \frac{\pi}{3} \ rad/s$. Simultaneously at $t = 0$,a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ m$ from $O$. If the pebble hits the block at $t = 1 \ s$,the value of $v$ is (take $g = 10 \ m/s^2$):