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 $s$)

$Assertion :$ In simple harmonic motion,the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$.

$A$ body executing simple harmonic motion has a maximum acceleration equal to $24 \, m/s^2$ and a maximum velocity equal to $16 \, m/s$. The amplitude of the simple harmonic motion is:

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:

Match the following physical quantities for a particle executing Simple Harmonic Motion $(SHM)$ given by $y = A \sin(\omega t)$:
$(a)$ Velocity $(v)$
$(b)$ Potential Energy $(PE)$
$(c)$ Total Energy $(TE)$
$(d)$ Acceleration $(a)$
$(i)$ Constant
(ii) $A\omega \cos(\omega t)$
(iii) $\frac{1}{2} k A^2 \sin^2(\omega t)$
(iv) $-\omega^2 y$

Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system (see figure). Because of the torsional constant $k$,the restoring torque is $\tau = k\theta$ for an angular displacement $\theta$. If the rod is rotated by $\theta_0$ and released,the tension in it when it passes through its mean position will be