$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation is

$A$ particle is exhibiting simple harmonic motion and has its displacement $x$ and velocity $v$ related as $4v^2 = 25 - x^2$. The time period of the $SHM$ is

$A$ spring executes $S.H.M.$ with mass $10 \ kg$ attached to it. The force constant of the spring is $10 \ N/m$. If at any instant its velocity is $40 \ cm/s$,the displacement at that instant is (Amplitude of $S.H.M.$ $= 0.5 \ m$) (in $m$)

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

The displacement of a particle executing $SHM$ is given by $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$,where $x$ is in $m$ and $t$ is in $s$. The amplitude and maximum speed of the particle are: