Maximum speed of a particle in simple harmonic motion is $v_{max}.$ Then average speed of a particle in one complete oscillation is equal to

The force ($F$ in newton) acting on a particle of mass $90 \text{ g}$ executing simple harmonic motion is given by $F + 0.04 \pi^2 y = 0$, where $y$ is the displacement of the particle in meters. If the amplitude of the particle is $\frac{6}{\pi} \text{ m}$, then the maximum velocity of the particle is: (in $\text{ m/s}$)

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

$A$ 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

$A$ particle is executing simple harmonic motion. If the force acting on the particle at a position is $86.6 \%$ of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is

In the figure,what will be the sign of the velocity of the point $P_1$,which is the projection of the velocity of the reference particle $P$? $P$ is moving in a circle of radius $R$ in an anti-clockwise direction.