$A$ particle is executing $SHM$ with an amplitude $A$ and angular frequency $\omega$. The ratio of the maximum acceleration to the maximum velocity of the particle is:

Two springs with negligible masses and force constants $K_1 = 200\, Nm^{-1}$ and $K_2 = 160\, Nm^{-1}$ are attached to a block of mass $m = 10\, kg$ as shown in the figure. Initially,the block is at rest at the equilibrium position where both springs are neither stretched nor compressed. At time $t = 0$,a sharp impulse of $50\, Ns$ is given to the block with a hammer.

$A$ mass $m$ oscillates with simple harmonic motion with frequency $f = \frac{\omega}{2\pi}$ and amplitude $A$ on a spring with constant $K$. Therefore:

$A$ system of two identical rods ($L$-shaped) of mass $m$ and length $l$ are resting on a peg $P$ as shown in the figure. If the system is displaced in its plane by a small angle $\theta$,find the period of oscillations:

The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$,where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$. Choose the incorrect statement.

$A$ 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 ... $cm/s$.