$A$ mass attached to a spring performs $S.H.M.$ whose displacement is $x = 3 \times 10^{-3} \cos(2\pi t) \text{ m}$. The time taken to obtain maximum speed for the first time is (in $s$):

On a frictionless horizontal plane,a bob of mass $m=0.1 \text{ kg}$ is attached to a spring with natural length $l_0=0.1 \text{ m}$. The spring constant is $k_1=0.009 \text{ N/m}$ when the length of the spring $l > l_0$ and is $k_2=0.016 \text{ N/m}$ when $l < l_0$. Initially,the bob is released from $l=0.15 \text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) \text{ s}$,then the integer closest to $n$ is:

Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in the figure.

In the situation as shown in the figure,the time period of vertical oscillation of the block for small displacements will be:

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The amplitude of oscillations is

$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.