$A$ rectangular block of mass $5\,kg$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1\,m$ and time period $3.14\,s$. The maximum force exerted by the spring on the block is $.......N$.

$A$ spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio of $T_2 / T_1$ is

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

$A$ stiff spring having spring constant $k = 400 \text{ N/m}$ is attached to the floor vertically. $A$ mass $m = 10 \text{ kg}$ is placed on top of the spring. The block oscillates if it is pressed downward and released. Find the extension in the spring at which the block loses contact with the spring. (Take $g = 10 \text{ m/s}^2$) (in $\text{ cm}$)

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m \text{ kg}$ and the spring constant is $k \text{ Nm}^{-1}$. At a given instant, the extension of the spring is $x \text{ m}$ and the speed of the particle is $v \text{ ms}^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

Two particles $A$ and $B$ of masses $m$ and $2m$ are suspended from massless springs of force constants $K_1$ and $K_2$. During their oscillation, if their maximum velocities are equal, then the ratio of amplitudes of $A$ and $B$ is