$A$ coin is placed on a horizontal platform. The platform performs vertical simple harmonic motion with an angular frequency $\omega$. The amplitude of oscillation is gradually increased. At what condition will the coin first lose contact with the platform?

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:

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

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_{1}$ and $K_{2}$ respectively. If the maximum velocities during oscillations are equal,the ratio of the amplitude of $A$ and $B$ is

$A$ block of mass $2\,kg$ is attached to two identical springs,each with a spring constant of $20\,N/m$. The block is placed on a frictionless surface,and the outer ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position,it executes simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in $SI$ units. The value of $x$ is $..........$

When a body of mass $1.0 \,kg$ is suspended from a certain light spring hanging vertically, its length increases by $5 \,cm$. By suspending a $2.0 \,kg$ block to the spring and if the block is pulled through $10 \,cm$ and released, the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 \,m/s^2$)