$A$ system consisting of three masses attached to a spring is in equilibrium. When the $700\,g$ mass is removed,the system oscillates with a period of $3\,s$. When the $500\,g$ mass is also removed,what will be the new time period of the system? (in $\text{seconds}$)

Two bodies $A$ and $B$ of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$,respectively. If the bodies oscillate vertically such that their amplitudes are equal,the ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is

The springs in the figure are identical,but the length of the spring in $A$ is three times that in $B$. The ratio of the time periods $T_A/T_B$ is:

The force-deformation equation for a nonlinear spring fixed at one end is $F = 4x^{1/2}$,where $F$ is the force (expressed in newtons) applied at the other end and $x$ is the deformation expressed in meters.

$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards 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 $= 10 \,m/s^2$)