$A$ 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

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

$A$ block of mass $100 \,g$ is suspended vertically between two massless springs, each with a spring constant $k=1 \,N/m$. The block is hit from above to impart an impulse of $2 \,Ns$. Calculate the maximum displacement from the equilibrium position of the block. (Take $g=10 \,m/s^2$) (in $\,m$)

Two identical springs of spring constant $2k$ are attached to a block of mass $m$ and to fixed supports (see figure). When the mass is displaced from the equilibrium position on either side,it executes simple harmonic motion. The time period of oscillations of this system is ...... .

$A$ mass hangs from a spring and oscillates vertically. The top end of the spring is attached to the top of a box,and the box is placed on a scale,as shown in the figure. The reading on the scale is largest when the mass 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 $= 10 \,m/s^2$)