Two springs,of force constants $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

$A$ body of mass $1 \,kg$ is suspended from a spring of negligible mass. Another body of mass $500 \,g$ moving vertically upwards hits the suspended body with a velocity of $3 \,ms^{-1}$ and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is $\frac{10}{\pi} \,Hz$,the amplitude of the motion and the spring constant are respectively,

The frequency of oscillations of a mass $m$ connected horizontally by a spring of spring constant $k$ is $4 \ Hz$. When the spring is replaced by two identical springs connected in series as shown in the figure,the effective frequency is:

$A$ vertical spring oscillates with a period of $6 \ s$ when a mass $m$ is suspended from it. When the mass is at rest,the spring is stretched through a distance of (Take acceleration due to gravity $g = \pi^2 = 10 \ m/s^2$): (in $m$)

$A$ block of mass $5 \ kg$ moves along the $x$-direction subject to the force $F = (-20x + 10) \ N$,with the value of $x$ in metre. At time $t = 0 \ s$,it is at rest at position $x = 1 \ m$. The position and momentum of the block at $t = (\pi / 4) \ s$ are

$A$ spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming the spring to be made of a uniform wire,the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is