Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal,the ratio of the amplitude of $M$ to that of $N$ is

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:

$A$ particle connected to the end of a spring executes $S$.$H$.$M$. with period $T_1$. While the corresponding period for another spring is $T_2$. If the period of oscillation with two springs in series is $T$,then

$A$ 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 constant $K$ are connected in series and parallel as shown in the figure. $A$ mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

$A$ block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 \text{ cm}$ from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The kinetic energy of the block at $x = 5 \text{ cm}$ is $0.25 \text{ J}$. The spring constant of the spring is nearly (in $\text{Nm}^{-1}$)